Math 273 Solutions to Review Problems for Exam 1 1. True or False ...

Math 273Solutions to Review Problems for Exam 1

1. True or False? Circle ONE answer for each. Hint: For effective study, explain why if ‘true’ and give acounterexample if ‘false.’

(a) T or F©: If a⊥b and b⊥c, then a⊥c.Solution:

Let a = i,b = j, and c =< 1, 0, 1 > . Then a · b = i · j = 0 and b · c =< 0, 1, 0 > · < 1, 0, 1 >= 0. However,

a · c =< 1, 0, 0 > · < 1, 0, 1 >= 1.

(b) T© or F: If a · b = 0, then ||a× b|| = ||a||||b||.Solution:

Assume both a and b are nonzero. (If either is zero the result is obvious). If θ is the angle between a and

b, then θ = π2

since a and b are orthogonal. ||a× b|| = ||a||||b|| sin θ = ||a||||b|| sinπ/2 = ||a||||b||.

(c) T© or F: For any vectors u,v in R3, ||u× v|| = ||v × u||.Solution:

If θ is the angle between u and v, then ||u×v|| = ||u||||v|| sin θ = ||v||||u|| sin θ = ||v×u||. (Or, ||u×v|| =|| − v × u|| = | − 1|||v × u|| = ||v × u||.)

(d) T or F©: The vector < 3,−1, 2 > is parallel to the plane 6x− 2y + 4z = 1.Solution:

A normal vector to the plane is n =< 6,−2, 4 > . Because < 3,−1, 2 >= 12n, the vector is parallel to n

and hence perpendicular to the plane.

(e) T or F©: If u · v = 0, then u = 0 or v = 0.Solution:

For example i · j = 0 but i 6= 0 and j 6= 0.

(f) T or F©: If u× v = 0, then u = 0 or v = 0.Solution:

For example i× i = 0 but i 6= 0.

(g) T© or F: If u · v = 0 and u× v = 0, then u = 0 or v = 0.Solution:

If both u and v are nonzero, then u · v = 0 implies u and v are orthogonal. But u × v = 0 implies that

u and v are parallel. Two nonzero vectors can’t be both parallel and orthogonal, so at least one of them

must be 0.

(h) T© or F: The curve r(t) =⟨0, t2, 4t

⟩is a parabola.

Solution:

Parametric equations for the curve are x = 0, y = t2, z = 4t, and since t = z4

we have y = t2 =(z

4

)2or

y =z2

16, x = 0. This is an equation of a parabola in the yz−plane.

(i) T© or F: If κ(t) = 0 for all t, the curve is a straight line.Solution:

Notice that κ(t) = 0 ⇐⇒ ||T′(t)|| = 0 ⇐⇒ T′(t) = 0 for all t. But then T(t) = C, a constant vector,

which is true only for a straight line.

(j) T or F©: Different parameterizations of the same curve result in identical tangent vectors at a givenpoint on the curve.Solution:

For example, r1(t) =< t, t > and r2(t) =< 2t, 2t > both represent the same plane curve (the line y = x),

but the tangent vector r1(t) =< 1, 1 > for all t, while r1(t) =< 2, 2 > . In fact, different parametrizations

give parallel tangent vectors at a point, but their magnitudes may differ.

2. Which of the following are vectors? Circle all that apply.

(a) [(a · b)c]× aSolution: A vector.

(b) c× [(a · b)× c]Solution: Nonsense.

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(c) c× [(a · b)c]Solution: A vector.

(d) (a× b) · cSolution: A scalar.

3. Which of the following are meaningful? Circle all that apply.

(a) ||w||(u× v)Solution: Meaningful.

(b) (u · v)×wSolution: Nonsense.

(c) u · (v ×w)Solution: Meaningful.

4. Find the values of x such that the vectors < 3, 2, x > and < 2x, 4, x > are orthogonal.For the two vectors to be orthogonal, we need < 3, 2, x > · < 2x, 4, x >= 0. That is, 3(2x) + 2(4) + x(x) = 0, or

x = −2 or x = −4.

5. Let a =

⟨1√2,

1√2

⟩,b =

⟨1√2,−1√

2

⟩, and u =< 3, 0 > .

(a) Show that a and b are orthogonal unit vectors.

Solution: a · b = 0 and ||a|| =√

12+12√2

= 1, and ||b|| =√

12+(−1)2√2

= 1.

(b) Find the decomposition of u along a.

