Vectors Directed Line Segments and Vectors A directed line segment is defined as an initial point, P, and a terminal point Q. Example P = (2,3) and Q = (-1,4) Definition of a Vector A vector is the equivalence class of all directed segments of the same length and direction. We can represent a vector by writing the unique directed line segment that has its initial point at the origin. Example P = (2,3) and Q = (-1,4) is equivalent to the directed line segment "Q - P" = <-3, 1> When we write the <> we mean that the vector has initial point at the origin and terminal point at (-3,1). This notation is called the component form of the vector. The length of the vector <x,y> is called the norm or magnitude.
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Vectors
Directed Line Segments and Vectors
A directed line segment is defined as an initial point, P, and a terminal point Q.
Example
P = (2,3) and Q = (-1,4)
Definition of a Vector
A vector is the equivalence class of all directed segments of the same length and direction.
We can represent a vector by writing the unique directed line segment that has its initial point at the origin.
Example
P = (2,3) and Q = (-1,4)
is equivalent to the directed line segment
"Q - P" = <-3, 1>
When we write the <> we mean that the vector has initial point at the origin and terminal point at (-3,1). This notation is called the component form of the vector.
The length of the vector <x,y> is called the norm or magnitude.
We can find it by the formula:
Length of a Vector
Example:
We also use the notation
-3i + j
to denote the vector <-3,1>.
Example
A vector that has length 10 makes an angle of /6 with the x-axis. Find its components.
Solution:
x = r cos , y = r sin
So that
x = (10)( / 2), y = 10 (1/2) = 5
We can write the vector as
5 i + 5j
Unit Vectors in the Direction of v
A vector is called a unit vector if it has magnitude = 1. If
v = <a,b>
then the unit vector in the direction of v can be found by
The Unit Vector in the Direction of v
1 u = v || v ||
Example:
The unit vector in the direction of <-3,1> is
We can use the <> notation and the i j notation interchangeably.
Algebra of Vectors
If
v = <a,b> and w = <c,d>
and k is a constant, then we can define the sum an scalar multiplication as follows
Sum and Scalar Multiplication of Vectors
v + w = <a + c,b + d>
and
kv = <ka, kb>
Example
3 <2,1> - 2<-1,3> = <6 + 2,3 - 6>
= <8,-3> = 8i - 3j
Geometrically v + w is the vector that corresponds to the diagonal of the parallelogram with two sides v and w.
The appropriate diagram can also be drawn to show how
v - w = v + (-w).
Prope
rties of Vector Addition and Subtraction
We have the following four properties of vectors: If u, v ,and w are vectors and a and b are numbers then
1. (u + v) + w = u + (v + w)2. a(u + v) = au + av
3. a(bv) = (ab)v
4. u + v = v + u
Applications
A boat captain wants to travel due south at 40 knots. If the current is moving northwest at 16 knots, in what direction and magnitude should he work the engine?
Solution
We have
u = v + w
where u corresponds to the velocity vector of the boat, v corresponds to the engine's vector, and w corresponds to the velocity of the current. We have
u = -40j and w = -8 i + 8 j
Hence
v = u - w = -40j - (-8 i + 8 j) = 8 i - (40+8 )j
The magnitude is
[(8 )2 + (40+8 )2]1/2 = 52.5
The direction is
3 Dimensional Coordinates
To generalize the plane to 3 dimensions, we draw a third axis, called the z-axis at a right angle from the plane so that if you grab on to the z-axis with your right hand your hand will curl from the positive x-axis to the positive y-axis. To plot a point in the xyz-space
We first plot a point in the xy-plane and then draw a segment parallel to the z-axis of length equal to the z coordinate.
Example:
Plot (1,2,3)
Solution:We first draw the x,y, and z-axes. Then we plot the point (1,2) in the xy-plane. Finally move up three units and plot the point.
Exercise
Plot (2,4,3)
The Distance Formula
The distance formula is derived from the three dimensional version of the Pythagorean theorem, which is displayed below.
The distance between two points (x1,y1,z1) and (x2,y2,z2) and is given by
Distance Formula in Three Dimensions
Algebra of vectors in 3D
A vector in space is given by
<x,y,z> = xi + yj + zk
The algebra rules are similar to those in two dimensions.
The Dot and Cross Product
The Dot Product
Definition We define the dot product of two vectors
v = ai + bj and w = ci + dj
to be
v . w = ac + bd
Notice that the dot product of two vectors is a number and not a vector. For 3 dimensional vectors, we define the dot product similarly:
Dot Product in R3
If
v = ai + bj + ck and w = di + ej + fk
then v . w = ad + be + cf
Examples:
If
v = 2i + 4j
and
w = i + 5j
then
v . w = (2)(1) + (4)(5) = 22
Exercise
Find the dot product of
2i + j - k and i + 2j
The Angle Between Two Vectors
We define the angle theta between two vectors v and w by the formula
v . w cos = ||v|| ||w||
so that
v . w = ||v|| ||w|| cos
Two vectors are called orthogonal if their angle is a right angle.
We see that angles are orthogonal if and only if
v . w = 0
Example
To find the angle between
v = 2i + 3j + k
and
w = 4i + j + 2k
we compute:
and
and
v . w = 8 + 3 + 2 = 13
Hence
1 13 cos14 21
Direction Angles
Definition of Direction Cosines
Let
v = ai + bj + ck
be a vector, then we define the direction cosines to be the following:
1. acos = ||v||
2. bcos = ||v||
3. ccos = ||v||
Projections and Components
Suppose that a car is stopped on a steep hill, and let g be the force of gravity acting on it. We can split the vector g into the component that is pushing the car down the road and the component that is pushing the car onto the road. We define
Definition
Let u and v be the vectors. Then u can be broken up into two components, r and s such that s is parallel to v and r is perpendicular to v. s is called the projection of u onto v and r is called the component of u perpendicular to v.
We see that
||u|| ||v|| ||s|| u . v = ||u|| ||v|| cos = = ||v|| ||s|| ||u||
hence
u . v ||s|| = ||v||
We can calculate the projection of u onto v by the formula:
u . vprojvu = v ||v||2
Work
The work done by a constant force F along PQ is given by
W = F . PQ
Example
Find the work done against gravity to move a 10 kg baby from the point (2,3) to the point (5,7)?
Solution
We have that the force vector is
F = ma = (10)(-9.8j) = -98j
and the displacement vector is
v = (5 - 2) i + (7 - 3) j = 3i + 4j
The work is the dot product
W = F . v = (-98j)
. (3i + 4j)
= (0)(3) + (-98)(4) = -392
Notice the negative sign verifies that the work is done against gravity. Hence, it takes 392 J of work to move the baby.
Torque
Suppose you are skiing and have a terrible fall. Your body spins around and you ski stays in place (do not try this at home). With proper bindings your bindings will release and your ski will come off. The bindings recognize that a force has been applied. This force is called torque. To compute it we use the cross produce of two vectors which not only gives the torque, but also produces the direction that is perpendicular to both the force and the direction of the leg.
The Cross Product Between Two Vectors
Definition Let u = ai + bj + ck and v = di + ej + fk be vectors then we define the cross product v x w by the determinant of the matrix:
We can compute this determinant as
= (bf - ce) i + (cd - af) j + (ae - bd) k
Example
Find the cross product u x v if
u = 2i + j - 3k v = 4j + 5k
Solution
We calculate
= 17i - 10j + 8k
Exercises
Find u x v when
A. u = 3i + j - 2k, v = i - kB. u = 2i - 4j - k, v = 3i - j + 2k
Notice that since switching the order of two rows of a determinant changes the sign of the determinant, we have
u x v = -v x u
Geometry and the Cross Product
Let u and v be vectors and consider the parallelogram that the two vectors make. Then
||u x v|| = Area of the Parallelogram
and the direction of u x v is a right angle to the parallelogram that follows the right hand rule
Note: For i x j the magnitude is 1 and the direction is k, hence i x j = k.
Exercise
Find j x k and i x k
Torque Revisited
We define the torque (or the moment M of a force F about a point Q) as
M = PQ x F
Example
A 20 inch wrench is at an angle of 30 degrees with the ground. A force of 40 pounds that makes and angle of 45 degrees with the wrench turns the wrench. Find the torque.
Solution
We can write the wrench as the vector
20 cos 30 i + 20 sin 30 j = 17.3 i + 10 j
and the force as
-40 cos 75 i - 40 sin 75 j = -10.3 i - 38.6 j
hence, the torque is the magnitude of their cross product:
= -564 inch pounds
Parallelepipeds
To find the volume of the parallelepiped spanned by three vectors u, v, and w, we find the triple product:
Volume = u . (v x w)
This can be found by computing the determinate of the three vectors:
Example
Find the volume of the parallelepiped spanned by the vectors
u = <1,0,2> v = <0,2,3> w = <0,1,3>
Solution
We find
Lines and Planes
Lines
Our goal is to come up with the equation of a line given a vector v parallel to the line and a point (a,b,c) on the line. The figure (shown in 2D for simplicity) shows that if P is a point on the line then
<x,y> = P + tv
for some number t.
The picture is the same for 3D. The formula is given below.
Parametric Equations of a Line
The parametric equations for the line through the point (a,b,c) and parallel to the vector v are
<x,y,z> = <a,b,c> + tv
Example:
Find the parametric equations of the line that passes through the point (1, 2, 3) and is parallel to the vector <4, -2, 1>
Find the parametric equations of the line through the two points (2,1,7) and (1,3,5).
Hint: a vector parallel to the line has tail at (2,1,7) and head at (1,3,5).
Planes
If S is a plane then a vector n is normal (perpendicular) to the plane if it is orthogonal to every vector that lies on the plane. Suppose that n is a normal vector to a plane and (a,b,c) is a point on the plane. Let (x,y,z) be a general point on the plane, then
<x - a, y - b, z - c>
is parallel to the plane, hence
n . <x - a, y - b, z - c> = 0
this defines the equation of the plane.
Example:
Find the equation of the plane that contains the point (2,1,0) and has normal vector <1,2,3>
Solution:
We have
<1,2,3> . <x - 2,y - 1,z - 0> = 0
so that
1(x - 2) + 2(y - 1) + 3z = 0
or
x + 2y + 3z = 4
Example
Find the equation of the plane through the points
P = (0,0,1) Q = (2,1,0) and R = (1,1,1)
Solution
Let
v = Q - P = <2, 1, -1>
and
w = R - P = <1, 1, 0>
then to find a vector normal to the plane, we find the cross product of v and w:
or
<1, -1, 1>
We can now use the formula:
<1, -1, 1> . <x, y, z - 1> = 0
or
x - y + z - 1 = 0
or
x - y + z = 1
Distance Between a Point and a Plane
Let P be a point and Q be a point on a plane with normal vector n, then the distance between P and the plane is given by
Distance Between a Point P and a Plane With Normal Vector n
Let Q be a point on the plane with normal vector n. Then the distance from the point P to this plane is given by
||PQ . n|| ProjnPQ = ||n||
Example
Find the distance between the point (1,2,3) and the plane
2x - y - 2z = 5
Solution
The normal vector can be read off from the equation as
n = <2, -1, -2>
Now find a convenient point on the plane such as Q = (0, -5, 0). We have
PQ = <-1, -7, -3>
and
n . PQ = -2 + 7 + 6 = 11
We find the magnitude of n by taking the square root of the sum of the squares. The sum is
4 + 1 + 4 = 9
so
|| n || = 3
Hence the distance from the point to the plane is 11/3.
The Angle Between 2 Planes
The angle between two planes is given by the angle between the normal vectors.
Example
Find the angle between the two planes
3x - 2y + 5z = 1 and 4x + 2y - z = 4
We have the two normal vectors are
n = <3,-2,5> and m = <4,2,-1>
We have
n . m = 3,
, 38 21 n m
hence the angle is
3arccos 1.46
38 21
Surfaces
Cylinders
Let C be a curve, then we define a cylinder to be the set of all lines through C and perpendicular to the plane that C lies in.
We can tell that an equation is a cylinder is it is missing one of the variables.
