Intro. to Differential Equation

8/9/2019 Intro. to Differential Equation

1/127

Chapter 8

Further Applications of

Integration


2/127

A differential equation is an equation that contains

an unknown function and some of its derivatives.

edx

d

dx

d xydx

dyyx

y =++ 22

2

3

3

Below are some

examples:

The orderof a differential equation is the order ofthe highest derivative that occurs in the equation.

Thus the above 3 equations are of the order 1, 2, and

3, respectively.

xyy =' 0'2" =++ yyy

In each of these differential equationsy is an unknown

function ofx.

8.1 Differential Equations


3/127

To solve a differential equation means to find all

possible solutions of the equation. For example,

any solution of the equation y"+ y=0 is of theform y =Asinx+Bcos x, where both A and B are

constants. So it is called the general solution of

the differential equation.

A functionfis called asolution of a differential

equation if the equation is satisfied wheny =f(x)

and its derivatives are substituted into the equation.For example,fis a solution of equationy'=xy if

f '(x) =xf(x).


4/127

Any particular solutions are obtained by

substituting values for the arbitrary constantsA andB. For instance,y = sinx is a particular

solution of the above differential equation by

choosingA = 1,B = 0 in the general solution.

In general, solving a differential equation is

not an easy matter.


5/127

Separable equation

)(

)(

yh

xg

dx

dy=

Aseparable equation is a first-order differential

equation that can be written in the form dy/dx =

g(x)f(y). The nameseparable comes from the fact

that the expression on the right side can be

separated into a function ofx and a function ofy.

Equivalently, we could write


6/127

so that allys are on one side of the equation

and all xs are on the other side. Then weintegrate both sides of the equation:

= dxxgdyyh )()(

To solve this equation we rewrite it in thedifferential form

h(y)dy =g(x)dx

It definesy implicitly as a function ofx.

In some cases we may be able to solve fory in

terms ofx


7/127

The justification of the above last step comes form

the Substitution Rule:

= dxdx

dyxyhdyyh ))(()(

=

=

dxxg

dxxyhxgxyh

)(

))(()())((


8/127

Solve the differential equationyy

x

dx

dy

cos2

6 2

+=

Cxyy

dxxdyyy

dxxdyyy

+=+ =+

=+

32

2

2

2sin

6)cos2(

6)cos2(

Solution

Writing the equation in differential form and

integrating both sides, we have

Example 1


9/127

where Cis an arbitrary constant. (We could have used

a constant C1on the left side and another constant C

2

on the right side, but then we could combine these

constants by writing C= C2- C

1

The above general solution is in implicit form. In this

case it is impossible to expressy explicitly as a

function ofx.


10/127

Solve the differential equation yxy 2'=

SolutionRewrite the equation using Leibniz notation:

Ify 0, we can rewrite it in differential notation andintegrate:

dy/dx =x2y.

Example 2

dy/y = x2dx, y 0, ln|y| =x3/3 + C


11/127

Note that the functiony = 0 is also a solution

of the given differential equation. So thegeneral solution is in the form 33x

Aey =

In this case, we can solve explicitly fory:

333ln333

,x

Cx

CCx

y

eeyeeeey ====+

whereA is an arbitrary constant (A = eCor 0).


12/127

Initial-value problem

The problem of finding a solution of the differential

equation that satisfies the initial condition is called

an initial-value problem.

In many physical problems we need to find theparticular solution that satisfies a condition of the

formy(x0) =y0. This is called an initial condition.


13/127

Example 1

Solve the differential equation

.2)4(,0,' =>= yxyxySolution

Write the differential equation as:

Integrate both sides:

xdy/dx = -y or dy/y = -dx/x.

ln|y| = -ln|x| + C, |y| = 1/|x| eC

= xdxydy //


14/127

To determineKwe putx = 4 andy = 2 in thisequation:

2 =K/4 K= 8

So the solution of the initial-value problem is

y = 8/x , x > 0


15/127

Example 2

Find the solution of dy/dx = 6x2/(2y + cosy) that

satisfiesy(1) = .

Solution

From Example 1 in the last part, we know that the

general solution isy2 + siny = 2x3 + C


16/127

Therefore, the solution is given implicitly by

y2 + siny = 2x3+ 2 2


17/127

Example 3 Solve y'= 1 + y2 - 2x - 2xy2 , y(0) = 0,

and graph the solution.

Substitutingx = 0 andy = 0 in this equation, we

get C = 0. So

Solution

Factor the right side as the product of a function ofx

and a function ofy:

tan-1y =x -x2


18/127

To graph this equation, notice that it is equivalent

to y = tan(x -x2)

provided that - /2


19/127

Example 4

A tank contains 20kg of salt dissolved in 5000L of

water. Brine that contains 0.03kg of salt per liter ofwater enters the tank at the rate of 25L/min. The

solution is kept thoroughly mixed and drains from

the tank at the same rate. How much salt remains in

the tank after half an hour?


20/127

Solution

Let y(t) be the amount of salt (in kilograms)

aftertminutes. We are given thaty(0) = 20and we want to find y(30). We do this byfinding a differential equation satisfied byy(t).Note that dy/dtis the rate of change in

the amount of salt, sody/dt= (rate in) (rate out)

where (rate in) is the rate at which salt entersthe tank and (rate out) is the rate at whichsalt leaves the tank.