Solution: u||a = (u · a)a =3√2

⟨1√2,

1√2

⟩=

⟨3

2,

3

2

⟩. Thus, u⊥a =< 3, 0 > −

⟨32, 32

⟩=⟨32,− 3

2

⟩.

(c) Find the decomposition of u along b.

Solution: Similarly, u||b = (u · b)b =

(< 3, 0 > ·

⟨1√2,− 1√

2

⟩)⟨1√2,− 1√

2

⟩=

⟨3

2,−3

2

⟩. Thus,

u⊥b =< 3, 0 > −⟨32, 32

⟩=⟨32,− 3

2

⟩= u⊥a.

6. (a) Find an equation of the sphere that passes through the point (6, -2, 3) and has center (-1,2,1).Solution: Use the distance formula to find the distance between (6, -2, 3) and (-1,2,1). Then, theequation for the circle is (x+ 1)2 + (y − 2)2 + (z − 1)2 = 69.

(b) Find the curve in which this sphere intersects the yz-plane.Solution:

The intersection of this sphere with the yz−plane is the set of points on the sphere whose x−coordinateis 0. Putting x=0 in to the equation, (y − 2)2 + (z − 1)2 = 68, which represents a circle in theyz−plance with center (0, 2, 1) and radius

√68.

7. For each of the following quantities (cos θ, sin θ, x, y, z, and w) in the picture below, fill in the blank withthe number of the expression, taken from the list to the right, to which it is equal.Solutions: cos θ= 5; sin θ = 4; x = 2; y= 3; z= 7 ; w= 6

8. Find an equation for the line through (4,−1, 2) and (1, 1, 5).Solution:

The line has direction v =< −3, 2, 3 > . Letting P0 = (4,−1, 2), parametric equations are

x = 4− 3t, y = −1 + 2t, z = 2 + 3t.

9. Find an equation for the line through (−2, 2, 4) and perpendicular to the plane 2x− y + 5z = 12.Solution:

A direction vector for the line is a normal vector for the plane, n =< 2,−1, 5 >, and parametric equationsfor the line are

x = −2 + 2t, y = 2− t, z = 4 + 5t.

10. Find an equation of the plane through (2, 1, 0) and parallel to x+ 4y − 3z = 1.Solution:

Since the two planes are parallel, they will have the same normal vectors. Then we can take n =<1, 4,−3 > and an equation of the plane is 1(x− 2) + 4(y − 1)− 3(z − 0) = 0.

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11. (a) Find an equation of the plane that passes through the pointsA(2, 1, 1), B(−1,−1, 10), and C(1, 3,−4.)Solution:

The vector−−→AB =< −3,−2, 9 > and

−→AC =< −1, 2,−5 > lie in the plane, so n =

−−→AB ×

−→AC =<

−8,−24,−8 > or equivalently, < 1, 3, 1 > is a normal vector to the plane. The point A(2, 1, 1) lieson the plane so an equation for the plane is 1(x− 2) + 3(y − 1) + 1(z − 1) = 0.

(b) A second plane passes through (2, 0, 4) and has normal vector < 2,−4,−3 > . Find an equation forthe line of intersection of the two planes.Solution:

The point (2,0,4) lies on the second plane, but the point also satisfies the equation of the firstplane, so the point lies on the line of intersection of the planes. A vector v in the direction of thisintersecting line is perpendicular to the normal vectors of both planes, so take v =< 1, 3, 1 > × <2,−4,−3 >=< −5, 5,−10 > or just use < 1,−1, 2 > . Parametric equations for the line are

x = 2 + t, y = −t, z = 4 + 2t.

12. Find an equation of the plane through the line of intersection of the planes x − z = 1 and y + 2z = 3and perpendicular to the plane x+ y − 2z = 1.Solution:

n1 =< 1, 0,−1 > and n2 =< 0, 1, 2 > . Setting z = 0, it is easy to see that (1, 3, 0) is a point on the line of

intersection of x − z = 1 and y + 2z = 3. The direction of this line is v1 = n1 × n2 =< 1,−2, 1 > . A second

vector parallel to the desired plane is v2 =< 1, 1,−2 >, since it is perpendicular to x + y − 2z = 1. Therefore,

the normal of the plane in question is n = v1×v2 =< 3, 3, 3 >= 3 < 1, 1, 1 > . Taking (x0, y0, z0) = (1, 3, 0), the

equation we are looking for is (x− 1) + (y − 3) + z = 0 ⇐⇒ x+ y + z = 4.