Quadric Surfaces
Recall that the quadrics or conics are lines, hyperbolas, parabolas, circles, and ellipses. In three dimensions, we can combine any two of these and make a quadric surface. For example
is a paraboloid since for constant z we get a circle and for constant x or y we get a parabola. We use the suffix -oid to mean ellipse or circle. We have:
x2 y2 z2
+ + = 1 is an ellipsoid a2 b2 c2
x2 y2 z2
- - + = 1 is a hyperboloid of 2 sheets while a2 b2 c2
x2 y2 z2
+ - = 1 is a hyperboloid of 1 sheet a2 b2 c2
Surface of Revolution
Let y = f(x) be a curve, then the equation of the surface of revolution abut the x-axis is
y2 + z2 = f(x)2
Example
Find the equation of the surface that is formed when the curve
y = sin x 0 < x < /2
is revolved around the y-axis.
Solution
This uses a different formula since this time the curve is revolved around the y-axis. The circular cross section has radius sin-1 y and the circle is perpendicular to the y-axis. Hence the equation is
x2 + z2 = (sin-1 y)2
Cylindrical Coordinates
We can extend polar coordinates to three dimensions by
x = rcosy = rsinz = z
Example
We can write (1,1,3) in cylindrical coordinates. We find
and
so that the cylindrical coordinates are
( , /4, 3)
Spherical Coordinates
An alternate coordinate system works on a distance and two angle method called spherical coordinates. We let denote the distance from the point to the origin, represent the same as in cylindrical coordinates, and denote the angle from the positive z-axis to the point. The picture tells us that
r = sin
and that
z = cos
From this we can find
x = rcos = sin cos y = r sin = sin sin z = cos
Immediately we see that
x2 + y2 + z2 = 2
We use spherical coordinates whenever the problem involves a distance from a source.
Example
convert the surface
z = x2 + y2
to an equation in spherical coordinates.
Solution
We add z2 to both sides
z + z2 = x2 + y2 + z2
Now it is easier to convert
cos + 2 cos2 = 2
Divide by to get
cos + cos2 =
Now solve for .
cos cos = = = csc cot 1 - cos2 sin2
Vector Valued Functions
Definition of a Vector Valued Function
A vector valued function is a function where the domain is a subset of the real numbers and the range is a vector.
In two dimensions
r(t) = x(t)i + y(t)j
In three dimensions
r(t) = x(t)i + y(t)j + z(t)k
You will notice the strong resemblance to parametric equations. In fact there is an equivalence between vector valued functions and parametric equations.
Example
r(t) = 3i + tj + (sin t) k
To graph a vector valued function we can just graph the parametrically defined function
Example
Sketch the graph of
r(t) = (t - 1)i + t2 j
Solution
We draw vectors for several values of t and connect the dots. Notice that the graph is the same as
y = (x + 1)2
Limits
We define the limit of a vector valued function by taking the limit of each of the components. Formally
The Limit of a Vector Valued Function
Example
Find the limit
if
sin t r(t) = et i + j + (t ln t) k t
Solution
We take the three limits one at a time
The first function is continuous at t = 0, so we can just plug in to get
e0 = 1
For the second function, we get 0/0, so we use L'Hospital's rule to get
cos t 1
Now plug in to get
1/1 = 1
For the kth component, we rewrite as
ln t 1/t
Now use L'Hospital's rule to get
1/t = t -1/t2
Plugging in 0 gives 0. Finally, gathering our results gives a limit of
i + j
Continuity
We define continuity of vector valued functions in a similar way to how continuity of real valued functions was defined.
Definition of Continuity
A vector valued function is continuous at t0 if it is defined at t0 and
The practical way to investigate continuity is to look at each of the components.
Example
Determine where the following vector valued function is continuous.
r(t) = ln(1 - t)i + 1/t j + 3t k
Solution
The first component is continuous for all values of t less than 1, the second component is continuous for t nonzero, and the third component is continuous for all real numbers. We can conclude that r(t) is continuous for all t less than 1 but not equal to 0.
Differentiation and Integration of Vector Valued Functions
Calculus of Vector Valued Functions
The formal definition of the derivative of a vector valued function is very similar to the definition of the derivative of a real valued function.
The Derivative of a Vector Valued Function
Let r(t) be a vector valued function, then
Because the derivative of a sum is the sum of the derivative, we can find the derivative of each of the components of the vector valued function to find its derivative.
Examples
d/dt (3i + sintj) = costj
d/dt (3t2 i + cos(4t) j + tet k) = 6t i -4sin(t)j + (et + tet) k
Properties of Vector Valued Functions
All of the properties of differentiation still hold for vector values functions. Moreover because there are a variety of ways of defining multiplication, there is an abundance of product rules.
Properties of Vector Valued Functions
Suppose that v(t) and w(t) are vector valued functions, f(t) is a scalar function, and c is a real number then
1. d/dt(v(t) + w(t)) = d/dt(v(t)) + d/dt(w(t))
2. d/dt(cv(t)) = c d/dt(v(t))
3. d/dt(f(t) v(t)) = f '(t) v(t) + f(t) v'(t)
4. (v(t) . w(t))' = v'(t)
. w(t)+ v(t) . w'(t)
5. (v(t) x w(t))' = v'(t) x w(t)+ v(t) x w'(t)
6. d/dt(v(f(t))) = v'(f(t)) f '(t)
Example
Show that if r is a differentiable vector valued function with constant magnitude, then
r . r' = 0
Solution
Since r has constant magnitude, call it k,
k2 = ||r||2 = r . r
Taking derivatives of the left and right sides gives
0 = (r . r)' = r'
. r + r
. r'
= r . r' + r
. r' = 2r
. r'
Divide by two and the result follows
Integration of vector valued functions
We define the integral of a vector valued function as the integral of each component. This definition holds for both definite and indefinite integrals.
Example
Evaluate
(sin t)i + 2t j - 8t3 k dt
Solution
Just take the integral of each component
( (sin t)dt i) + ( 2t dt j) - ( 8t3 dt k)
= (-cost + c1)i + (t2 + c2)j + (2t4 + c3)k
Notice that we have introduce three different constants, one for each component.
Velocity and Acceleration
Definition of Velocity and Speed
In single variable calculus the velocity is defined as the derivative of the position function. For vector calculus, we make the same definition.
Definition of Velocity
Let r(t) be a differentiable vector valued function representing the position vector of a particle at time t. Then the velocity vector is the derivative of the position vector.
v(t) = r'(t) = x'(t)i + y'(t)j + z'(t)k
Example
Find the velocity vector v(t) if the position vector is
r(t) = 3ti + 2t2j - sin t k
Solution
We just take the derivative
v(t) = 3i + 4tj + cos t k
When we think of speed, we think of how fast we are going. Speed should not be negative. In one variable calculus, speed was the absolute value of the velocity. For vector calculus, it is the magnitude of the velocity.
Definition of Speed
Let r(t) be a differentiable vector valued function representing the position of a particle. Then the speed of the particle is the magnitude of the velocity vector.
Speed = || v(t) || = || r'(t) ||
Example
Let
r(t) = 3i + 2t j + cos t k
Find the speed after /4 seconds.
Solution
We first find the velocity vector
v(t) = r'(t) = 2 j - sin t k
We have
v(p/4) = 2 j - /2 k
Its magnitude is the square root of the sum of the squares or
Speed = || v || =
Acceleration
In one variable calculus, we defined the acceleration of a particle as the second derivative of the position function. Nothing changes for vector calculus.
Definition of Acceleration
Let r(t) be a twice differentiable vector valued function representing the position vector of a particle at time t. Then the acceleration vector is the second derivative of the position vector.
a(t) = r''(t) = x''(t)i + y''(t)j + z''(t)k
Example
Find the velocity and acceleration of the position function
r(t) = 4t i + t2 j
when t = -1. Then sketch the vectors.
Solution
The velocity vector is
v(t) = r'(t) = 4 i + 2t j
Plugging in -1 for t gives
v(-1) = 4 i - 2j
Take another derivative to find the acceleration.
a(t) = v'(t) = 2j
Below is a picture of the vectors.
Projectile Motion
Since the velocity and acceleration vectors are defined as first and second derivatives of the position vector, we can get back to the position vector by integrating.
Example
You are a anti-missile operator and have spotted a missile heading towards you at the position
re = 1000i + 500j
with velocity
ve = -30i + 3j
You can fire your anti-missile at 100 meters per second. At what angle should you fire it so that you intercept the missile. Assume that gravity is the only force acting on the projectiles.
Solution
The acceleration vector of the enemy missile is
ae(t) = -9.8 j
Integrating, we get the velocity vector
ve(t) = v1 i + (v2 - 9.8t) j
Setting t = 0 and using the initial velocity of the enemy missile gives
ve(t) = -30 i + (3 - 9.8t) j
Now integrate again to find the position function
re(t) = (-30t + r1) i + (-4.9t2 + 3t + r2) j
Again setting t = 0 and using the initial conditions gives
re(t) = (-30t + 1000) i + (-4.9t2 + 3t + 500) j
The acceleration of your anti-missile-missile is also
ay(t) = -9.8 j
Integrating, we get the velocity vector
vy(t) = v1 i + (v2 - 9.8t) j
Since the magnitude of our velocity is 100, we can say
vy(0) = 100 cos i + 100 sin j
So that
vy(t) = 100 cos i + (100 sin - 9.8t) j
Now integrate again to find the position function
ry(t) = (100t cos + r1) i + (-4.9t2 + 100t sin + r2) j
Our anti-missile-missile starts out at base, so the initial position is the origin. All the constants are zero.
ry(t) = (100t cos ) i + (-4.9t2 + 100t sin ) j
Since we want to intercept the enemy missile, we set the position vectors equal to each other.
(100t cos ) i + (-4.9t2 + 100t sin ) j = (-30t + 1000) i + (-4.9t2 + 3t + 500) j
Equating coefficients gives
100t cos = -30t + 1000
-4.9t2 + 100t sin = -4.9t2 + 3t + 500
The first equation gives
1000 t = 100cos + 30
Simplifying the second equation and substituting gives
100000 sin 3000 = + 500 100cos + 30 100cos + 30
Clear denominators to get
100000 sin = 3000 + 50000 cos + 15000
At this point we use a calculator to solve for to
= .62535 radians
The Unit Tangent and the Unit Normal Vectors
The Unit Tangent Vector
The derivative of a vector valued function gives a new vector valued function that is tangent to the defined curve. The analogue to the slope of the tangent line is the direction of the tangent line. Since a vector contains a magnitude and a direction, the velocity vector contains more information than we need. We can strip a vector of its magnitude by dividing by its magnitude.
Definition of the Unit Tangent Vector
Let r(t) be a differentiable vector valued function and v(t) = r'(t) be the velocity vector. Then we define the unit tangent vector by as the unit vector in the direction of the velocity vector.
v(t) T(t) = ||v(t)||
Example
Let
r(t) = t i + et j - 3t2 k
Find the T(t) and T(0).
Solution
We have
v(t) = r'(t) = i + et j - 6t k
and
To find the unit tangent vector, we just divide
To find T(0) plug in 0 to get
The Principal Unit Normal Vector
A normal vector is a perpendicular vector. Given a vector v in the space, there are infinitely many perpendicular vectors. Our goal is to select a special vector that is normal to the unit tangent vector. Geometrically, for a non straight curve, this vector is the unique vector that point into the curve. Algebraically we can compute the vector using the following definition.
Definition of the Principal Unit Normal Vector
Let r(t) be a differentiable vector valued function and let T(t) be the unit tangent vector. Then the principal unit normal vector N(t) is defined by
T'(t) N(t) = ||T'(t)||
Comparing this with the formula for the unit tangent vector, if we think of the unit tangent vector as a vector valued function, then the principal unit normal vector is the unit tangent vector of the unit tangent vector function. You will find that finding the principal unit normal vector is almost always cumbersome. The quotient rule usually rears its ugly head.
Example
Find the unit normal vector for the vector valued function
r(t) = ti + t2 j
and sketch the curve, the unit tangent and unit normal vectors when t = 1.
Solution
First we find the unit tangent vector
Now use the quotient rule to find T'(t)
Since the unit vector in the direction of a given vector will be the same after multiplying the vector by a positive scalar, we can simplify by multiplying by the factor
The first factor gets rid of the denominator and the second factor gets rid of the fractional power. We have
Now we divide by the magnitude (after first dividing by 2) to get
Now plug in 1 for both the unit tangent vector to get
The picture below shows the graph and the two vectors.