21/127

We have

rate in = (0.03kg/L)(25L/min) = 0.75 kg/min

The tank always contains 5000L of liquid, so theconcentration at time tisy(t)/5000 (kg/L). Since the

brine flows out at a rate of 25L/min, we have

rate out = (y

(t)/5000 kg/L)(25L/min) =[y(t)/200 ]kg/min

Thus dy/dt= 0.75 [y(t)/200] = [150 -y(t)]/200

Solve the separable differential equation byintegrating

dy/(150 -y) = dt/200

-ln|150 y | = t/200 + C.


22/127

Sincey(0) = 20, we have ln130 = C, so

-ln|150 y | = t/200 ln130.

Therefore

Since y(t) is continuous andy(0) = 20 and the right

side is never 0, we deduce that 150 y(t) is always

positive. Thus |150 y | = 150 y and

The amount of salt after 30 min is

.130150 200t

ey

=

.130150)( 200t

ety

=

.kg1.38130150)30( 20030

=

ey


23/127

Logistic growth

Under the conditions of unlimited environment and

food supply, the rate of population growth isproportional to the size of the population. This can

be described by the differential equation

dy/dt= ky


24/127

Solve the separable equation:

Where or 0 is an arbitrary constant.

kt

ktCCkt

Aey

eeey

Ckty

ykdtdyy

=

==+=

=

+

ln

0,1

CeA =

I i d i d i h li i d f d


25/127

In a restricted environment and with limited food

supply, the population cannot exceed a maximal

size Mat which it consumes its entire food supply.

If we make the assumption that the rate of growthof population is jointly proportional to the size of

the population y and the amount by which y falls

short of the maximal size (M-y), then we have theequation

dy/dt= ky (M-y)

where k is a constant. This equation is called thelogistic differential equation.


26/127

The logistic equation is separable, so we write it in

the form

Using the partial fraction, we have 1/[y(M-y)] =1/M[1/y +1/(M-y)]

and so

1/M[ dy/ y + dy/(M-y)] = kdt = kt + C

1/M(ln|y| - ln|M-y|) = kt + C

=

kdtdyyMy )(

1

Si 0 < < M | | d |M | M h


27/127

Since 0 < y < M, |y| = y and |M-y| = M-y, so we have

ln(y/M-y) = M(kt + C)

y/(M-y) =AekMt (A = eMC)

If the population at time t= 0 isy(0) = y0, thenA =y0/(M-

y0), so

y/(M-y) =y0/(M-y0)ekMt

Solve this equation for y, we get

y=y0MekMt

/(M-y0 +y0ekMt

)=y0M/[y0+(M-y0)e-kMt

]We can see that

which is to be expected.

Mtyt

=

)(lim


28/127

The graph of the logistic growth function is shown

here. At first the graph is concave upward and the

growth curve appears to be almost exponential,

but then it becomes concave downward and

approaches the limiting population M.


29/127

Direction fields

),(' yxFy =

whereF(x,y) is some expression inx andy. Evenif it is impossible to find a formula for the solution,

we can still visualize the solution curves by means

of a direction field.

Suppose we are given a first-order differential

equation of the form


30/127

If a solution curve passes through a point (x,y),

then its slope at that point isy', which is equal to

F(x,y). If we draw short line segments with slope

F(x,y) at several points (x,y), the result is called a

direction field(orslope field). These line segments

indicate the direction in which a solution curve isheading, so the direction field helps us visualize

the general shape of these curves.

E l


31/127

(a) We start by computing the slope at several

points as in the chart

Example

(a) Sketch the direction field for the differential

equationy'= x2

+y2

1.(b) Use part (a) to sketch the solution curve that passes

through the origin.

Solution


32/127

(b) Now we draw short line segments with these slopes

at these points. The result is the direction fieldshown in the figure (on the next slide).


33/127

(b) We start at the origin and move to the right in the direction

of the line segment (which has slope 1). We continue to draw

the solution curve so that it moves parallel to the nearby line

segments. The resulting solution curve is shown in the figure.Returning to the origin, we draw the solution curve to the left as

well.


34/127

The more line segments we draw in a direction

field, the clearer the picture becomes. Of course, it

is tedious to compute slopes and draw linesegments for a huge number of points by hand, but

computers are well suited for this task. This

enables us to draw the solution curves with

reasonable accuracy.

The idea of direction fields is adapted to find

numerical approximations to the values of solutionsof differential equations. This technique is called

Eulers method.


35/127

we take a partitionPof [a, b] determined by pointsxiwith a= x0< x1


36/127


37/127

Therefore, we define the length L of the curve C

with equationy = f(x), , as the limit of the

lengths of these inscribed polygons (if the limitexists):

Notice that the procedure for defining arc length is

very similar to the procedure we used for defining

area and volume. We divided the curve into a largenumber of small parts. We then found the

approximate lengths of the small parts and added

them. Finally we took the limit as ||P|| 0.

.||lim10||||

1=

=n

iPPP iiL

bxa


38/127

The definition of arc length given above is not

very convenient for computational purposes,but we can derive an integral formula forL in

the case wheref has a continuous derivative.

[Such a functionfis calledsmooth because asmall change inx produces a small change in

f '(x).]


39/127

If we let yi =yi -yi-1 , then

By the Mean Value Theorem tofon the interval

[xi-1, xi], there is a numberxi* betweenxi-1 and xi suchthat

f(xi) -f(xi-1 ) =f '(xi*) (xi -xi-1 ),

i.e. yi =f '(xi*) xi .

Thus we have

.|| )()(22

1 yxPP iiii +=

)()(|| 221 yxPP iiii +=

.1 )]('[

])('[)(

*

*

2

22

xxf

xxfx

ii

iii

+=

+=


40/127


41/127

By the definition of a definite integral, we

recognize the above expression as being equal to

This integral exists because the function

is continuous.