13. Provide a clear sketch of the following traces for the quadratic surface y =√x2 + z2 + 1 in the given

planes. Label your work appropriately.

x = 0;x = 1; y = 0; y = 2; z = 0.

14. Match the equations with their graphs. Give reasons for your choices.

(a) 8x+ 2y + 3z = 0 Solution: II

(b) z = sinx+ cos y Solution: I

(c) z = sin

(π

2 + x2 + y2

)Solution: IV

(d) z = ey Solution: III

15. Describe the set of all points P = (x, y, z) satisfying x2 + y2 ≤ 4 in

a. cylindrical coordinatesSolution:

In cylindrical coordinates we have x2 + y2 = r2, hence the inequality x2 + y2 ≤ 4 becomes r2 ≤ 4 or r ≤ 2

and 0 ≤ θ ≤ 2π. That is, {(r, θ, z) : r ≤ 2, 0 ≤ θ ≤ 2π}. This is a solid cylinder of radius 2.

b. spherical coordinatesSolution:

In spherical coordinates we have x2+y2+z2 = ρ2 and z = ρ cosφ. Therefore, x2+y2 = ρ2−z2 = ρ2−ρ2 cos2 φ =

ρ2 sin2 φ. the inequality x2+y2 ≤ 4 in spherical coordinates is, thus, ρ2 sin2 φ ≤ 4. Notice that since 0 ≤ φ ≤ π,we have sinφ ≥ 0. Also ρ ≥ 0, therefore ρ sinφ ≥ 0, hence inequality ρ2 sin2 φ ≤ 4 is equivalent to ρ sinφ ≤ 2.

We obtain the following description in spherical coordinates: {(ρ, θ, φ) : ρ sinφ ≤ 2, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π}.

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16. Find a vector function that represents the curve of intersection of the cylinder x2 + y2 = 16 and theplane x+ z = 5.Solution:

The projection of the curve C of intersection onto the xy−plane is the circle x2+y2 = 16, z = 0. So we canwrite x = 4 cos t, y = 4 sin t, 0 ≤ t ≤ 2π. From the equation of the plane, we have z = 5−x = 5−4 cos t, soparametric equations for C are x = 4 cos t, y = 4 sin t, z = 5− 4 cos t, 0 ≤ t ≤ 2π, and the correspondingvector function is r(t) =< 4 cos t, 4 sin t, z = 5− 4 cos t >, 0 ≤ t ≤ 2π.

17. Find an equation for the tangent line to the curve x = 2 sin t, y = 2 sin 2t, and z = 2 sin 3t at the point(1,√

3, 2).Solution:

The curve is given by r(t) =< 2 sin t, 2 sin 2t, 2 sin 3t >, so r′(t) =< 2 cos t, 4 cos 2t, 6 cos 3t > . The point(1,√

3, 2) corresponds to t = π/6, so the tangent vector there is r′(π/6) =<√

3, 2, 0 > . Then thetangent line has direction vector <

√3, 2, 0 > and includes the point (1,

√3, 2), so parametric equations

are x = 1 +√

3t, y =√

3 + 2t, z = 2.

18. A helix circles the z−axis, going from (2, 0, 0) to (2, 0, 6π) in one turn.

(a) Parameterize this helix.Solution:

r(t) =< 2 cos t, 2 sin t, 3t > . (Note that 1 revolution is 2π, so 2π(3) = 6π.)

(b) Calculate the length of a single turn.Solution:

For 0 ≤ t ≤ 2π, ||r′(t)|| =√

4 + 9 =√

13. Thus s =

∫ 2π

0

√13dt = 13(2π).

(c) Find the curvature of this helix.Solution:

The unit tangent vector is T(t) = 1√13

< −2 sin t, 2 cos t, 3 >, so T′(t) = −2√13

< cos t, sin t, 0 >. Thus,

||T′(t)|| = 2√13. Since ||r′(t)|| =

√13, κ = 2

13

19. (a) Sketch the curve with vector function r(t) = 〈t, cosπt, sinπt〉 , t ≥ 0.Solution: The corresponding parametric equations for the curve are x = t, y = cosπt, z = sinπt. Since

y2 + z2 = 1, the curve is contained in a circular cylinder with axis the x−axis. Since x = t, the curve is a

helix.

(b) Find r′(t) and r′′(t).Solution: Since r(t) = 〈t, cosπt, sinπt〉 , r′(t) = 〈1,−π sinπt, π cosπt〉 , and r′′(t) =

⟨0,−pi2 cosπt,−π2 sinπt

⟩.