Tangential and Normal Components of Acceleration
Imagine yourself driving down from Echo Summit towards Myers and having your brakes fail. As you are riding you will experience two forces (other than the force of terror) that will change the velocity. The force of gravity will cause the car to increase in speed. A second change in velocity will be caused by the car going around the curve. The first component of acceleration is called the tangential component of acceleration and the second is called the normal component of acceleration. As you may guess the tangential component of acceleration is in the direction of the unit tangent vector and the normal component of acceleration is in the direction of the principal unit normal vector. Once we have T and N, it is straightforward to find the two components. We have
Tangential and Normal Components of Acceleration
The tangential component of acceleration is
and the normal component of acceleration is
and
a = aNN + aTT
Proof
First notice that
v = ||v|| T and T' = ||T'|| N
Taking the derivative of both sides gives
a = v' = ||v||' T + ||v|| T' = ||v||' T + ||v|| ||T' || N
This tells us that the acceleration vector is in the plane that contains the unit tangent vector and the unit normal vector. The first equality follows immediately from the definition of the component of a vector in the direction of another vector. The second equalities will be left as exercises.
Example
Find the tangential and normal components of acceleration for the prior example
r(t) = ti + t2 j
Solution
Taking two derivatives, we have
a(t) = r''(t) = 2j
We dot the acceleration vector with the unit tangent and normal vectors to get
For a parametrically defined curve we had the definition of arc length. Since vector valued functions are parametrically defined curves in disguise, we have the same definition. We have the added benefit of notation with vector valued functions in that the square root of the sum of the squares of the derivatives is just the magnitude of the velocity vector.
Definition of Arc Length
Let
r(t) = x(t) i + y(t) j + z(t) k
be a differentiable vector valued function on [a,b]. Then the arc length s is defined by
Example
Suppose that
r(t) = 3t i + 2j + t2k
Set up the integral that defines the arc length of the curve from 2 to 3. Then use a calculator or computer to approximate the arc length.
Solution
We use the arc length formula
Notice that we could do this integral by hand by letting t = 9/2 tan , however the question only asked us to use a machine to approximate the integral. A TI 89 calculator gives
Recall that like parametric equations, vector valued function describe not just the path of the particle, but also how the particle is moving. Among all representations of a curve there is a "simplest" one. If the particle travels at the constant rate of one unit per second, then we say that the curve is parameterized by arc length. We have seen this concept before in the definition of radians. On a unit circle one radian is one unit of arc length around the circle. When we say "simplest" we in no way mean that the equations are simple to find, but rather that the dynamics of the particle are simple. To aid us in parameterizing by arc length, we define the arc length function.
Definition of the Arc Length Function
If r(t) is a differentiable vector valued function, then the arc length function is defined by
Remark: By the second fundamental theorem of calculus, we have
s'(t) = ||v(t)||
If a vector valued function is parameterized by arc length, then
s(t) = t
If we have a vector valued function r(t) with arc length s(t), then we can introduce a new variable
s = s-1(t)
So that the vector valued function r(s) will have arc length equal to
s(s-1(t)) = t
so that r(s) will be parameterized by arc length. This process is usually impossible for two reasons. Firstly, the integral that defines arc length involves a square root in the integrand. This integral is usually impossible to determine. Secondly, even if the integral is possible to evaluate, finding the inverse of a function is often impossible. There are a few special curves that can be parameterized by arc length. We will demonstrate one of them.
Example: Parameterizing by arc length
Find the arc length parameterization of the helix defined by
r(t) = cos t i + sin t j + t k
Solution
First find the arc length function
Solving for t gives
t = s /
Now substitute back into the position equation to get
r(s) = cos(s / ) i + sin(s / ) j + s / k
Definition of Curvature
Consider a car driving along a curvy road. The tighter the curve, the more difficult the driving is. In math we have a number, the curvature, that describes this "tightness". If the curvature is zero then the curve looks like a line near this point. While if the curvature is a large number, then the curve has a sharp bend.
More formally, if T(t) is the unit tangent vector function then the curvature is defined at the rate at which the unit Tangent vector changes with respect to arc length.
Curvature = k = ||d/ds (T(t)) || = ||r''(s)||
As we stated previously, this is not a practical definition, since parameterizing by arc length is typically impossible. Instead we use the chain rule to get
||d/ds (T(t)) || = ||T'(t) dt/ds||
||T'(t)|| ||T'(t)|| = = ||ds/dt|| ||r'(t)||
This formula is more practical to use, but still cumbersome. T'(t) is typically a mess. Instead we can borrow from the formula for the normal vector to get the curvature
||r'(t) x r''(t)|| K(t) = ||r'(t)||3
Example
Find the curvature at /2 if
r(t) = cos t i - 1/t j + sin t k
Solution
We take derivatives
r'(t) = -sin t i + 1/t2 j + cos t k
r''(t) = -cos t i - 2/t3 j - sin t k
Plugging in /2 gives
r'(/2) = -i + 4/2 j
r''(/2) = -16/3 j - k
Now take the cross product to get
r'(/2) x r''(/2) = -4/2 i - j + 16/3 k
Finally, we plug this information into the curvature formula to get
Curvature of a Plane Curve
If a curve in the xy-plane is defined by the function y = f(t) then there is an easier formula for the curvature. We can parameterize the curve by
r(t) = t i + f(t) j
We have
r'(t) = i + f '(t) j
r''(t) = f ''(t) j
Their cross product is just
r'(t) x r''(t) = f ''(t) k
which has magnitude
||r'(t) x r''(t)|| = |f ''(t)|
The curvature formula gives
Curvature for a Plane Curve
Example
Find the curvature for the curve
y = sin x
Solution
We have
f '(x) = cos x
f ''(x) = -sin x
Plugging into the curvature formula gives
The Osculating Circle
In first year calculus we saw how to approximate a curve with a line , parabola, etc. Instead we can find the best fitting circle at the point on the curve. If P is a point on the curve, then the best fitting circle will have the same curvature as the curve and will pass through the point P. We will see that the curvature of a circle is a constant 1/r, where r is the radius of the circle. The center of the osculating circle will be on the line containing the normal vector to the circle. In particular the center can be found by adding
OP + 1/K N
Exercise
Find the equation of osculating circle to y = x2 at x = -1.
How is the normal component of acceleration related to the curvature. If you remember, the normal component the acceleration tells us how fast the particle is changing direction. If a curve has a sharp bend (high curvature) then the directional change will be faster. We now show that there is a definite relationship between the normal component of acceleration and curvature.
A function of several variables is a function where the domain is a subset of Rn and range is R.
Example:
f(x,y) = x - y
is a function of two variables
x - y g(x,y,z) = y - z
is a function of three variables.
Finding the Domain
To find the domain of a function of several variables, we look for zero denominators and negatives under square roots:
Example
Find the domain of
First, the inside of the square root must be positive, that is
x - y > 0
second, the denominator must be nonzero, that is
x + y 0
hence we need to stay off the line
y = -x
Putting this together gives
{(x,y) | x - y > 0 and y -x}
The graph to the right shows the domain as the shaded green region.
.
Exercise
Find the domain of the function
Level Curves
The topographical map shown below is of the Rubicon Trail. It represents the function that maps a longitude and latitude to an altitude.
Each curve represents a path where the z-coordinate (altitude) is a constant. Crossing many topo lines in a short distance represents a path that is very steep.
Now lets make our own contour map of the function.
f(x,y) = y - x2
by setting constant values for z:
z Equation
1 y = x2 + 1
2 y = x2 + 2
We see that each topo line is a parabola and that the y-intercept gives the height. Below is a contour diagram of this function.
Names for the curves drawn are level curves, isotherms (for temperature), isobars (for pressure), and equipotential lines (for electric potential fields) depending on what the two variable function represents.
Let P be a point in the plane then a neighborhood (ball) of P is the set of points that are less than units away from P. If R is a region, then a point P is called an interior point of R if there is a neighborhood totally contained in R. If every point of R is an interior point of R then R is called open. A point P is called a boundary point of R if every neighborhood of R contains both points in R and not in R. R is called closed if it contains all of its boundary points.
The Definition of a Limit
Definition
Let f(x,y) be a function defined near the point P, then
if there is a such that f(x,y) is close to L for all points (except possibly P) in the neighborhood of P.
Equivalently, the limit is L if for all paths that lead to P, the function also tends towards P. (Recall that for the one variable case we needed to check only the path from the left and from the right.) To show that a limit does not exist at a point we need only find two paths that both lead to P such that f(x,y) tends towards different values.
Techniques For Finding Limits
Example
Show that
Does not exist
Solution
First select the path along the x-axis. On this path
y = 0
so the function becomes:
0 f(x,0) = = 0 x2
Now choose the path along the y = x line:
x2 1 f(0,y) = = 2x2 2
Hence the function tends towards two different values for different paths. We can conclude that the limit does not exist. The graph is pictured below.
Example
Find
We could try the paths from the last example, but both paths give a value of 0 for the limit. Hence we suspect that the limit exists. We convert to polar coordinates and take the limit as r approaches 0:
We have
r3cos3 + r3sin3
f(r,) = = rcos3 + rsin3 r2
as r approaches 0 the function also approaches 0 no matter what is. Hence the limit is 0.
Below is the graph of this function.
Exercises: Find the limit if it exists
A.
B.
Continuity
We make the following definition for continuity.
Definition
A function of several variables is continuous at a point P if the limit exists at P and the function defined at P is equal to this limit.
As with functions of one variable, polynomials are continuous, sums, products, and compositions of continuous functions are continuous. Quotients of continuous functions are continuous. A function is continuous if it is continuous at every point.
Let f(x,y) be a function of two variables. Then we define the partial derivatives as
Definition of the Partial Derivative
if these limits exist.
Algebraically, we can think of the partial derivative of a function with respect to x as the derivative of the function with y held constant. Geometrically, the derivative with respect to x at a point P represents the slope of the curve that passes through P whose projection onto the xy plane is a horizontal line. (If you travel due East, how steep are you climbing?)
Example
Let
f(x,y) = 2x + 3y
then
We also use the notation fx and fy for the partial derivatives with respect to x and y respectively.
Exercise:
Find fy for the function from the example above.
Finding Partial Derivatives the Easy Way
Since a partial derivative with respect to x is a derivative with the rest of the variables held constant, we can find the partial derivative by taking the regular derivative considering the rest of the variables as constants.
Example Let
f(x,y) = 3xy2 - 2x2y
then
fx = 3y2 - 4xy
and
fy = 6xy - 2x2
Exercises
Find both partial derivatives for
A. f(x,y) = xy sin x
B. x + yf(x,y) = x - y
Higher Order Partials
Just as with function of one variable, we can define second derivatives for functions of two variables. For functions of two variables, we have four types:
fxx, fxy fyx and fyy
Example
Let
f(x,y) = y ex
then
fx = yex
and
fy = ex
Now taking the partials of each of these we get:
fxx = y ex fxy = ex fyx = ex and fyy = 0
Notice that
fxy = fyx
Theorem
Let f(x,y) be a function with continuous second order derivatives, then
fxy = fyx
Functions of More Than Two Variables
Suppose that
f(x,y,z) = xy - 2yz
is a function of three variables, then we can define the partial derivatives in much the same way as we defined the partial derivatives for three variables.
We have
fx = y fy = x - 2z and fz = -2y
Application: The Heat Equation
Suppose that a building has a door open during a snowy day. It can be shown that the equation
Ht = c2Hxx
models this situation where H is the heat of the room at the point x feet away from the door at time t. Show that
Suppose you are given a topographical map and want to see how steep it is from a point that is neither due West or due North. Recall that the slopes due north and due west are the two partial derivatives. The slopes in other directions will be called the directional derivatives. Formally, we define
Definition
Let f(x,y) be a differentiable function and let u be a unit vector then the directional derivative of f in the direction of u is
Note that if u is i then the directional derivative is just fx and if u is j the it is fy.
Just as there is a difficult and an easy way to compute partial derivatives, there is a difficult way and an easy way to compute directional derivatives.
Theorem
Let f(x,y) be a differentiable function, and u be a unit vector with direction , then
( , ) , cos ,sinu x yD f x y f f
Example:
Let
f(x,y) = 2x + 3y2 - xy
and
v = <3,2>
Find
Dv f(x,y)
Solution
We have
fx = 2 - y
and
fy = 6y - x
and
Hence
Dv f(x,y) = <2 - y, 6y - x> . <3/ , 2/ >
2 3 = (2 - y) + (6y - x)
Exercise
Let
Find Dv f(x,y)
The Gradient
We define
grad f = <fx, fy>
Notice that
Du f(x,y) = (grad f) . u
The gradient has a special place among directional derivatives. The theorem below states this relationship.
Theorem
1. If grad f(x,y) = 0 then for all u, Du f(x,y) = 02. The direction of grad f(x,y) is the direction with
maximal directional derivative.