Thus we have proved

.1 )]('[2

+b

a dxxf

)]('[1)(2

xfxg +=

The arc length formula


42/127

Iff 'is continuous on [a,b], then the length of the

curvey = f(x), , isbxa

.1 )]('[2

+=b

adxL xf

.1 )(2

+= ba dxLdx

dy

The arc length formula

Use the Leibniz notation for derivatives,

Example 1:


43/127

Example 1:

Find the length of the arc of the semicubical parabolay2 =x3 between the

point (1,1) and (4,8).

Solution:


44/127

,2

32/1

xdx

dy=

and so the arc length formula gives

.4

911

4

1

4

1

2

)( +=+= dxxdxLdx

dy

Substitute u = 1+9x/4, then du = 9dx/4. Whenx = 1, u = 13/4;whenx = 4, u = 10.

Therefore += +=

4

1

4

1

2

4

91)(1 dxxdx

dx

dyL

.

27

13131080

][278 )

413(10

2

3

23

=

=

For the top half of the curve we havey =x3/2,


45/127

If a curve has the equation x = g(y), , then

by the interchanging the roles ofx and y in theformula, we obtain

.11 )()]('[

2

2

+=+=d

c

d

c dydyLdydxyg

dyc

Example 2:


46/127

Example 2:

Find the length of the arc of the parabolay2 =x from (0,0) to (1,1).

S l i


47/127

Solution:

Sincex =y2, we have dx/dy = 2y, and so

.111

0

21

0

2

4)( +=+= dydyL ydydx

Make the trigonometric substitution tan21=y

ddy 221 sec= sectan141 22 =+=+ yand

Wheny = 0, = 0, so ; when , sotan 0= 2ta

nsay.,2arctan ==

Thus

, which gives


48/127

= =aa

ddL0

32

0 sec2

1sec

2

1sec

[ ]

( ).|tansec|lntansec41

2

1

2

1|tansec|lntansec

0

++=

= ++ a

.4

)25ln(

2

5 ++=L

Since , we have2tan =

and5secso,5tan1sec22

==+=


49/127

Because of the presence of the square root sign in

the arc length formula, the calculation of an arc

length often leads to an integral that is very difficultor even impossible to evaluate explicitly.

Thus we sometimes have to be content with finding

an approximation to the length of a curve as in thefollowing example.

Example 3


50/127

Example 3

(a) Set up an integral for the length of the arc of the

hyperbolaxy = 1 from the point (1,1) to (2,1/2).

(b) Use Simpsons Rule with n = 10 to estimate the

arc length.

Solution

(a)We have y = 1/x, dy/dx = -1/x2.

and so the arc length is

.11

112

1 2

4

2

1 4

2

1

2

)( +

=+=+= dxdxdxLx

x

xdx

dy

(b) Using Simpsons Rule with a=1, b=2, n=10, x=0.1, ./11)(4

xxf +=


51/127

We have

+=2

1

4/11 dxxL

[ ]

.1321.1

]1

11

141

12

114

112

114

11[

3

1.0

)2()9.1(4)8.1(2)3.1(4)2.1(2)1.1(4)1(3

2)9.1()8.1(

)3.1()2.1()1.1(1

444

4444

+++++++

+++++++=

+++++++

fffffffx

Th l h f i


52/127

The arc length function

If a smooth curve Chas the equationy = f(x),

, lets(x) be the distance along Cfrom the initialpointP

0(a

,f(a)) to the point Q(x

,f(x)). Thens is a

function, called the arc length function, and

Use Part 1 of the Fundamental Theorem of

Calculus to differentiate the above equation (since

the integrand is continuous):

.1)( )]('[2

+=x

adtxs tf

.11 )()]('[2

2

dx

dyxf

dx

ds +=+=

bxa


53/127

The above equation shows that the rate of change

ofs with respect tox is always at least 1 and isequal to 1 whenf '(x), the slope of the curve, is 0.

Th diff ti l f l th i2

dy


54/127

The differential of arc length is

This equation is sometimes written in the symmetric

form

.1 )( dxdsdx

dy+=

.)()()(222

dydxds +=The geometric interpretation of it is shown in the figure

The symmetric form can be used as a mnemonic


55/127

The symmetric form can be used as a mnemonic

device for remembering the arc length formula. If

we write , then from the symmetric

form, we either solve to get

which gives

= dsL

dxdx

dyds )(1

2

+=

.1 )(

2

+=b

a dxL dxdy

Or we can solve to get

which givesdydy

dx

ds )(1

2

+=

.1 )(

2

+=

d

c dyL dy

dx

E l


56/127

Example

Find the arc length function for the curve y = x2

(lnx)/8 takingP0 (1,1) as the starting point.Solution

Thus the arc length function is given by

.8

1

211 8

1

28

1

2)]('[

22

2

xxxxxxxf +==+=+

+

xxxf

8

12)(' =

+= +=xx

dtttdttfxs 112

)8

1

2()]('[1)(

1ln8

1ln

8

1 2

1

2 +=

+= xxtt

x

8 3 A f S f f R l i


57/127

8.3 Area of a Surface of Revolution

A surface of revolution is formed when a curve isrotated about a line. Such a surface is the lateral

boundary of a solid of revolution.

We want to define the area of a surface of

revolution. We can think of peeling away a very thin

outer layer of the solid of revolution and laying it out

flat so that we can measure its area.


58/127

Some simple surfaces

(a) The lateral surface area of a circularcylinder with radius rand height h is taken to

be A = 2 rh because we can imagine cuttingthe cylinder and unrolling it to obtain a

rectangle with dimensions 2 rand h.