20. Which curve below is traced out by r(t) =

⟨sinπt, cosπt,

1

4t2⟩, 0 ≤ t ≤ 2.

Solution: I. Note that r(0) =< 0, 1, 0 > .

21. Find a point on the curve r(t) =⟨t+ 1, 2t2 − 2, 5

⟩where the tangent line is parallel to the plane

x+ 2y − 4z = 5.Solution:The plane has normal vector n =< 1, 2,−4 > . Since r′(t) =< 1, 4t, 0 >, we want < 1, 4t, 0 > · < 1, 2,−4 >= 0.That is, 1 + 8t = 0, and r(− 1

8) =< 7

8, −63

42, 5 > .

22. Let r(t) =⟨√

2− t, (et − 1)/t, ln(t+ 1)⟩.

4

(a) Find the domain of r.Solution:

The expressions√

2− t, (et − 1)/t, and ln(t + 1) are all defined when 2 − t ≥ 0. Thus, t ≤ 2, t 6= 0, and

t+ 1 > 0⇒ t > −1. Finally, the domain of r is (−1, 0) ∪ (0, 2].

(b) Find limt→0

r(t).

Solution:

limt→0

r(t) =⟨√

2, 1, 0⟩. Note that in the y− component we use l’Hospital’s Rule.

(c) Find r′(t).Solution:

r′(t) =⟨− 1

2√

2−t ,tet−et+1

t2, 1t+1

⟩.

23. Suppose that an object has velocity v(t) =⟨3√

1 + t, 2 sin(2t), 6e3t⟩

at time t, and position r(t) =<0, 1, 2 > at time t = 0. Find the position, r(t), of the object at time t.Solution:

r(t) =

∫v(t)dt =

⟨∫3√

1 + tdt,

∫2 sin(2t)dt,

∫6e3tdt

⟩=⟨

2(1 + t)3/2 + c1,− cos(2t) + c2, 2e3t + c3

⟩. Thus,

r(0) =< 2+c1,−1+c2, 2+c3 >=< 0, 1, 2 >⇒ c1 = −2, c2 = 2, c3 = 0. So, r(t) =⟨

2(1 + t)3/2 − 2,− cos(2t) + 2, 2e3t⟩.

24. If r(t) =⟨t2, t cosπt, sinπt

⟩, evaluate

∫ 1

0

r(t)dt.

Solution:∫ 1

0

r(t)dt =

⟨∫ 1

0

t2dt,

∫ 1

0

t cosπtdt,

∫ 1

0

sinπtdt

⟩=

⟨1

3,− 2

π2,

2

π

⟩.

25. Find the length of the curve: x = 2 cos(2t), y = 2t3/2, and z = 2 sin(2t); 0 ≤ t ≤ 1.Solution:

s =

∫ 1

0

√(dx

dt

)2

+

(dy

dt

)2

+

(dz

dt

)2

dt =

∫ 1

0

√16 sin2(2t) + 16 cos2(2t) + 9t dt =

∫ 1

0

√16 + 9tdt =

122

27.

26. Reparameterize the curve r(t) =< et, et sin t, et cos t > with respect to arc length measured from thepoint (1, 0, 1) in the direction of increasing t.Solution:

The parametric value corresponding to the point (1, 0, 1) is t = 0. Since r′(t) =< et, et(cos t + sin t), et(cos t −sin t) >, ||r′(t)|| = et

√1 + (cos t+ sin t)2 + (cos t− sin t)2 =

√3et, and s(t) =

∫ t0eu√

3du =√

3(et − 1) ⇒ t =

ln(

1 + 1√3s). Therefore, r(t(s)) =

⟨1 + 1√

3s,(

1 + 1√3s)

sin ln(

1 + 1√3s),(

1 + 1√3s)

cos ln(

1 + 1√3s)⟩

.

27. Find the tangent line to the curve of intersection of the cylinder x2 + y2 = 25 and the plane x = z atthe point (3, 4, 3).Solution:

Let x(t) = 5 cos t, y(t) = 5 sin t, and z(t) = 5 cos t, so that r(t) =< 5 cos t, 5 sin t, 5 cos t > and r′(t) =<

−5 sin t, 5 cos t,−5 sin t > . When (x, y, z) = (3, 4, 3), x(t) = z(t) = 5 cos t = 3, and y(t) = 5 sin t = 4, so

r′(t0) =< −4, 3,−4 > is a direction vector for the line. The tangent line has parametric equations x(t) =

3 − 4t, y(t) = 4 + 3t, and z(t) = 3 − 4t. ANOTHER SOLUTION: Let x(t) = t, y =√

25− x2 =√

25− t2, and

z(t) = x(t) = t. Then r(t) =⟨t, (25− t)1/2, t

⟩. In this case, r′(t) =

⟨1, −t√

25−t2, 1

⟩. When x = 3, t = 3, and

r′(3) =< 1, −34, 1 > . So the tangent line has parametric equations x(t) = 3 + t, y(t) = 4− 3

4t, and z(t) = 3 + t.