3. The direction of -grad f(x,y) is the direction with the minimal directional derivative.
Proof:
1. If
gradf(x,y) = 0
then
Du f(x,y) = grad f . u = 0 . u = 02. Du f(x,y) = grad f . u = ||grad f || cos
This is a maximum when = 0 and a minimum when = . If = 0 then grad f and u point in the same direction. If = then u and grad f point in opposite directions. This proves 2 and 3.
Example:
Suppose that a hill has altitude
w(x,y) = x2 - y
Find the direction that is the steepest uphill and the steepest downhill at the point (2,3).
Solution
We find
grad w = <2x, -y> = <4, -3>
Hence the steepest uphill is in the direction
<4,-3>
while the steepest downhill is in the direction
-<4,-3> = <-4,3>
The Gradient and Level Curves
If f is differentiable at (a,b) and grad f is nonzero at (a,b) then grad f is perpendicular to the level curve through (a,b).
Let z = f(x,y) be a function of two variables. We can define a new function F(x,y,z) of three variables by subtracting z. This has the condition
F(x,y,z) = 0
Now consider any curve defined parametrically by
x = x(t), y = y(t) z = z(t)
We can write,
F(x(t), y(t), z(t)) = 0
Differentiating both sides with respect to t, and using the chain rule gives
Fx(x, y, z) x' + Fy(x, y, z) y' + Fz(x, y, z) z' = 0
Notice that this is the dot product of the gradient function and the vector <x',y',z'>,
GradF . <x', y', z'> = 0
In particular the gradient vector is orthogonal to the tangent line of any curve on the surface. This leads to
Definition
Let F(x,y,z) define a surface that is differentiable at a point (x0,y0,z0), then the tangent plane to F ( x, y, z ) at ( x0 , y0 , z0 ) is the plane with normal vector
Grad F(x0,y0,z0)
that passes through the point (x0,y0,z0). In Particular the equation of the tangent plane is
Grad F(x0,y0,z0) . < x - x0 , y - y0 , z - z0 > = 0
Example
Find the equation of the tangent plane to
z = 3x2 - xy
at the point (1,2,1)
Solution
We let
F(x,y,z) = 3x2 - xy - z
then
Grad F = <6x - y, -x, -1>
At the point (1,2,1), the normal vector is
Grad F(1,2,1) = <4, -1, -1>
Now use the point normal formula for a plan
<4, -1, -1> . <x - 1, y - 2, z - 1> = 0
or
4(x - 1) - (y - 2) - (z - 1) = 0
Finally we get
4x - y - z = 1
Normal Lines
Given a vector and a point, there is a unique line parallel to that vector that passes through the point. In the context of surfaces, we have the gradient vector of the surface at a given point. This leads to the following definition.
DefinitionLet F(x,y,z) define a surface that is differentiable at a point (x0,y0,z0), then the normal line to F(x,y,z) at ( x0 , y0 , z0 ) is the line with normal vector
GradF(x0,y0,z0)
that passes through the point (x0,y0,z0). In Particular the equation of the normal line is
x(t) = x0 + Fx(x0,y0,z0) t
y(t) = y0 + Fy(x0,y0,z0) t
z(t) = z0 + Fz(x0,y0,z0) t
Example
Find the parametric equations for the normal line to
x2yz - y + z - 7 = 0
at the point (1,2,3).
Solution
We compute the gradient
Grad F = <2xyz, x2z - 1, x2y + 1> = <12, 2, 3>
Now use the formula to find
x(t) = 1 + 12t y(t) = 2 + 2t z(t) = 3 + 3t
The diagram below displays the surface and the normal line.
Angle of Inclination
Given a plane with normal vector n the angle of inclination, is defined by
|n . k| cos = ||n||
More generally, if
F(x,y,z) = 0
is a surface, than the angle of inclination at the point (x0, y0, z0) is defined by the angle of inclination of the tangent plane at the point.
Find the tangent line to the curve of intersection of the sphere
x2 + y 2 + z2 = 30
and the paraboloid
z = x2 + y2
at the point (1,2,5).
Solution
We find the Grad of the two surfaces at the point
Grad (x2 + y 2 + z2) = <2x, 2y, 2z> = <2, 4,10>
and
Grad (x2 + y 2 - z) = <2x, 2y, -1> = <2, 4, -1>
These two vectors will both be perpendicular to the tangent line to the curve at the point, hence their cross product will be parallel to this tangent line. We compute
Hence the equation of the tangent line is
x(t) = 1 - 44t y(t) = 2 + 22t z(t) = 5
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MATH 202 PRACTICE MIDTERM 1
Please work out five of the given six problems and indicate which problem
you are omitting. Credit will be based on the steps that you show towards
the final answer. Show your work.
Printable Key
PROBLEM 1 Please answer the following true or false. If false, explain why or provide
a counter example. If true explain why
A) (12 Points) If r(t) is parameterized by arclength, then a and N are parallel.
Solution
B) (13 Points) If r(t) is a differentiable vector valued function then
Solution
PROBLEM 2 (25 Points)
Let
r(t) = 2t i - 4t2 j
A. Find T(-1).
Solution
B. Find N(-1).
Solution
C. Find the equation of the circle of curvature for r(t) at t = -1.
Just as we can take partial derivative by considering only one of the variables a true variable and holding the rest of the variables constant, we can take a "partial integral". We indicate which is the true variable by writing "dx", "dy", etc. Also as with partial derivatives, we can take two "partial integrals" taking one variable at a time. In practice, we will either take x first then y or y first then x. We call this an iterated integral or a double integral.
Definition of a Double Integral
Let f(x,y) be a function of two variables defined on a region R bounded below and above by
y = g1(x) and y = g2(x)
and to the left and right by
x = a and x = b
then the double integral (or iterated integral) of f(x,y) over R is defined by
Example
Find the double integral of f(x,y) = 6x2 + 2y over R where R is the region between y = x2 and y = 4.
Solution
First we have that the inside limits of integration are x2 and 4. The region is bounded from the left by x = -2 and from the right by x = 2 as indicated by the picture below.
If a region is bounded from the left by x = h1(y) and the right by x = h2(y) and below and above by y = c and y = d, then we can find the double integral of "dxdy" by first integrating with respect to x then with respect to y. Sometimes there is a choice to make as to whether to integrate first with respect to x and then with respect to y. We do whatever is easier.
Example
Find the double integral of f(x,y) = 3y over the triangle with vertices (-1,1), (0,0), and (1,1).
Solution
If we try to integrate with respect y first, we will have to cut the region into two pieces and perform two iterated integrals. Instead we integrate with respect to x first. The region is bounded on the left and the right by x = -y and x = y. The lowest the region gets is y = 0 and the highest is y = 1. The integral is
Example
Evaluate the integral
Solution
Try as you may, you will not find an antiderivative of and we don't want to get into
power series expansions. We have another choice. The picture below shows the region.
We can switch the order of integration. The region is bounded above and below by y = 1/3 x and y = 0. The double integral with respect to y first and then with respect to x is
The integrand is just a constant with respect to y so we get
This integral can be performed with simple u-substitution.
u = x2 du = 2x dx
and the integral becomes
Area
Recall from first year calculus, if a region R is bounded below by y = g1(x) and above by y = g2(x), and a < x < b, the area is given by
There is another way of achieving this expression. If we let the integrand by 1 then the double integral over the region R is
This gives us another way of finding area.
Theorem: Area and Double Integrals
If a region R is bounded below by y = g1(x) and above by y = g2(x), and a < x < b, then the area is given by
Remark: If the region if bounded on the left by x = h1(y) and the right by h2(y) with c < y < d, then the double integral of 1 dxdy can also be used to find the area.
Example
Set up the double integral that gives the area between y = x2 and y = x3. Then use a computer or calculator to evaluate this integral.
Solution
The picture below shows the region
We set up the integral
A computer gives the answer of 1/12.
An animation of the "Little Man"
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Double Integrals and Volume
Double Riemann Sums
In first year calculus, the definite integral was defined as a Riemann sum that gave the area under a curve. There is a similar definition for the volume of a region below a function of two variables. Let f(x,y) be a positive function of two variables and consider the solid that is bounded below by f(x,y) and above a region R in the xy-plane.
For a two dimensional region, we approximated the area by adding up the areas of many approximating rectangles. For the volume of a three dimensional solid, we take a similar approach. Instead of rectangles, we use rectangular solids for the approximation. We cut the region R into rectangles by drawing vertical and horizontal lines in the xy-plane. Rectangles will be formed. We let the rectangles be the base of the solid, while the height is the z-coordinate of the lower left vertex. One such rectangular solid is shown in the figure.
Taking the limit as the rectangle size approaches zero (and the number of rectangles approaches infinity) will give the volume of the solid. If we fix a value of x and look at the rectangular solids that contain this x, the union of the solids will be a solid with constant width x. The face will be approximately equal to the area in the yz-plane of the (one variable since x is held constant) function z = f(x,y).
If we add up all these slices and take the limit as x approaches zero, we get
Which is just the double integral defined in the last section
Instead of fixing the variable x we could have held y constant. The picture below illustrates the resulting wedge.
By a similar argument, the volume of the wedge is
Adding up all the wedge areas gives the total volume
This shows that the volume is equal to the iterated integral no matter which we integrate first. This is called Fubini's Theorem. Technically the volume is defined as the double Riemann sum of f(x,y) where we sum over the partition of R in the xy-plane. We state it below.
Fubini's Theorem
Let f, g1, g2, h1, and h2 be defined and continuous on a region R. Then the double integral equals
Notice that all the typical properties of the double integral hold. For example, constants can be pulled out and the double integral of the sum of two functions is the sum of the double integrals of each function.
Finding Volume
Example
Set up the integral to find the volume of the solid that lies below the cone
and above the xy-plane.
Solution
The cone is sketched below
We can see that the region R is the blue circle in the xy-plane. We can find the equation by setting z = 0.
Solving for y (by moving the square root to the left hand side, squaring both sides, etc)
gives
The "-" gives the lower limit and the "+" gives the upper limit. For the outer limits, we can see that
-4 < x < 4
Putting this all together gives
Either by hand or by machine we can obtain the result
Volume = 64 /3
Notice that this agrees with the formula
Volume = r2h/3
Exercise
Set up the double integral for this problem with dxdy instead of dydx. Then show that the two integrals give the same result.
Example
Set up the double integral that gives the volume of the solid that lies below the sphere
x2 + y2 + z2 = 6
and above the paraboloid
z = x2 + y2
Solution
The picture below indicated that the region is the disk that lies inside that circle of intersection of the two surfaces. We substitute
x2 + y2 + (x2 + y2)2 = 6
or
x2 + y2 + (x2 + y2)2 - 6 = 0
Now factor with x2 + y2 as the variable to get
(x2 + y2 - 2)(x2 + y2 + 3) = 0
The second factor has no solution, while the first is
x2 + y2 = 2
Solving for y gives
and
- < x <
Just as we did in one variable calculus, the volume between two surfaces is the double integral of the top surface minus the bottom surface. We have
Again we can perform this integral by hand or by machine and get
Volume = 7.74
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Double Integration in Polar Coordinates
Polar Double Integration Formula
Many of the double integrals that we have encountered so far have involved circles or at least expressions with x2 + y2. When we see these expressions a bell should ring and we should shout, "Can't we use polar coordinates." The answer is, "Yes" but only with care. Recall that when we changed variables in single variable integration such as u = 2x, we needed to work out the stretching factor du = 2dx. The idea is similar with two variable integration. When we change to polar coordinates, there will also be a stretching factor. This is evident since the area of the "polar rectangle" is not just as one may expect . The picture is shown below.
Even if r and are very small the area is not the product (r)(). This comes from the definition of radians. An arc that extends radians a distance r out from the origin has length r. If both r and are very small then the polar rectangle has area
Area = r r
This leads us to the following theorem
Theorem: Double Integration in Polar Coordinates
Let f(x,y) be a continuous function defined over a region R bounded in polar coordinates by
This is definitely a case for polar coordinates. The region R is the part of the xy-plane that is inside the cylinder. In polar coordinates, the cylinder has equation
r2 = 4
Taking square roots and recalling that r is positive gives
r = 2
The inside of the cylinder is thus the polar rectangle
0 < r < 2 0 < < 2
The equation of the parabola becomes
z = 9 - r2
We find the integral
This integral is a matter of routine. It evaluates to 28.