(b) F i l ith b di d l t


59/127

(b) For a circular cone with base radius r and slant

height l, cut it along the broken line (see the figure) ,

and flatten it to form a sector of a circle with radius land central angle = 2 r/l.


60/127

In general, the area of a sector of a circle with

radius land angle is

l2

21

Therefore, we define the lateral surface area of a

cone to beA = rl

rllrllA =

== 2

21

21 22

In this case, it is

(c) The area of the band (or frustum of a cone) with


61/127

( ) ( )

slant height l and upper and lower radii r1 and r2 is

found by subtracting the areas of two cones:

From similar triangles we have

which gives or Putting them together, we get

or

where r = ()(r1 + r2) is the

average radius of the band.

( )[ ]lrlrrlrllrA 21121112 )( +=+=

r

ll

r

l

2

1

1

1 +=

lrlrlr 11112 += lrrr 112 )( =

)( 21 lr

lr

A +=

rl

A2

The definition of surface area


62/127

Consider the surface obtained by rotating the curve y = f(x), ,

about thex-axis, wherefis positive and has a continuous derivative.

Take a partitionPof [a,b] determined by pointsxi with a = x0< x1


63/127

When xi is small, we have yi f(xi) f(xi ) andyi-1 f(xi-1 ) f(xi ),

sincefis continuous. Therefore

and so an approximation to what we think of as the area of the

complete surface of revolution is

This approximation appears to become better as ||P|| 0 and

( ) ( )[ ] xxfxPPyy

iiii

ii

if ++

*'2

*

1

1 12

2

2

( ) ( )[ ] xxfxf in

i

i i +

+1

2* *'12

( ) ( )[ ] [ ] dxxfxfxxfxfb

aii

n

in

i )(1)(212lim'*'

22*

1

+=+=

Therefore in the case where f is positive and has a continuous


64/127

Therefore, in the case where f is positive and has a continuous

derivative, we define the surface area of the surface obtained by

rotating the curvey = f(x), , about thex-axis as

With the Leibniz notation for derivatives,

If the curve is described byx = g(y), , then

[ ] dxxfxfSb

a )(1)(2' 2+=

dxdx

dy

ySb

a

+=

2

12

dyxSd

c

dy

dx

+=

2

12

bxa

dyc

Using the notation for arc length given in Section 8 2


65/127

Using the notation for arc length given in Section 8.2,

where

For rotation about they-axis, the surface area formula is

where

ydsS = 2

xdsS = 2

dx

dx

dyds

+=2

1

dydydxds +=

2

1

Example 1:


66/127

The curve is an arc of the circle .

Find the area of the surface obtained by rotating this arcabout the

x-axis. (The surface is a portion of a sphere of radius 2.)

Solution

so

===

=

+=

+=

1

1

2

1

1

2

1

1 2

2

2

2

1

1

8)2(414

4

242

4142

12

dx

dx

dx

dxyS

xx

x

x

x

dx

dy

x

xxx

dx

dy2

2/1

4)2()4(

2

12

==

11,4 2 = xxy 422 =+ yx

Example 2


67/127

The arc of the parabolay = x2 from (1,1) to (2,4) is rotated about the y-

axis. Find the area of the resulting surface.

Solution1

Using y = x2 and dy/dx = 2x, we have

Substituting u = 1+4x2, we have du = 8xdx.

Remembering to change the limits of

integration, we have

dxx

dxxxdsS

x

dx

dy

22

1

2

2

1

412

122

+=

+==

( )5517176

3

2

442/3

17

5

17

5

=

==

uduuS

Solution 2dx 1


68/127

Using and

we have

( )5517176

4

144

112

122

17

5

4

1

4

1

2

4

1

=

=

+=+=

+==

duu

dyydyy

y

dyxxdsSdy

dx

yx =ydy

dx

2

1=

Example 3


69/127

Find the area of the surface generated by rotating the curve

y =ex, , about thex-axis.

Solution y = ex

so dy/dx = ex

, we have

Since tan = e, sec2 = 1+ tan2 = 1 + e2 so

[ ]

[ ])12ln(2)tanln(sectansec2

1

2

2

12

12

12

|tansec|lntansec

sec

4/

4/

3

1

2

21

0

1

0

2

+++==

=

+=

+=

+=

++

d

du

dx

dxyS

e

xx

u

ee

dx

dy

) l 2 ) 1 l 122

e e e e S

10 x

8.4 Application to Economics


70/127

pp

(1) Consumer surplus

The demand functionp(x) is the price that a company has to

charge in order to sellx units of a commodity.

The demand function and demand curve

Usually, selling larger quantities

requires lowering prices, so thedemand function is a decreasing

function.The graph of a typical demand

function, called a demandcurve is shown in the figure. If

X is the amount of the

commodity that is currently

available, thenP= X is the current sellin rice.

The consumer surplus


71/127

The consumer surplus

Partition the interval [0, X] into n subintervals, each of length x =

X/n, withxibe the right endpoint of the ith subinterval.

For the consumers between xi-1 and xi, the price they are willing to

pay is aboutp(xi), the price they actually pay isP. Therefore, we can

consider they have saved an amount of

(savings per unit)(number of units) = [p(xi) P] x

C id i i il f illi f th


72/127

Considering similar groups of willing consumers for the

other subintervals, adding the savings, we get the total

savings:

Let , this Riemann sum approaches the integral

which is called the consumer surplus of the commodity.

xPxpn

ii

=])([

0

n

dxPxpX ])([0


73/127

The consumer surplus represents the amount of money

saved by consumers by purchasing the commodity at price

P, corresponding to an amount demanded ofX.