28. For the curve given by r(t) =⟨13 t

3, t2, 2t⟩, find

(a) the unit tangent vector

Solution: T(t) =r′(t)

||r′(t)|| =< t2, 2t, 2 >√t4 + t2 + 1

=< t2, 2t, 2 >

(t2 + 2).

(b) the unit normal vectorSolution:

T′(t) =< 4t, 4− 2t2,−4t >

(t2 + 2)2⇒ ||T′(t)|| =

√4(t4 + 4t2 + 4)

(t2 + 2)2=

2

t2 + 2and N(t) =

< 2t, 2− t2,−2t >

t2 + 2

5

(c) the curvatureSolution:

κ(t) =||T′(t)||||r′(t)|| =

2

(t2 + 2)2

29. A particle moves with position function r(t) =< t ln t, t, e−t >. Find the velocity, speed, and accelerationof the particle.Solution:

v(t) = r′(t) =< 1 + ln t, 1,−e−t > . ν(t) = ||v(t)|| =√

(1 + ln t)2 + 1 + (−e−t)2 =√

2 + 2 ln t+ (ln t)2 + e−2t.

a(t) = v′(t) =<1

t, 0, e−t > .

30. A particle starts at the origin with initial velocity< 1,−1, 3 > and its acceleration is a(t) =< 6t, 12t2,−6t >. Find its position function.Solution:

v(t) =

∫a(t)dt =

⟨∫6t dt,

∫12t2 dt,

∫−6t dt

⟩=⟨3t2, 4t3,−3t2

⟩+ C, but < 1,−1, 3 >= v(0) + C, so

C =< 1,−1, 3 > and v(t) =⟨3t2 + 1, 4t3 − 1,−3t2 + 3

⟩. r(t) =

∫v(t)dt =

⟨t3 + t, t4 − t, 3t− t3

⟩= D. But

r(0) = 0, so D = 0, and r(t) =⟨t3 + t, t4 − t, 3t− t3

⟩.

31. Find the tangential and normal components of the acceleration vector of a particle with position functionr(t) =< t, 2t, t2 > .Solution:

r′(t) =< 1, 2, 2t >, r′′(t) =< 0, 0, 2 >, ||r′(t)|| =√

1 + 4 + 4t2 =√

4t2 + 5. Then aT =r′(t) · r′′(t)||r′(t)|| =

4t√4t2 + 5

and aN =||r′(t)× r′′(t)||||r′(t)|| =

2√

5√4t2 + 5

.

32. A flying squirrel has position r(t) =

⟨t+

t2

2, 1− t, 2 + t2

⟩at time t. Compute the following at time t = 1:

(a) The velocity at time t = 1, v(1) = < 2,−1, 2 >.Solution:

v(t) = r′(t) =< 1 + t,−1, 2t > .

(b) The speed at time t = 1, ν(1) = 3 .Solution:

ν = ||v(t)||. ||v(1)|| =√

4 + 1 + 4 = 3.

(c) The acceleration at time t = 1, a(1) = < 1, 0, 2 >.Solution:

a(t) = v′(t) =< 1, 0, 2 > .

(d) The tangential component of acceleration at time t = 1, aT(1) = 2 .Solution:

aT =a · v||v|| , aT(1) = <1,0,2>·<2,−1,2>

3= 2+4

3= 2.

(e) The normal component of acceleration at time t = 1, aN(1) = 1 .Solution:

aN =√||a|| − a2T =

√5− 4 = 1 OR aN =

||a× v||||v|| =

|| < 2, 2,−1 > ||3

=3

3.

(f) The curvature of the squirrel’s path of motion at the point(32 , 0, 3

), κ = 1/9 .

Solution:

1 = aN = ν2κ = 9κ, κ = 19

33. Consider the vector valued function r(t) = describing the curve shown below. Put the curvature of r atA,B and C in order from smallest to largest. Draw the osculating circles at those points.

Solution: B, A, C.

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