Example
Find the volume of the part of the sphere of radius 3 that is left after drilling a cylindrical hole of radius 2 through the center.
Solution
The picture is shown below
The region this time is the annulus (washer) between the circles r = 2 and r = 3 as shown below.
The sphere has equation
x2 + y2 + z2 = 9
In polar coordinates this reduces to
r2 + z2 = 9
Solving for z by subtracting r2 and taking a square root we get top and bottom surfaces of
We saw before that the double integral over a region of the constant function 1 measures the area of the region. If the region has uniform density 1, then the mass is the density times the area which equals the area. What if the density is not constant. Suppose that the density is given by the continuous function
Density = (x,y)
In this case we can cut the region into tiny rectangles where the density is approximately constant. The area of mass rectangle is given by
Mass = (Density)(Area) = (x,y) x y
You probably know where this is going. If we add all to masses together and take the limit as the rectangle size goes to zero, we get a double integral.
Mass
Let (x,y) be the density of a lamina (flat sheet) R at the point (x,y). Then the total mass of the lamina is the double integral
Example
A rectangular metal sheet with 2 < x < 5 and 0 < y < 3 has density function
(x,y) = x + y
Set up the double integral that gives the mass of the metal sheet.
Solution
We just have the integral
Moments and Center of Mass
We have seen in first year calculus that the moments about an axis are defined by the product of the mass times the distance from the axis.
Mx = (Mass)(y) My = (Mass)(x)
If we have a region R with density function (x,y), then we do the usual thing. We cut the region into small rectangles for which the density is constant and add up the moments of each of these rectangles. Then take the limit as the rectangle size approaches zero. This will give us the total moment.
Definition of Moments of Mass and Center of Mass
Suppose that (x,y) is a continuous density function on a lamina R. Then the moments of mass are
and if M is the mass of the lamina, then the center of mass is
Example
Set up the integrals that give the center of mass of the rectangle with vertices (0,0), (1,0), (1,1), and (0,1) and density function proportional to the square of the distance from the origin. Use a calculator or computer to evaluate these integrals.
Solution
The mass is given by
The moments are given by
These evaluate to
Mx = 5k/12 and My = 5k/12
It should not be a surprise that the moments are equal since there is complete symmetry with respect to x and y. Finally, we divide to get
(x,y) = (5/8,5/8)
This tells us that the metal plate will balance perfectly if we place a pin at (5/8,5/8)
Moments of Inertia
We often call Mx and My the first moments. They have first powers of y and x in their definitions and help find the center of mass. We define the moments of inertia (or second moments) by introducing squares of y and x in their definitions. The moments of inertia help us find the kinetic energy in rotational motion. Below is the definition
Moments of Inertia
Suppose that (x,y) is a continuous density function on a lamina R. Then the moments of inertia are
Exercise
Find the moments of inertia for the previous square metal plate.
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Surface Area
Definition of Surface Area
In first year calculus we have seen how to find the surface area of revolution. Now that we how the power of double integration, we are ready to take on the surface area for more general surfaces. We can think of a smooth surface as a quilt flapping in the wind. It consists of many rectangles patched together. More generally and more accurately, let z = f(x,y) be a surface in R3 defined over a region R in the xy-plane. cut the xy-plane into rectangles. Each rectangle will project vertically to a piece of the surface as shown in the figure below. Although the area of the rectangle in R is
Area = yx
The area of the corresponding piece of the surface will not be yx since it is not a rectangle. Even if we cut finely, we will still not produce a rectangle, but rather will approximately produce a parallelogram. With a little geometry we can see that the two adjacent sides of the parallelogram are (in vector form)
u = x i + fx(x,y)x k
and
v = fy(x,y)y i + y k
We can see this by realizing that the partial derivatives are the slopes in each direction. If we run x in the i direction, then we will rise fx(x,y)x in the k direction so that
rise/run = fx(x,y)
Which agrees with the slope idea of the partial derivative. A similar argument will confirm the equation for the vector v. Now that we know the adjacent vectors we recall that the area of a parallelogram is the magnitude of the cross product of the two adjacent vectors. We have
This is the area of one of the patches of the quilt. To find the total area of the surface, we add up all the areas and take the limit as the rectangle size approaches zero. This results in a double Riemann sum, that is a double integral. We state the definition below.
We could work this integral out, but there is a much easier way. The integral of a
constant is just the constant times the area of the region. Since the region is a circle, we
get
Surface Area = 9(16) = 144
In reality, since there is a square root in the formula, most surface area calculations
require intensive integration skills or the use of a machine. The prior example and the
next example are not meant to deceive, but rather to show how the essence of surface
area problems work without the integration difficulty clouding your understanding.
Example
Find the surface area of the part of the paraboloid
z = 25 - x
2
- y
2
that lies above the xy-plane.
Solution
We calculate partial derivatives
fx(x,y) = -2x fy(x,y) = -2y
so that
1 + fx2(x,y) + fy
2(x,y) = 1 + 4x2 + 4y2
At this point if we listen closely, we should hear a little voice pleading "Polar Coordinates". We listen to its call and realize that the region is just the circle
r = 5
Now convert the integrand to polar coordinates to get
2 52
0 0
1 4 r rdrd
Now let
u = 1 + 4r2 du = 8rdr
and substitute
1012 101 21/2 3/2
0 1 0 1
1 1 530.958 12
u dud u
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Triple Integrals
Definition of the Triple Integral
We have seen that the geometry of a double integral involves cutting the two dimensional region into tiny rectangles, multiplying the areas of the rectangles by the value of the function there, adding the areas up, and taking a limit as the size of the rectangles approaches zero. We have also seen that this is equivalent to finding the double iterated iterated integral.
We will now take this idea to the next dimension. Instead of a region in the xy-plane, we will consider a solid in xyz-space. Instead of cutting up the region into rectangles, we will cut up the solid into rectangular solids. And instead of multiplying the function value by the area of the rectangle, we will multiply the function value by the volume of the rectangular solid.
We can define the triple integral as the limit of the sum of the product of the function times the volume of the rectangular solids.
Instead of the double integral being equivalent to the double iterated integral, the triple integral is equivalent to the triple iterated integral.
Definition of the Triple Integral
Let f(x,y,z) be a continuous function of three variables defined over a solid Q. Then the triple integral over Q is defined as
where the sum is taken over the rectangular solids included in the solid Q and lim is taken to mean the limit as the side lengths of the rectangular solid.
This definition is only practical for estimating the triple integral when a data set is given. When we have a symbolically defined function, we use an extension of the fundamental theorem of calculus which is just Fubini's theorem for triple integrals.
Theorem for Evaluating Triple Integrals
Let f(x,y,z) be a continuous function over a solid Q defined by
a < x < b h1(x) < y < h2(x) g1(x,y) < z < g2(x,y)
Then the triple integral is equal to the triple iterated integral.
Remark: As with double integrals the order of integration can be changed with care.
and Q is the solid that lies in the first octant and below the plain
3x + 2y + z = 6
Solution
The picture of the region
The challenge here is to find the limits. We work on the innermost limit first which corresponds with the variable "z". Think of standing vertically. Your feet will rest on the lower limit and your head will touch the higher limit. The lower limit is the xy-plane or
z = 0
The upper limit is the given plane. Solving for z, we get
z = 6 - 3x - 2y
Now we work on the middle limits that correspond to the variable "y". We look at the projection of the surface in the xy-plane. It is shown below.
Now we find the limits just as we found the limits of double integrals. The lower limit is just
y = 0
If we set z = 0 and solve for y, we get for the upper limit
y = 3 - 3/2 x
Next we find the outer limits, corresponding to the variable "x". The lowest x gets is 0 and highest x gets is 2. Hence
0 < x < 2
The integral is thus
Example
Switch the order of integration from the previous example so that dydxdz appears.
Solution
This time we work on the "y" variable first. The lower limit for the y-variable is 0. For the upper limit, we solve for y in the plane to get
y = 3 - 3/2 x - 1/2 z
To find the "x" limits, we project onto the xz-plane as shown below
The lower limit for x is 0. To find the upper limit we set y = 0 and solve for x to get
x = 2 - 1/3 z
Finally, to get the limits for z, we see that the smallest z will get is 0 and the largest z will get is 6. We get
0 < z < 6
We can write
Mass, Center of Mass, and Moments of Inertia
For a three dimensional solid with constant density, the mass is the density times the volume. If the density is not constant but rather a continuous function of x, y, and x, then we can cut the solid into very small rectangular solids so that on each rectangular solid the density is approximately constant. The volume of the rectangle is
Mass = (Density)(Volume) = f(x,y,z) xyz
Now do the usual thing. We add up all the small masses and take the limit as the rectangular solids get small. This will give us the triple integral
We are often interested in the center of mass of a solid. For example when the NEAR satellite orbited around the asteroid Eros, NASA scientists needed to compute the center of mass of the asteroid. Kepler told us that a stable orbit will always orbit in an elliptical orbit with the center of mass as one of the foci.
The NEAR satellite orbiting around Eros
We find the center of mass of a solid just as we found the center of mass of a lamina. Since we are in three dimensions, instead of the moments about the axes, we find the moments about the coordinate planes. We state the definitions from physics below.
Definition: Moments and Center of Mass
Let (x,y,z) be the density of a solid Q. Then the first moments about the coordinate planes are
and the center of mass is given by
Notice that letting the density function being identically equal to 1 gives the volume
Exercise
Find the center of mass of the solid that lies below the paraboloid
z = 4 - x2 - y2
that lies above the xy-plane if the density of the region is given by
(x,y,z) = x2 + 2y2 + z
You may use your calculator or computer to evaluate the integrals.
Just as with lamina, there are formulas for moments of inertial about the three axes. They involve multiplying the density function by the square of the distance from the axes. We have
Definition: Moments of Inertia
Let (x,y,z) be the density of a solid Q. Then the first moments of inertia about the coordinate axes are
Triple Integrals in Cylindrical and Spherical Coordinates
Cylindrical Coordinates
When we were working with double integrals, we saw that it was often easier to convert to polar coordinates. For triple integrals we have been introduced to three coordinate systems. The rectangular coordinate system (x,y,z) is the system that we are used to. The other two systems, cylindrical coordinates (r,,z) and spherical coordinates (,,) are the topic of this discussion.
Recall that cylindrical coordinates are most appropriate when the expression
x2 + y2
occurs. The construction is just an extension of polar coordinates.
x = r cos y = r sin z = z
Since triple integration can be looked at as iterated integration we have
This leads us the the following theorem
Theorem: Integration With Cylindrical Coordinates
Let f(x,y,z) be a continuous function on a solid Q. Then
Find the moment of inertia about the z-axis of the solid that lies below the paraboloid
z = 25 - x2 - y2
inside the cylinder
x2 + y2 = 4
above the xy-plane, and has density function
(x,y,z) = x2 + y2 + 6z
Solution
By the moment of inertia formula, we have
The region, being inside of a cylinder is ripe for cylindrical coordinates. We get
Spherical Coordinates
Another coordinate system that often comes into use is the spherical coordinate system. To review, the transformations are
x = cos sin y = sin sin z = cos
In the next section we will show that
dzdydx = sin ddd
This leads us to
Theorem: Integration With Spherical Coordinates
Let f(x,y,z) be a continuous function on a solid Q. Then
Example
Find the volume of solid that lies inside the sphere
x2 + y2 + z2 = 2
and outside of the cone
z2 = x2 + y2
Solution
We convert to spherical coordinates. The sphere becomes
=
To convert the cone, we add z2 to both sides of the equation
2z2 = x2 + y2 +z2
Now convert to
22cos = 2
Canceling the 2 and solving for we get
= cos-1(1/ ) = /4 or 7/4
In spherical coordinates (since the coordinates are periodic)
7/4 = 3/4
To find the volume we compute
32 24
2
0 04
sin V d d d
Evaluating this integral should be routine at this point and is equal to
8 V = 3
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Jacobians
Review of the Idea of Substitution
Consider the integral
To evaluate this integral we use the u-substitution
u = x2
This substitution sends the interval [0,2] onto the interval [0,4]. We can see that there is stretching of the interval. The stretching is not uniform. In fact, the first part [0,0.5] is actually contracted. This is the reason why we need to find du.
du dx 1 = 2x or = dx du 2x
This is the factor that needs to be multiplied in when we perform the substitution. Notice for small positive values of x, this factor is greater than 1 and for large values of x, the factor is smaller than 1. This is how the stretching and contracting is accounted for.