The figure below shows the interpretation of the consumer

surplus as the area under the demand curve and above the

linep =P.

Example


74/127

Example

The demand for a product, in dollars, isp = 1200 0.2x

0.0001x2. Find the consumer surplus when the sales level is

500.

Solution

The corresponding price forX= 500 is

P= 1200 0.2(500) 0.0001(500)2= 7500

Therefore, the consumer surplus is

33.333,33$30001.01.0)500(125

)0001.02.0125(

)10750001.02.01200(])([

)500(

)500(

30001.01.0125

3

2

500

0

500

0

2

500

0

2500

0

32

==

==

=

xxxx

x

dxx

dxxdxPxp

(2) Present value of an income stream


75/127

(2) Present value of an income stream

Continuously compounded interest rate

With continuously compounded interest rate r, the valueof a savings account y(t) increases at a rate proportional

to that value, i.e.

Then at time t, the value ofy is

(see Example of section 8.1).

Present value of a future amountIfA0 is the amount that will grow toA in tyears, thenA0ert = A and soA0 = Ae

-rt .A0 is called thepresent value of

A.

eytyrt

)0()( =

.ry

dt

dy=

Present value of an income stream


76/127

ese v ue o co e s e

Suppose that income will be received over a period of

time from t = a to t = b at a rate off(t) dollars per year

at time t. This is referred to as an income stream.

To find the total present value of this income stream, we

partition the interval [a,b] into n subintervals of equal

length . From time t = ti-1 to time t = ti the incomereceived will be approximately dollars, with a

present value of . So an approximation to the

present value of the total income is

If we let , the Riemann sum approaches the integral

which is thepresent value of the income streamf(t).

t ttf i )(

ttfe itir )(

=

n

i

tir ttfe i1

)(

n

b

art dttfe )(

Example:


77/127

A trust fund pays $8000 a year for 10 years, starting 5

years from now, at a rate of 10% per year compounded

continuously.(a) Find the present value of the trust fund.

(b) Find the value 3 years from now.

Solution

(a) Here the income stream isf(t) = 8000. Using the

formula with a = 5, b = 15 and r= 0.1, the present value

of the trust fund is

= 80,000( e-0.5 e-1.5 ) = $30,672.04

(b) The value 3 years from now is

30,672.04 e(0.1)3 = $41,402.92

155

)1.0(15

5 )1.0( ]1.08000[)8000( =

eet

t dt

8 5 Curves Defined by Parametric


78/127

8.5 Curves Defined by Parametric

Equations

Suppose that x and y are both given as continuous functions of a

third variable t(called aparameter) by theparametric equations

x =f(t) y =g(t) .Each value of tdetermines a point (x, y), which we can plot in a

coordinate plane. As tvaries, the point (x, y) = (f(t), g(t) ) varies

and traces out a curve C.

If we interpret tas time and (x,y) = (f(t),g(t) ) as the position of a

particle at time t, then we can imagine the particle moving along

the curve C.

Example 1:


79/127

Identify the curve defined by theparametric equationsx = t22tand

y = t +1.

equations moves along the curve in the

direction of the arrows as tincreases.

Eliminate theparametertas follows:

Fromy = t +1we obtain t = y 1. Substitute

it intox = t22t, it gives

x = (y-1)22(y-1) = y24y +3

and so the curve represented by the given

parametric equations is aparabola.

Look at the figure: A particle whose position is given by theparametricSolution:

Example 2:


80/127

What curve is represented by theparametric equationsx = cos tandy =

sin t , 0 t 2 ?

Solution:

Eliminate tby noting that x2 +y2 = cos2t+ sin2t = 1Thus the point (x, y) moves on the unit

circlex2 +y2 = 1. Notice that theparametert

can be interpreted as the angle shown in thefigure.

As tincreases from 0 to 2 , the point (x,y) = (cos t, sin t) moves once around the

circle in the counterclockwise directionstarting from the point (1,0) .

Example 3:


81/127

What curve is represented by theparametric equationsx = sin2t and

y = cos2t , 0 t 2 ?

Thus again the point (x, y) moves on the

unit circlex2 +y2 = 1.

But as tincreases from 0 to 2 , the point(x, y) = (sin2t, cos2t) moves twice around

the circle in the clockwise direction starting

from the point (0,1) .

Solution:

Again, eliminate tby noting that x2 +y2 = sin22t + cos22t= 1

Example 4


82/127

Observe that y =x2 and so the point (x,y) moves on the parabola y =x2.

But note that since -1 sint 1, we have-1x 1, so the parametric equation

represent only the part of the parabola forwhich -1 x 1.

Since sin tis periodic, the point (x,y)=(sin t,

sin2t) moves back and forth infinitely often

along the parabola from (-1,1) to (1,1).

Sketch the curve represented by theparametric equationsx = sint

and y = sin2t.

Solution

Example 5:The curve traced out by a pointPon the circumference of acircle as the circle rolls along a straight line is called a cycloid If the


83/127

circle as the circle rolls along a straight line is called a cycloid. If the

circle has radius rand rolls along thex axis and if one of the positions of

Pis the origin, findparametric equations for the cycloid.

has rolled from the origin is |OT| = arcPT =

r .

Let the coordinates ofPbe (x,y), thenx = |OT| - |PQ| = r - rsin = r( - sin )

y = |TC| - |QC| = r- rcos = r(1 - cos )

This is also valid for other values of ( try it

Solution: Choose the angle for which the circle has rotated as the

parameter( =0 whenPis at the origin). For0< < /2, the distance

So theparametric equations of the cycloid are

( i ) (1 )


84/127

x = r( - sin ) y = r(1 - cos ) R

One arch of the cycloid comes from one rotation of the circle and so is

described by 0 2 .Some properties ofcycloid:

(1) A particle slides along the curve from point

A to a lower point B not directly beneath A.Among all possible curves joining A to B, the

particle will take the least time if the curve is

an inverted arch of a cycloid.

(2) No matter where a particle P is placed on

an inverted cycloid, it takes the same time to

slide to the bottom.

8.6 Tangents and Areas


85/127

By eliminating the parameter, some curves defined by parametricequationsx =f(t) and y =g(t) can be expressed in the form y =F(x). If

we substitutex =f(t)and y =g(t) in the equationy =F(x), we get

g(t) =F(f(t))

Ifg,Fandfare differentiable, the Chain Rule gives

g'(t) =F'(f(t))f '(t) =F '(x)f '(t)

Iff '(t) 0, we can solve forF '(x):

F '(x) =g '(t)/f '(t)

Since the slope of the tangent to the curvey =F(x) at (x,F(x)) isF '(x),

the above equation enables us to find tangent to parametric curves

without having to eliminate theparameter.

(1) Tangents

Using Leibniz notation, we can rewrite the above equation asd


86/127

if

It can be seen from the above equation that the curve has a

horizontal tangent when dy/dx = 0 (provided that dx/dt 0) and it

has a vertical tangent when dx/dy = 0 (provided that dy/dt 0). Thisinformation is useful when sketching parametric curves.

To obtain d2y/dx2, replacey by dy/dx in the above equation:

dt

dxdt

dy

dx

dy= 0

dt

dx

dt

dx

dx

dy

dt

d

dx

dy

dx

d

xd

yd

=

=

2

2

Example 1


87/127

(a) Find dy/dx and d2y/dx2 for the cycloidx = r( -sin ),y = r(1-cos ).

(b) Find the tangent to the cycloid at the point where = / 3.

(c) At what points is the tangent horizontal? When is it vertical?

(d) Discuss the concavity.

cos1

sin

)cos1(

sin

=== rr

d

dxd

dy

dx

dy

cos1

11cossinsin)cos1(cos

cos1

sin

)cos1()cos1(22

=

=

=

=

d

d

dx

dy

d

d

Solution

(a)


88/127

)cos1(

1

)cos1(cos1

1

22

2

=

=

=rr

d

dx

dx

dy

d

d

xd

yd


89/127

(b) When = / 3, we have

and

Therefore, the slope of the tangent is and the equation is

or

=

= 23

33sin3

rrx 23cos1

r

ry =

=

3=211

23=

31

3=

/

/

)/cos(

)/sin(

dx

dy

+= 2

3

332

rrx

ry

= 233

ryx

3

(c) The tangent is horizontal when dy/dx = 0, which occurs when sin =

0 d 1 0 h i (2 1) h i i Th


90/127

0 and 1-cos 0, that is, = (2n-1) , where n is an integer. Thecorresponding point on the cycloid is ((2n-1) r, 2r).

When = 2n , both dy/d and dx/d are 0. There are verticaltangents at these points. We can verify this by using lHospitals Rule:

==

=+++

sin

coslim

cos1

sinlimlim

222 nnn dx

dy

From part(a) we have d2y/dx2 = -1 / [r(1-cos )2]. Since r>0, this

s that d2y/dx2 < 0 except when cos =1. Thus the cycloid is concave

nward on the intervals (2n , 2(n+1) ).

similar computation shows that dy/dx- as 2n -, so indeed there

are vertical tangents when = 2n , that is, when x = 2n r.

Example 2 A curve Cis defined byx = t2 andy = t3-3t.

(a) Show that C has two tangents at (3 0) and find their equations


91/127

(a) Show that Chas two tangents at (3, 0) and find their equations.

(b) Find the points on Cwhere the tangent is horizontal or vertical.

(c) Determine where the curve rises or falls and where it is concaveupward ordownward.

(d) Sketch the curve.


92/127

=b

dxxFA )(

(2) AreasThe area under a curve y = F(x) from a to b is


93/127

If the curve is given byparametric equationsx =f(t)and y =g(t), t , then we can adapt the earlier formula by

using the Substitution Rule forDefinite Integrals as follows:

= a dxxFA )(

==b

adttftgydxA

)(')( dttftgor )(')(

))()((or

)()()())(()()(

=

=

==

dttftg

dttftgdttftfFdxxFAtfx

b

a

The area under a curvey F(x) from a to b is

whereF(x) 0.

Example 1

Find the area under one arch of the cycloidx = r( -sin ),y = r(1-cos ).


94/127

.20

Using the Substitution Rule withy = r(1-cos ) and dx =r(1-cos )d , we have

Solution

One arch of the cycloid is given by

=r

ydxA2

0


95/127

( ) rr 22 3223

==

y0 =

2

0)cos1()cos1( drr

=

2

0

22 )cos1( dr

+=

2

0

2 )2coscos21( dr

++=

2

0

2

)]2cos1(21cos21[ dr

+=

4sinsin241

2

32

0

2

r

Example 2

Find the area of the region enclosed by the loop of the curve


96/127

Find the area of the region enclosed by the loop of the curve

defined byx = t2 andy = t3-3t.(the same as that in Example 2

of the first part of this section ).

The point on the loop where the curve crosses itself is

Solution


97/127

The point on the loop where the curve crosses itself is

(3, 0), the corresponding parameter values are .

The area of the loop is obtained by subtracting the areaunder the bottom part of the loop from the area under the top

part of the loop.