Jacobians
We have seen that when we convert to polar coordinates, we use
dydx = rdrd
With a geometrical argument, we showed why the "extra r" is included. Taking the analogy from the one variable case, the transformation to polar coordinates produces stretching and contracting. The "extra r" takes care of this stretching and contracting. The goal for this section is to be able to find the "extra factor" for a more general transformation. We call this "extra factor" the Jacobian of the transformation. We can find it by taking the determinant of the two by two matrix of partial derivatives.
Definition of the Jacobian
Let
x = g(u,v) and y = h(u,v)
be a transformation of the plane. Then the Jacobian of this transformation is
Find the Jacobian of the polar coordinates transformation
x(r,) = r cos y(r,) = r sin
Solution
We have
This is comforting since it agrees with the extra factor in integration.
Double Integration and the Jacobian
Theorem: Integration and Coordinate Transformations
Let
given by
x = g(u,v), y = h(u,v)
be a transformation on the plane that is one to one from a region S to a region R. If g and h have continuous partial derivatives such that the Jacobian is never zero, then
Remark: A useful fact is that the Jacobian of the inverse transformation is the reciprocal of the Jacobian of the original transformation.
This is a consequence of the fact that the determinant of the inverse of a matrix A is the reciprocal of the determinant of A.
Idea of the Proof
As usual, we cut S up into tiny rectangles so that the image under T of each rectangle is a parallelogram.
We need to find the area of the parallelogram. Considering differentials, we have
T(u + u,v) T(u,v) + (xuu,yuu)
T(u,v + v) T(u,v) + (xvv,yvv)
Thus the two vectors that make the parallelogram are
P = guu i + huu j
Q = gvv i + hvv j
To find the area of this parallelogram we just cross the two vectors.
and the extra factor is revealed.
Example
Use an appropriate change of variables to find the volume of the region below
z = (x - y)2
above the x-axis, over the parallelogram with vertices (0,0), (1,1), (2,0), and (1,-1)
Solution
We find the equations of the four lines that make the parallelogram to be
y = x y = x - 2 y = -x y = -x + 2
or
x - y = 0 x - y = 2 x + y = 0 x + y = 2
The region is given by
0 < x - y < 2 and 0 < x + y < 2
This leads us to the inverse transformation
u(x,y) = x - y v(x,y) = x + y
The Jacobian of the inverse transformation is
Since the Jacobian is the reciprocal of the inverse Jacobian we get
The region is given by
0 < u < 2 and 0 < v < 2
and the function is given by
z = u2
Putting this all together, we get the double integral
Jacobians and Triple Integrals
For transformations from R3 to R3, we define the Jacobian in a similar way
Example
Find the Jacobian for the spherical coordinate transformation
x = cos sin y = sin sin z = cos
Solution
We take partial derivatives and compute
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MATH 202 PRACTICE MIDTERM 2
Please work out each of the given problems. Credit will be based on the steps that
you show towards the final answer. Show your work.
Printable Key
PROBLEM 1 Please answer the following true or false. If false, explain why or provide
a counter example. If true explain why.
A) If all six limits of integration of an integral written in spherical coordinates are
constants, then the region of integration is a sphere.
Solution
B)
Solution
PROBLEM 2 Set up integrals to evaluate the following. Use the coordinate system that
will most effectively solve the integral.
A. The mass of the tetrahedron that lies in the first octant and below the plane x + y +
2z = 2 such that the density function is f(x,y,z) = 3yz .
Solution
B. The surface area of the part of the paraboloid z = 9 - x2 - y2 that lies above the
plane z = 5 .
Solution
C. The moment of inertia about the z-axis of the solid between the cylinders x2 + y2 =
We have now seen many types of functions. They are characterized by the domain and the range. Below is a list of some of the functions that we have encountered so far.
Domain Range Name
R R One variable Function
R R2 Parametric Equations
R2 R Function of 2 Variables
R Vectors Vector Valued Function
By letting the domain be Rn and the range be n-dimensional vectors, we get a new type of functions called a vector field.
Definition of a Vector Field
A vector field is be a function where the domain is Rn and the range is n-dimensional vectors.
Example
An important vector field that we have already encountered is the gradient vector field. Let f(x,y) be a differentiable function then the function that take a point (x0,y0) to gradf(x0,y0) is a vector field since the gradient of a function at a point is a vector. For example, if
f(x,y) = 0.1xy - 0.2y
then
gradf(x,y) = 0.1yi + (0.1x - 0.2)j
The sketch of the gradient is pictured below.
The best way to sketch a vector field is to use the help of a computer, however it is important to understand how they are sketched. For this example, we pick a point, say (1,2) and plug it into the vector field
Next, sketch the vector that begins at (1,2) and ends at (1 + .2, .2 - .1). Notice that when we sketch vector fields, we use the definition that involves two points rather than the definition that assumes all vectors emanate from the origin.
Example
In physics, many vector fields satisfy the inverse square law. A vector field F satisfying the inverse square law has the property that if
r = xi + yj + zj
and u is the unit vector in the direction of r (u = r/||r||), then
Examples of force fields that satisfy the inverse square law are gravitational force fields and electric force fields. Below is a plot of a vector field that satisfies the inverse square law.
Notice that as the distance from the origin gets small, the vector become small quickly. For gravity, this tells us that as we fly away from the earth, we experience less gravity, until it seems like weightlessness.
Other examples of vector fields that occur in nature include velocity fields. For example, the currents in the ocean ensure movement of the water. The if we sketch the velocity vector of at each point of the ocean, we get a vector field. A similar velocity field is produced from wind in the atmosphere.
Conservative Vector Fields
Our first and most important example was the gradient vector field. Can we go backwards? That is, given a vector field F, can we construct a function f with the property that
F = gradf
The answer to this question is only for some very special vector fields, which we call conservative.
Definition of a Conservative Vector Field
Let F be a vector field. Then F is called conservative if there is a differentiable
function f such that
gradf = F
f is called the potential function for F.
Fortunately, we do not need to flounder about to determine if a vector field F is conservative. We use the following argument. If
F = <M,N>
is conservative, then
<M,Q> = gradf = <fx,fy>
So that
M = fx and N = fy
Now take partial derivatives with respect to y of the first equation and with respect to x of the second equation to get
My = fxy and Nx = fyx
Since mixed partial derivative are the same (for nice functions) we get
My = Nx
This turns out to be both a necessary and sufficient condition for a vector field to be conservative.
Theorem: Testing for Conservativeness
If M and N have continuous first order partial derivatives, then the vector field
F = Mi + Nj
is conservative if and only if
My = Nx
Example
Determine which of the two vector fields are conservative
A. F = 3xyi - x2j
B. G = (1 + 2xy)i + (x2 - 2)j
Solution
For part A. we find
My = 3x Nx = -2x
Since they are not equal the vector field is not conservative.
For part B. we find
My = 2x Nx = 2x
They are equal, so the vector field is conservative.
Once we know that a vector field is conservative, how do we find the potential function? We investigate the relations
M = fx and N = fy
Example
Find the potential function for the conservative vector field in the last example
G = (1 + 2xy)i + (x2 - 2)j
Solution
We have
M = 1 + 2xy = fx
Integrating both sides with respect to x we get
x + x2y + c(y) = f(x,y)
Notice that the constant of integration may involve y terms since we are treating y as a constant. Now differentiate with respect to y to get
x2 + c'(y) = fy = N = x2 - 2
Thus
c'(y) = -2
Integrating with respect to y, we get
c(y) = -2y
We do not need a constant of integration here since we just want "a" potential function not the general potential function. Putting it all together, we get the potential function
f(x,y) = x + x2y - 2y
The Curl of a Vector Field
If F is a vector field then we want to look at what operation we can do to it. One such operation is called the Curl of F. We define it as follows.
Curl of a Vector Field
If F is a differentiable vector field with
F = Mi + Nj + Pk
then
Notice that for a two dimensional vector field, where there is only a k component for a cross product, that if the curl is zero then the vector field field is conservative. Actually more is true. Remembering that all second order partial derivatives are independent of order for nice functions, we have
Theorem
Let F be a three dimensional differentiable vector field with continuous partial derivatives. Then
Curl F = 0
if and only if F is conservative
Example
Determine if the vector field
F = yz2i + (xz2 + 2)j + (2xyz - 1)k
is conservative. If it is, find a potential function.
Solution
We calculate
The vector field is conservative. Now equate
fx = yz2
and integrate to get
f = xyz2 + c(y,z)
Notice the constant now depends on both y and z, since they were held constant when we integrated. Now take a partial with respect to y to get
xz2 + cy(y,z) = fy = xz2 + 2
Hence
cy(y,z) = 2
Integrate with respect to y to get
c(y,z) = 2y + c(z)
Notice that z is the only "constant" remaining. We get
f = xyz2 + 2y + c(z)
Now take a derivative with respect to z to get
2xyz + c'(z) = fz = 2xyz - 1
so that
c'(z) = -1
finally integrate to get
c(z) = -z
The potential function is
f(x,y,z) = xyz2 + 2y - z
The Divergence of a Vector Field
The curl of a vector field was defined as the cross product of the "dell" operator with the vector field. A curious student may try to take a dot product instead and see where it leads.
Definition of the Divergence of a Vector Field
If F is a differentiable vector field with
F = Mi + Nj + Pk
then
Notice that the curl of a vector field is a vector field, while the divergence of a vector field is a real valued function.
Example
Find the divergence of the vector field
F = (x-3y)i + (x2z2 + cos(z3))j + (xyz2)k
Solution
We just take the three partial derivatives and add them up
(1) + (0) + (2xyz) = 1 + 2xyz
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Line Integrals
Definition of a Line Integral
By this time you should be used to the construction of an integral. We break a geometrical figure into tiny pieces, multiply the size of the piece by the function value on that piece and add up all the products. For one variable integration the geometrical figure is a line segment, for double integration the figure is a region, and for triple integration the figure is a solid.
The geometrical figure of the day will be a curve. If we have a function defined on a curve we can break up the curve into tiny line segments, multiply the length of the line segments by the function value on the segment and add up all the products. As always, we will take a limit as the length of the line segments approaches zero. This new quantity is called the line integral and can be defined in two, three, or higher dimensions.
Suppose that a wire has as density f(x,y,z) at the point (x,y,z) on the wire. Then the line integral will equal the total mass of the wire. Below is the definition in symbols.
Definition of the Line Integral
Let f be a function defined on a curve C of finite length. Then the line integral of f along C is
(for two dimensions)
(for three dimensions)
Evaluating Line Integrals
This definition is not very useful by itself for finding exact line integrals. If data is provided, then we can use it as a guide for an approximate answer. Fortunately, there is an easier way to find the line integral when the curve is given parametrically or as a vector valued function. We will explain how this is done for curves in R2. The case for R3 is similar.
Let
r(t) = x(t)i + y(t)j
be a differentiable vector valued function. Then
We are now ready to state the theorem that shows us how to compute a line integral.
be a differentiable vector valued function that defines a smooth curve C. Then
and for three dimensions, if
r(t) = x(t)i + y(t)j + z(t)k a < t < b
then
Example
Find the line integral
where C is the ellipse
r(t) = (2cos t)i + (3sin t)j 0 < t < 2
You may use a calculator or computer to evaluate the final integral.
Solution
We find
We have the integral
With the help of a machine, we get
15.87
Work
The main application of line integrals is finding the work done on an object in a force field. If an object is moving along a curve through a force field F, then we can calculate the total work done by the force field by cutting the curve up into tiny pieces. The work done W along each piece will be approximately equal to
dW = F . Tds
Now recall that
r'(t) T = ||r'(t)||
and that
ds = ||r'(t)||dt
Hence
dW = F . r'(t) dt
As usual, we add up all the small pieces of work and take the limit as the pieces get small to end up with an integral.
Definition of Work
Let F be a vector field and C be a curve defined by the vector valued function r. Then the work done by F on an object moving along C is given by
Example
Find the work done by the vector field
F(x,y,z) = xi + 3xyj - (x + z)k
on a particle moving along the line segment that goes from (1,4,2) to (0,5,1)
Notice that work done by a force field on an object moving along a curve depends on the direction that the object goes. In fact the opposite direction will produce the negative of the work done in the original direction. This is clear from the fact that everything is the same except the order which we write a and b.
Line Integrals in Differential Form
We can rewrite r'(t)dt as
dr dx dy dz dt = ( i+ j + k ) dt dt dt dt dt
= dx i + dyj + dzk
So that if
F = Mi + Nj + Pk
then
F . r'(t)dt = Mdx + Ndy + Pdz
This is called the differential form.