3=t

tdtttdttA tt 2)3(2)3(3

0

33

0

3 =


98/127

35

24

5

44 33

2/52/3 ==

tt )()( 00

tdttt 2)3(3

3

3

=

dttt23

3

4 62 =

[ ]tt 3552 23

3

=

8.7 Polar Coordinates


99/127

A coordinate system represents a point in the plane by an

ordered pair of numbers called coordinates. So far wehave been using Cartesian coordinates, which are directed

distances from two perpendicular axes. Now we describe

a coordinate system called the polar coordinate system,

which is more convenient for many purposes.

We choose a point in the plane that is called the pole (or

origin) and labeled O. Then we draw a ray (half-line)

starting at O called the polar axis. This axis is usually

drawn horizontally to the right and corresponds to the

positivex-axis in Cartesian coordinates.

IfP is any other point in the plane, let r be the distance

f O P d l b h l ( ll d i


100/127

from O to Pand let be the angle (usually measured inradians) between the polar axis and the line OPas in the

figure. Then the pointPis represented by the ordered pair(r, ) and r, are calledpolar coordinates ofP.

We use the convention that an angle is positive if measured

i h l k i di i f h l i d


101/127

in the counterclockwise direction from the polar axis and

negative in the clockwise direction.

IfP= O, then r= 0 and we agree that (0, ) represents thepole for any value of .

O

x

We extend the meaning of polar coordinates (r, ) to thei hi h i ti b i th t th i t (


102/127

case in which ris negative by agreeing that the points (-r,

) and (r, ) lies on the same line through O and at the

same distance |r| from O, but on the opposite sides ofO.Ifr> 0, the point (r, ) lies in the same quadrant as ; ifr< 0, the point (r, ) lies in the quadrant on the oppositeside of the pole.


103/127

Notice that (-r, ) represents the same point as (r,

+ ).In fact, since a complete counterclockwise rotation is

given by an angle 2 ,the point represented by polar

coordinates (r, ) is also represented by (r, +2n )and (-r, +(2n+1) )

Example 1

Plot the points whose polar coordinates are given


104/127

p p g

(a) (1, 5 /4) (b) (2, 3 ) (c) (2, -2 /3) (d) (-3, 3 /4)

Solution

The points are plotted in the figure. In part (d) the point (-3,

3 /4)is located three units from the pole in the fourthquadrant because the angle 3 /4 is in the second

quadrant and r = -3 is negative.

The connection between polar and Cartesian coordinates

can be seen from the figure in which the pole corresponds


105/127

can be seen from the figure, in which the pole corresponds

to the origin and the polar axis coincides with the positive

x-axis. If the point P has Cartesian coordinates (x, y) andpolar coordinates (r, ), then

cos =x/r, sin =y/r

and so

x = rcos , y = rsin

Although the above equations

were deduced from the figure,

which illustrates the case

where r> 0 and 0 < < /2, these equations are valid for

all values of r and .

We can use the above formula to find the Cartesian

coordinates of a point hen the polar coordinates are


106/127

coordinates of a point when the polar coordinates are

known. We can also use the below equations to find rand

if the Cartesian coordinates of a point are known:r2 = x2 + y2 tan = y/x

Notice that the above equations do not uniquely determine whenx andy are given, because as increasesthrough the interval 0


107/127

2o

x

y

Example 2

Convert the point (2, /3) from polar to Cartesian


108/127

Convert the point (2, /3) from polar to Cartesiancoordinates.

3

2

32

3

sin2sin

12

1

23cos2cos

====

====

ry

rx

Solution

ince r= 2 and = /3,

s.coordinateCartesianin)3(1,ispointtheTherefore,

Example 3

Represent the point with Cartesian coordinates (1 1) in terms


109/127

Since the point (1,-1) lies in the fourth quadrant, we canchoose =- /4 or =7 /4. Thus one possible answer is(2, - /4). Another is (2,7 /4).

( )

1tan

2112222

==

=+=+=

x

y

yxr

Represent the point with Cartesian coordinates (1,-1) in terms

of polar coordinates.

Solution If we choose rto be positive, then


110/127

The Graph of a Polar Equation

The graph of a polar equation r = f( ), or moregenerallyF(r, )=0, consists of all pointsPthat have atleast one polar representation (r, ) whose coordinatessatisfy the equation.

Example 1

Wh t i t d b th l ti 2


111/127

What curve is represented by the polar equation r= 2.

Solution

The curve consists of all points (r, ) with r= 2. Since rrepresents the distance from the point to the pole, the curve r

= 2 represents the circle with centerO and radius 2. In

general, the equation r= a represents a circle with centerOradius |a|.

Solution This curve consists of all points (r ) such

Example 2 Sketch the polar curve = 1.


112/127

Solution This curve consists of all points (r, ) suchthat the polar angle is 1 radian. It is the straight line that

passes through O and makes an angle of 1 radian with thepolar axis.

Notice that the points (r, 1) on the line with r> 0 are in the

first quadrant, whereas those with r< 0 are in the third

quadrant.

( , +1)2

Example 3(a) Sketch the curve with polar equationr= 2cos . (b) Find a Cartesian equation for this curve.


113/127

Solution

(a) We find the values ofrfor

some convenient values of and plot the corresponding

points (r, ). Then we jointhese points to sketch the

curve, which appears to be a

circle. We have used only

values of between 0 and , since if we let increases beyond , weobtain the same points again.

(b) Multiply rto both sides of the equation r= 2cos :2 2 2 + 2 2 2 + 2 2 0


114/127

r2 = 2 rcos , x2 +y2 = 2x, x2 +y2 - 2x = 0

Completing the square, we obtain

(x-1)2 +y2 = 1

which is the equation of a circle with center (1,0) and

radius 1.

The figure below shows a geometrical illustration that thecircle has the equation r=2cos . The angle OPQ is a rightangle and so

r/2 = cos .

Example 4 Sketch the curve r= 1 + sin .Solution


115/127

First sketch the graph of r= 1 + sin in Cartesian coordinates by shifting thesine curve up one unit. This enables us to read at a glance the values ofrthat

correspond to increasing values of . We see that as increases from 0 to /2, rincrease from 1 to 2; as increases from /2 to , rdecrease from 2to 1; as increases from to 3 /2, rdecrease from 1 to 0; as increasesfrom 3 /2 to 2 , rincrease from 0 to 1. If we let increases beyond 2 ordecrease beyond 0, we would simply retrace our path. Then we sketch out the

complete curve as in the figure. It is called a cardioidbecause it is shaped likea heart.

.

Example 5Sketch the curve with polar equation r= cos2 .Solution


116/127

We first sketch r= cos2 , 0 < 2 , in Cartesian coordinates. As increases from 0 to /4, rdecrease from 1 to 0, and so we draw thecorresponding portion of the polar curve. As increases from /4 to /2, rdecrease from 0 to 1. This means that the distance from Oincreases from 0 to 1, but instead of being in the first quadrant, this

portion of the polar curve lies on the opposite side of the pole in the

third quadrant. The remainder of the curve is drawn in a similar fashion.The resulting curve has four loops and is called afour-leaved rose.

When sketching polar curves it is sometimes helpful to take

advantage of symmetry The following are three rules


117/127

advantage of symmetry. The following are three rules.

(a) If a polar equation is

unchanged when is replaced by - . The curve issymmetric about the polar

axis.

(b)If a polar equation is

unchanged when r is replaced

by -r. The curve is symmetric

about thepole.(c) If a polar equation is

unchanged when is replacedby - . The curve iss mmetric about the vertical

The curve sketched in Examples 3 and 5 are symmetric


118/127

The curve sketched in Examples 3 and 5 are symmetric

about the polar axis. The curves in Example 4 and 5 are

symmetric about = /2. The four-leaved rose is alsosymmetric about the pole.

These symmetry properties could be used in sketchingcurves. We only need to plot a part of the curve and then

apply the symmetry.

Tangents to polar curves

To find a tangent line to a polar curve r = f( ) we regard


119/127

To find a tangent line to a polar curve r f( ) we regard as a parameter and write its parametric equations

x=f( )cos ,y =f( )sin .

Then using the method for finding slopes ofparametric

curves we have

We locate horizontal tangents by finding the points whendy/d =0 (provided that dx/d 0).

We locate vertical tangents at the points when dx/d = 0(provided that dy/d 0).

sincos

cossin

rd

drrd

dr

d

dxd

dy

dx

dy

+==

Notice that if we are looking for tangent lines at the pole,

then r = 0 and the above equation simplifies to


120/127

then r 0 and the above equation simplifies to

if

For instance, in Example 5, we found that r= cos2 = 0when = /4 or 3 /4. This means that the lines = /4 and = 3 /4 or (y =x andy = -x are tangent linesto r= cos2 at the origin.

tan=dxdy 0d

dx

Example(a) For the cardioidr= 1 + sin , find the slope of the tangent line when

= /3


121/127

= /3.

(b) Find the points on the cardioid where the tangent line is horizontal or

vertical.

)sin21)(sin1(

)sin21(cos

sin1

)sin21(cos

sin)sin1(coscos

cos)sin1(sincos

sincos

cossin

sin22

++

=

+=

+++

=

+=

r

d

dr

rd

dr

dx

dy

or

Solution

Using the derived formula with r= 1 + sin , we have

Instead of memorizing the formula, we could employ the


122/127

Instead of memorizing the formula, we could employ the

method used to derive the formula:

2cossin

2sincos

2cossin

cossin2cos

sinsinsin)sin1(sin

2sin2

1coscos)sin1(cos

2

++=

++==

+=+==

+=+==

ddx

ddy

dx

dy

ry

rx

(a) The slope of the tangent at the point where = /3 is


123/127

(a) The slope of the tangent at the point where = /3 is

131

31

)31)(32(

31

)31)(231(

)31(21

))3

sin(21))(3

sin(1(

))3sin(21)(3cos(

3

=

+=

++

=+

+=

++=

=

dx

dy

(b) Observe that

6

11,

6

7,

2

3,

2,0)sin21(cos

==+= when

d

dy


124/127

Therefore, there are horizontal tangents at the points (2, /2), (1/2, 7 /6), (1/2, 11 /6) and vertical tangents at (3/2, /6). (3/2, 5 /6).When =3 /2, both and are 0, so we must be careful. Using

lHospitals Rule, we have

By symmetry,

Thus there is a vertical tangent line at the pole.

6

5,

6,

2

3,0)sin21)(sin1(

==+= when

d

dx

=

=

+=

cos

sin

lim3

1

sin1

coslim

3

1lim

)23(

)23()23( dx

dy

=+ dx

dylim

)23(

d

dy

d

dx

r= 1 + sin

Page558 17233 tt


125/127

g

33 1

3,

1

3

t

ty

t

tx

+=

+=

Solution

).,22(attangentverticalaand),,22(and(0,0)

attangenthorizontalhascurvethetherefore,

,21

)2(,

)1(

63,

)1(

)2(3

31

32

32

31

3

3

23

3

23

3

t

tt

dx

dy

t

t

dt

dx

t

tt

dt

dy

=+

=+

=

>

Intro. to Differential Equation

Documents