Example
Find
where C is the part of the helix
r(t) = sin t i + cos t j + t k 0 < t < 2
Solution
We have
r'(t) = cos t i - sin t j + k
so that
ydx + zdy = (cos2t - t sin t)dt
This leads us to the integral
with a little bit of effort (using integration by parts) we get
3
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Conservative Vector Fields and Independence of Path
The Fundamental Theorem of Line Integrals
Consider the force field representing the wind shown below
You are a pilot attempting to minimize the work your engines need to do. Does it matter which path you take? Clearly the red path goes with the wind and the green path goes against the wind. With this vector field, work is dependent on the path that is taken.
Next consider the vector field
F(x,y) = yi + xj
shown below
It turns out that going from point A to point B, every path leads to the same amount of work done. What is special about this vector field?
The key here, as you can quickly check, is that the vector field F is conservative. (My = Nx ). Since for a conservative vector field, all paths produce the same amount of work, we seek a formula that gives this work quantity.
The theorem below shows us how to find this quantity. Notice the strong resemblance to the fundamental theorem of calculus.
The Fundamental Theorem of Line Integrals
Let F be a conservative vector field with potential function f, and C be any smooth curve starting at the point A and ending at the point B. Then
The next example demonstrates the power of this theorem.
Example
Find the work done by the vector field
F(x,y) = (2x -3y)i + (3y2 - 3x)j
along the curve indicated in the graph below
Solution
First notice that
My = -3 = Nx
We can use the fundamental theorem of line integrals to solve this. There are two approaches.
Approach 1
We find the potential function. We have
fx = 2x - 3y
Integrating we get
f(x,y) = x2 - 3xy + c(y)
Now take the derivative with respect to y to get
fy = -3x + c'(y) = 3y2 - 3x
Hence
c'(y) = 3y2
and
c(y) = y3
The potential function is
f(x,y) = x2 - 3xy + y3
Now use the fundamental theorem of line integrals to get
f(B) - f(A) = f(1,0) - f(0,0) = 1
Approach 2
We since the vector field is conservative any path from point A to point B will produce the same work. Hence the work over the easier line segment from (0,0) to (1,0) will also give the correct answer. We parameterize by
r(t) = ti 0 < t < 1
we have
ri(t) = i
so that
F . dr = ((2x -3y)i + (3y2 - 3x)j) . i = 2x - 3y = 2t
Now just integrate
Proof of the Fundamental Theorem of Line Integrals
To prove the fundamental theorem of line integrals we will use the following outcome of the chain rule:
Since the vector field is conservative, we can use the fundamental theorem of line integrals. Notice that the curve begins and ends at the same place. We do not even need to find the potential function, since whatever it is, say f, we have
f(A) - f(A) = 0
In general, the work done by a conservative vector field is zero along any closed curve. The converse is also true, which we state without proof.
Theorem: Conservative Vector Fields and Closed Curves
Let F be a vector field with components that have continuous first order partial derivatives and let C be a piecewise smooth curve. Then the following three statements are equivalent
1. F is conservative.
2. is independent of path.
3. for all closed curves C.
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Green's Theorem
A Little Topology
Before stating the big theorem of the day, we first need to present a few topological ideas. Consider a closed curve C in R2 defined by
r(t) = x(t)i + y(t)j a < t < b
We say C is simple if it does not intersect itself. A curve intersects itself if
r(u) = r(v)
for two distinct values u and v. A circle is a simple curve while a figure eight is not simple.
A region is called simply connected if it boundary is a single simple closed curve. Another way of thinking about simply connected regions is that their complement (the space minus the region) consists of only one piece. Below are examples of simply connected and non-simply connected regions.
Our final topological definition is orientation. We have seen that if we traverse a curve in the opposite direction, then the line integral will be the negative of the original. We want to have a way to define a positive orientation. We define it as follows.
Let R be a simply connected region with boundary curve C. Then C is called positively oriented if facing the direction that the curve is sketched, the region lies to the left of the curve. Otherwise the curve is said to be negatively oriented.
One way to remember this is to recall that in the standard unit circle angles are measures counterclockwise, that is traveling around the circle you will see the center on your left.
then the line integral over any smooth closed curve is zero. What can we do if the above quantity is nonzero. Green's theorem states that the line integral is equal to the double integral of this quantity over the enclosed region. Precisely, we have
Green's Theorem
Let R be a simply connected region with smooth boundary C, oriented positively and let M and N have continuous partial derivatives in an open region containing R, then
Sketch of the Proof
First we can assume that the region is both vertically and horizontally simple. Otherwise we can carefully cut the region into parts so that each of the parts are both vertically simple and horizontally simple. Below is an example of such a cut. Notice that the line where the regions is cut is drawn once upwards and once downwards. Thus the two line integrals over this line will cancel each other out.
We can assume that the region is as in the figure below
We will show that
The proof for the M part is similar. We will compute both sides and show they are the same. First we break the curve into its left and right half. Call the left half C1 and the right half C2. We have
Now we show that the double integral leads to the same expression. We have
2
1
2
1
( )
( )
( )
2 1( )
( , ) ( ( ), ) ( ( ), )
g yb
x xaR g y
g yb b
a ag y
N dxdy N dxdy
N x y dy N g y y N g y y dy
And the two expressions are equal.
Using Green's Theorem
Example
Determine the work done by the force field
F = (x - xy) i + y2 j
when a particle moves counterclockwise along the rectangle with vertices (0,0), (4,0), (4,6), and (0,6).
Solution
We could do this with a line integral, but this would involve four parameterizations (one for each side of the rectangle). Instead, we use Green's Theorem. We find
Nx - My = 0 - (-x) = x
The region is just a rectangle, so the limits are the constants. We have
Example
Calculate the line integral
Where C is the union of the unit circle centered at the origin oriented negatively and the circle of radius 2 centered at the origin oriented positively.
Solution
We cannot use Green's Theorem directly, since the region is not simply connected. However, if we think of the region as being the union its left and right half, then we see that the extra cuts cancel each other out.
In this light we can use Green's Theorem on each piece. We have
Nx - My = 1 - 0 = 1
Hence the line integral is just the double integral of 1, which is the area of the region. This area is
(22 - 12) = 4
Green's Theorem and Area
The example above showed that if
Nx - My = 1
then the line integral gives the area of the enclosed region. There are three special vector fields, among many, where this equation holds. We state the following theorem which you should be easily able to prove using Green's Theorem.
Theorem: Using Green's Theorem to Find Area
Let R be a simply connected region with positively oriented smooth boundary C. Then the area of R is given by each of the following line integrals.
1.
2. 3.
Example
Use the third part of the area formula to find the area of the ellipse
x2 y2 + = 1 4 9
Solution
To compute the line integral, we parameterize the curve
r(t) = 2 cos t i + 3 sin t j
r'(t) = -2 sin t i + 3 cos t j
We have
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Parametric Surfaces
Definition of a Parametric Surface
We have now seen many kinds of functions. When we talked about parametric curves, we defined them as functions from R to R2 (plane curves) or R to R3 (space curves). Because each of these has its domain R, they are one dimensional (you can only go forward or backward). In this section, we investigate how to parameterize two dimensional surfaces. Below is the definition.
Definition of Parametric Surfaces
A parametric surface is a function with domain R2 and range R3.
Remark: We typically use the variables u and v for the domain and x, y, and z for the range. We often use vector notation to exhibit parametric surfaces.
Example
A sphere of radius 7 can be parameterized by
r(u,v) = 7cos u sin v i + 7sin u sin v j + 7 cos v k
Notice that we have just used spherical coordinates with the radius held at 7.
We can use a computer to graph a parametric surface. Below is the graph of the surface
The idea is similar to parametric curves. We just let x = u and y = v, to get
r(u,v) = u i + v j + eu cos(u - v) k
Example
A surface is created by revolving the curve
y = cos x
about the x-axis. Find parametric equations for this surface.
Solution
For a fixed value of x, we get a circle of radius cos x. Now use polar coordinates (in the yz-plane) to get
r(u,v) = u i + r cos v j + r sin v k
Since u = x and r = cos x, we can substitute cos u for r in the above equation to get
r(u,v) = u i + cos u cos v j + cos u sin v k
Normal Vectors and Tangent Planes
We have already learned how to find a normal vector of a surface that is presented as a function of tow variables, namely find the gradient vector. To find the normal vector to a surface r(t) that is defined parametrically, we proceed as follows.
The partial derivatives
ru(u0,v0) and rv(u0,v0)
will lie on the tangent plane to the surface at the point (u0,v0). This is true, because fixing one variable constant and letting the other vary, produced a curve on the surface through
(u0,v0). ru(u0,v0) will be tangent to this curve. The tangent plane contains all vectors tangent to curves passing through the point.
To find a normal vector, we just cross the two tangent vectors.
Example
Find the equation of the tangent plane to the surface
r(u,v) = (u2 - v2) i + (u + v) j + (uv) k
at the point (1,2).
Solution
We have
ru(u,v) = (2u) i + j + v k
rv(u,v) = (-2v) i + j + u k
so that
ru(1,2) = 2 i + j + 2 k
rv(1,2) = -4 i + j + k
r(1,2) = -3 i + 3 j + 3 k
Now cross these vectors together to get
We now have the normal vector and a point (-3,3,2). We use the normal vector-point equation for a plane
-1(x + 3) - 10(y - 3) + 6(z - 2) = 0
-x - 10y + 6z = -15 or x + 10y - 6z = 15
Surface Area
To find the surface area of a parametrically defined surface, we proceed in a similar way as in the case as a surface defined by a function. Instead of projecting down to the region in the xy-plane, we project back to a region in the uv-plane. We cut the region into small rectangles which map approximately to small parallelograms with adjacent defining vectors ru and rv. The area of these parallelograms will equal the magnitude of the cross product of ru and rv. Finally add the areas up and take the limit as the rectangles get small. This will produce a double integral.
Area of a Parametric Surface
Let S be a smooth surface defined parametrically by
r(u,v) = x(u,v) i + y(u,v) j + z(u,v) k
where u and v are contained in a region R. Then the surface area of S is given by
Since the magnitude of a cross product involves a square root, the integral in the surface area formula is usually impossible or nearly impossible to evaluate without power series or by approximation techniques.
Example
Find the surface area of the surface given by
r(u,v) = (v2) i + (u - v) j + (u2) k 0 < u < 2 1 < v < 4
Solution
We calculate
ru(u,v) = j + 2u k
rv(u,v) = (2v) i - j
The cross product is
The surface area formula gives
This integral is probably impossible to compute exactly. Instead, a calculator can be used
to obtain
a surface area of 70.9.
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Surface Integrals
Surface Integrals for Parametric Surfaces
In the last section, we learned how to find the surface area for parametric surfaces. We cut the region in the uv-plane into tiny rectangles and added up the area of the corresponding tiny parallelograms in the xy-plane. The area of these parallelograms was
If we think of the surface as having varying density f(x,y,z), then the mass of this parallelogram will be
and adding up all these masses and taking the limit as the rectangle sizes approach zero, gives the definition of the surface integral.
Definition of the Surface Integral
Let S be a smooth surface given by the vector valued function
r(u,v) = x(u,v)i + y(u,v)j + z(u,v)k
and f(x,y,z) be a continuous function. Then the surface integral of f over S is
As with finding the surface area the integral typically results in an impossible integral.
f(x(u,v),y(u,v),z(u,v)) = x(u,v) +2z(u,v) = u +2(u + v) = 3u + v
We find
Although this integral is possible, its solution is quite involved. You can verify that the surface integral evaluates to approximately 525.27.
Surface Integrals for Surfaces that are Functions of Two Variables
We have seen before that if
z = g(x,y)
is a surface such that g has continuous first order partial derivatives, then the parameterization
r(u,v) = ui + vj + g(u,v)k
has the property that
This leads to the formula for surface integrals.
Theorem: The Surface Integral for Surfaces of the Form z = g(x,y)
Let S be a surface given by
z = g(x,y)
over a region R such that both first order partial derivatives of g are continuous and let f(x,y,z) be a continuous function. Then the surface integral of f over S is
Example
Find
where S is the part of the paraboloid
z = x
2
+ y
2
that lies inside the cylinder
x
2
+ y
2
= 1
and
f(x,y,z) = z
Solution
We have
and
f(x,y,z) = z =
x
2
+ y
2
At this point, you should be thinking, "This looks like a job for polar coordinates." And
we get
Let
u = 1 + 4r
2
du = 8r dr r
2
= 1/4 u - 1/4
and the substitution gives us
Oriented Surfaces and Flux
We have seen how a region R with boundary curve C can be oriented. Traveling along
C, we look to see if the region is on the right or left. Unfortunately, this definition does
not work will for surfaces in three dimensions. The idea of right and left is not well
defined. In fact not all surfaces can be oriented.
We say that a surface is orientable if a unit normal vector can be defined on the surface
such that it varies continuously over the surface.
Below is an example of a non-orientable surface (called the Mobius Strip)
You can see that there is no front or back of this surface.
Recall that a unit normal vector to a surface can be given by
There is another choice for the normal vector to the surface, namely the vector in the opposite direction, -N.
By this point, you may have noticed the similarity between the formulas for the unit
normal vector and the surface integral. This idea leads us to the definition of the Flux
Integral
Consider a fluid flowing through a surface S. The Flux of the fluid across S measures the
amount of fluid passing through the surface per unit time. If the fluid flow is represented
by the vector field F, then for a small piece with area S of the surface the flux will equal
to
Flux = F
.
N S
Adding up all these together and taking a limit, we get
Definition of the Flux Integral
Let F be a differentiable vector field on a surface S oriented by a unit normal vector N. The flux integral of F across N is given by
Notice that the denominator of N and the formula for dS both involve ||r
u
x r
v
||.
Canceling, we get
NdS = ru
x rv
dvdu
for a surface that is defined by the function z = g(x,y), we get the nice formula
NdS =
-gx(x,y)i - gy(x,y)j + k (oriented upward)
or
NdS =
gx(x,y)i + gy(x,y)j - k (oriented downward)
Example
Find the flux of
F(x,y,z) = xi + 2yj + zk
across the part of the surface
z = x + y2
with upward pointing normal that lies within the box
0 < x < 3 2 < y < 5
Solution
We compute
NdS = -i - 2yj + k dydx
and
F . N dS = -x - 4y2 + x + y2 = -3y2
The flux integral is
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Divergence Theorem
The Divergence Theorem
When we looked at Greens Theorem, we saw that there was a relationship between a region and the curve that encloses it. This gave us the relationship between the line integral and the double integral. Moving to three dimensions, the divergence theorem provides us with a relationship between a triple integral over a solid and the surface integral over the surface that encloses the solid.
The Divergence Theorem
Let Q be a solid region bounded by a closed surface oriented with outward pointing unit normal vector N, and let F be a differentiable vector field (components have continuous partial derivatives). Then
Example
Find
Where
F(x,y,z) = y2i + ex(1-cos(x2 + z2))j + (x + z)k
and S is the unit sphere centered at the point (1,4,6) with outwardly pointing normal vector.
Solution
This seemingly difficult problem turns out to be quite easy once we have the divergence theorem. We have
divF = 0 + 0 + 1 = 1
Now recall that a triple integral of the function 1 is the volume of the solid. Since the solid is a sphere of radius 1 we get .
Part of the Proof of the Divergence Theorem
As usual, we will make some simplifying remarks and then prove part of the divergence theorem.
Notice that the outward pointing normal vector is upward on the top surface and downward for the bottom region. We also note that the divergence theorem can be written as
We will show that
We have on the top surface
Pk . N dS = Pk . ( (-g2)x i - (g2)y j + k) = P(x,y,g2(x,y))
On the bottom surface, we get
Pk . N dS = Pk . ( (g1)x i + (g1)y j - k) = -P(x,y,g1(x,y))
Putting these together we get
For the triple integral, the fundamental theorem of calculus tell us that
An Interpretation Of Divergence
We have seen that the flux is the amount fluid flow per unit time through a surface. If the surface is closed, then the total flux will equal the flow out of the solid minus the flow in. Often in the solid there is a source (such as a star when the flow is electromagnetic radiation) or a sink (such as the earth collecting solar radiation) If we have a small solid S(P) containing a point P, then the divergence of the vector field is approximately constant, which leads to the approximation
The divergence theorem expresses the approximation
Flux through S(P) divF(P) (Volume)
Dividing by the volume, we get that the divergence of F at P is the Flux per unit volume.
If the divergence is positive, then the P is a source.
If the divergence is negative, then P is a sink.
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Stokes' Theorem
Stokes' Theorem
The divergence theorem is used to find a surface integral over a closed surface and Green's theorem is use to find a line integral that encloses a surface (region) in the xy-plane. The theorem of the day, Stokes' theorem relates the surface integral to a line integral.
Since we will be working in three dimensions, we need to discus what it means for a curve to be oriented positively.
Let S be a oriented surface with unit normal vector N and let C be the boundary of S. Then C is positively oriented if its orientation follows the right hand rule, that is if you right hand curls around N in the direction of C's orientation, then your thumb will be pointing in the direction of N.
Now we are ready to state Stokes' Theorem. The proof will be left for a more advanced course.
Stokes' Theorem
Let S be an oriented surface with unit normal vector N and C be the positively oriented boundary of S. If F is a vector field with continuous first order partial derivatives then
Example
Let S be the part of the plane
z = 4 - x - 2y
with upwardly pointing unit normal vector. Use Stokes' theorem to find
First notice that without Stokes' theorem, we would have to parameterize three different line segments. Instead we can find this with just one double integral.
We have
and
N dS = i + 2j + k
So that
Curl F . N dS = 1 + x + 2y - 1 = x 2y
We integrate
Curl and Circulation
Just as the divergence theorem assisted us in understanding the divergence of a function at a point, Stokes' theorem helps us understand what the Curl of a vector field is. Let P be a point on the surface and Ce be a tiny circle around P on the surface. The
measures the amount of circulation around P. You can see this by noticing that if F flows in the direction of the tangent vector, then F . dr will be positive. If it flows in the opposite direction, then it will be negative. The stronger the force field in the direction of the tangent vector, the greater the circulation.
Since the region enclosed by Ce is tiny, the surface integral can be approximated by
or
Curl F . N = Circulation per unit area
So the curl tell us how much the force field rotates around the point.
We can see that if this is a small piece of the surface containing P, then
Curl F . N > 0
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Math 202 Practice Midterm 3
Please work out each of the given problems. Credit will be based on the steps that you show towards the final answer. Show your work.
Printable Key
Problem 1 Please answer the following true or false. If false, explain why or provide a counter example. If true explain why.
A. Let f(x,y,z) be a function with continuous second order partial derivatives and let F(x,y,z) be the gradient of f(x,y,z). If S is the ellipsoid
x2 + y2 + z2 = 1 4
oriented outward, then
Solution
B. Let F(x,y) be a conservative vector field, then
Solution
Problem 2
Show that
for any closed surface S.
Solution
Problem 3
A fish starting at the origin swims in a straight path to the point (0.2,0.1,0.3), then changes direction and swims along the circular path through the point (0.5,0.2,0.6) and the point (0.7,0.5,0.8), and finally changes directions heading straight to the point (1.5,1,2). The current can by represented by the vector field
Please work out each of the given problems. Credit will be based on the steps that you show towards the final answer. Show your Work
Printable key
Problem 1 Please answer the following true or false. If false, explain why or provide a counter example. If true, explain why.
A. Let Q be a three dimensional solid and let
F(x,y,z) = (x2y + sin z)i+ (cos x - xy2)j + (3xy + z)k
and let S be the boundary of Q with outwardly pointing normal. The the volume so Q is given by
Solution
B. Let F = 3xy i + cosx j and let C1 and C2 be as shown below. Then
Solution
C. A new particle, the fluxon, has been discovered to be emitted from the sun. The particle emits a force field
F(x,y,z) = (y2 - z) i + (x2 - z) j + (x2 + y2) k
where the origin represents the center of the sun. If the total flux through the earth's northern hemisphere has been calculated as 10,000, then the total flux through the earth's southern hemisphere must also be 10,000.
Problem 2 You are the captain of the spaceship Potential that you have programmed to follow the vector-valued function
r(t) = (t2 + 5) i + (t - 3) j + t3 k
where t is measured in hours. However, at time t = 2, your engines fail and your ship begins drifting in deep space. There is a deep space station located at (6,2,38).
A. Find the vector-valued function that describes the Potential's flight after the engines failed. Use t = 2 to represent the time at which your engines first shut down. (Hint: This should be a linear vector valued function.)
Solution
B. Will your ship make it to the station, or will you float helplessly for eternity?
Solution
Problem 3 Show that the helix
r(t) = (R cos t) i + (R sin t) j + t k
where R is a positive constant, has the property that N(t) . r(t) is a constant. Find this
constant.
Solution
Problem 4 The probability density function for an event is given by
where R is the square with vertices (4,0), (6,2), (4,4), and (2,2).
A. Use the appropriate change of variables (Jacobians) to find k that is solve
Solution
B. Find the probability that 0 < x - y < 1
Solution
Problem 5 Switch the order of integration and write as one double integral
B. Let F(x,y) be a conservative vector field, then
Solution
True,
The first line integral traces out the line segment from (0,1) to (1,0) and the second traces out the quarter-circle from (0,1) to (1,0). Notice that in the first integral
By the fundamental theorem of line integral, the integral is independence of path, hence the two integrals are equal.
Problem 2
Show that
for any closed surface S.
Solution
We use Stokes' Theorem. We have
Where C is the boundary of the surface S. But since S is a closed surface, it has not boundary. Hence C is a curve of zero length and the right hand integral is zero.
Problem 3
A fish starting at the origin swims in a straight path to the point (0.2,0.1,0.3), then changes direction and swims along the circular path through the point (0.5,0.2,0.6) and the point (0.7,0.5,0.8), and finally changes directions heading straight to the point (1.5,1,2). The current can by represented by the vector field
and S is the rectangular solid with vertices (0,0,0), (1,0,0), (1,2,0), (0,2,0), (0,0,3), (1,0,3), (1,2,3), (0,2,3).
Solution
We use the divergence theorem. We have
divF = 1 - 1 + 2 = 2
We have
Since the integrand of the right hand side is just a constant, its value is equal to the constant times the volume of the solid. Since the solid is a rectangular solid with side lengths 1, 2, and 3, we have
2(Volume E) = 2(1)(2)(3) = 12
Problem 5
Find the work done by sailing a ship from the point (2,3) the the point (-1,2) against the wind with velocity field
F(x,y) = yi + (3x + 2y)j
Solution
First notice that F is not a conservative vector field. We need to parameterize the curve and perform the line integral. The curve can be parameterized by
in the first octant with upwardly pointing unit normal.
Solution
We use Stokes' theorem. We have
We can write the surface as
z = 1 - x - y
Using Stokes theorem we get
Math 202 Practice Final
Please work out each of the given problems. Credit will be based on the steps that you show towards the final answer. Show your Work
Problem 1 Please answer the following true or false. If false, explain why or provide a counter example. If true, explain why.
A. Let Q be a three dimensional solid and let
F(x,y,z) = (x2y + sin z)i+ (cos x - xy2)j + (3xy + z)k
and let S be the boundary of Q with outwardly pointing normal. The the volume so Q is given by
Solution
True,
We have
divF = 2xy - 2xy + 1 = 1
Using the divergence theorem, we see that
which is just the volume of the solid.
B. Let F = 3xy i + cosx j and let C1 and C2 be as shown below. Then
Solution
C. A new particle, the fluxon, has been discovered to be emitted from the sun. The particle emits a force field
F(x,y,z) = (y2 - z) i + (x2 - z) j + (x2 + y2) k
where the origin represents the center of the sun. If the total flux through the earth's northern hemisphere has been calculated as 10,000, then the total flux through the earth's southern hemisphere must also be 10,000.
Solution
False, since divF = 0, the total flux must be zero. If the flux through the northern hemisphere is 10,000, then the flux through the southern hemisphere must be -10,000.
Problem 2 You are the captain of the spaceship Potential that you have programmed to follow the vector-valued function
r(t) = (t2 + 5) i + (t - 3) j + t3 k
where t is measured in hours. However, at time t = 2, your engines fail and your ship begins drifting in deep space. There is a deep space station located at (6,2,38).
A. Find the vector-valued function that describes the Potential's flight after the engines failed. Use t = 2 to represent the time at which your engines first shut down. (Hint: This should be a linear vector valued function.)
Solution
The flight will go in a linear path in the direction of the unit tangent vector with speed equal to the speed when the engines fail. We have