Chapter 8

Further Applications of Integration

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 order of a differential equation is the order of the 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 equations y is an unknown

function of x.

8.1 Differential Equations

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 the form y =Asinx+Bcos x, where both A and B are constants. So it is called the general solution of the differential equation.

A function f is called a solution of a differential equation if the equation is satisfied when y = f(x) and its derivatives are substituted into the equation. For example, f is a solution of equation y'= xy if

f '(x) = xf(x).

Any particular solutions are obtained by substituting values for the arbitrary constants A and B. For instance, y = sin x is a particular solution of the above differential equation by choosing A = 1,B = 0 in the general solution.

In general, solving a differential equation is not an easy matter.

• Separable equation

)(

)(

yh

xg

dx

dy

A separable equation is a first-order differential equation that can be written in the form dy/dx = g(x)f(y). The name separable comes from the fact that the expression on the right side can be “separated” into a function of x and a function of y.

Equivalently, we could write

so that all y’s are on one side of the equation and all x’s are on the other side. Then we integrate both sides of the equation:

dxxgdyyh )()(

To solve this equation we rewrite it in the differential form

h(y)dy = g(x)dx

It defines y implicitly as a function of x.

In some cases we may be able to solve for y in

terms of x

The justification of the above last step comes form the Substitution Rule:

dxdx

dyxyhdyyh ))(()(

dxxg

dxxyh

xgxyh

)(

))((

)())((

Solve the differential equationyy

x

dx

dy

cos2

6 2

Cxyy

dxxdyyy

dxxdyyy

32

2

2

2sin

6)cos2(

6)cos2(

SolutionWriting the equation in differential form and

integrating both sides, we have

Example 1

where C is an arbitrary constant. (We could have used

a constant C1 on the left side and another constant C2

on the right side, but then we could combine these

constants by writing C = C2 - C1

The above general solution is in implicit form. In this

case it is impossible to express y explicitly as a

function of x.

Solve the differential equation yxy 2'Solution

Rewrite the equation using Leibniz notation:

If y 0, we can rewrite it in differential notation and integrate:

dy/dx = x2y.

Example 2

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

Note that the function y = 0 is also a solution of the given differential equation. So the general solution is in the form 3

3x

Aey

In this case, we can solve explicitly for y:

333ln

333

,x

Cx

CC

xy eeyeeeey

where A is an arbitrary constant (A = eC or 0).

•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 the particular solution that satisfies a condition of the form y(x0) = y0. This is called an initial condition.

Example 1

Solve the differential equation .2)4(,0,' yxyxy

SolutionWrite 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 //

To determine K we put x = 4 and y = 2 in this equation:

2 = K/4 K = 8

So the solution of the initial-value problem is

y = 8/x , x > 0

Example 2

Find the solution of dy/dx = 6x2/(2y + cosy) that satisfies y(1) = .

Solution

From Example 1 in the last part, we know that the general solution is

y2 + sin y = 2x3 + C

Therefore, the solution is given implicitly by

y2 + sin y = 2x3+ 2 – 2

Example 3 Solve y' = 1 + y2 - 2x - 2xy2 , y(0) = 0, and graph the solution.

Substituting x = 0 and y = 0 in this equation, we get C = 0. So

Solution

Factor the right side as the product of a function of x

and a function of y:

tan-1y = x - x2

To graph this equation, notice that it is equivalent to y = tan(x - x2)

provided that -/2<x - x2 < /2. Solving these inequalities, we find that

211211 21

21 x

This enables us to graph the solution as in the following figure.

Example 4

A tank contains 20kg of salt dissolved in 5000L of water. Brine that contains 0.03kg of salt per liter of water 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?

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

after t minutes. We are given that y(0) = 20 and we want to find y(30). We do this by finding a differential equation satisfied by y(t). Note that dy/dt is the rate of change in the amount of salt, so

dy/dt = (rate in) – (rate out)where (rate in) is the rate at which salt enters

the tank and (rate out) is the rate at which salt leaves the tank.

We have

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

The tank always contains 5000L of liquid, so the concentration at time t is y(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 by integrating

dy/(150 - y) = dt/200

-ln|150 – y | = t/200 + C.

Since y(0) = 20, we have –ln130 = C, so

-ln|150 – y | = t/200 – ln130.

Therefore

Since y(t) is continuous and y(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 200

t

ey

.130150)( 200

t

ety

.kg 1.38130150)30( 200

30

ey

•Logistic growth

Under the conditions of unlimited environment and food supply, the rate of population growth is proportional to the size of the population. This can be described by the differential equation

dy/dt = ky

Solve the separable equation:

Where or 0 is an arbitrary constant.

kt

ktCCkt

Aey

eeey

Ckty

ykdtdyy

ln

0 ,1

CeA

In a restricted environment and with limited food supply, the population cannot exceed a maximal size M at which it consumes its entire food supply. If we make the assumption that the rate of growth of 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 the equation

dy/dt = ky (M-y)

where k is a constant. This equation is called the logistic differential equation.

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

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 is y(0) = y0, then A = 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

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.

•Direction fields

),(' yxFy

where F(x, y) is some expression in x and y. Even if 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

If a solution curve passes through a point (x, y), then its slope at that point is y', 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 (or slope field). These line segments indicate the direction in which a solution curve is heading, so the direction field helps us visualize the general shape of these curves.

(a) We start by computing the slope at several points as in the chart

Example

(a) Sketch the direction field for the differential

equation y' = x2+y2–1.

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

through the origin.Solution

(b) Now we draw short line segments with these slopes at these points. The result is the direction field shown in the figure (on the next slide).

(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.

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 line segments 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 solutions

of differential equations. This technique is called

Euler’s method.

we take a partition P of [a, b] determined by points xi with a = x0< x1<…< xn= b. If yi = f(xi), then the point Pi (xi , yi) lies on C and the polygon with vertices P0, P1, …, Pn is an approximation to C. The length of this polygonal approximation is .||

11 PP i

n

ii

.bxa

8.2 Arc Length• The definition of arc length

Suppose that a curve C is defined by the equation y = f(x),

where To obtain a polygonal approximation to C,

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

Therefore, we define the length L of the curve C with equation y = f(x), , as the limit of the lengths of these inscribed polygons (if the limit exists):

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 large number 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

The definition of arc length given above is not very convenient for computational purposes, but we can derive an integral formula for L in the case where f has a continuous derivative. [Such a function f is called smooth because a small change in x produces a small change in

f ' (x).]

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

By the Mean Value Theorem to f on the interval

[xi-1, xi], there is a number xi* between xi-1 and xi such

that

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

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

Thus we have

.|| )()(22

1 yxPP iiii

)()(||22

1 yxPP iiii

.1 )]('[

])('[)(

*

*

2

22

xxf

xxfx

ii

iii

Therefore,

n

iii

PPPL

11

0||||||lim

n

ii

Pxxf i

1

2

0||||)]('[ *1lim

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)( 2xfxg

If f ' is continuous on [a,b], then the length of the curve y = f(x), , is bxa

.1 )]('[2

b

a dxL xf

.1 )(2

b

a dxLdx

dy

•The arc length formula

Use the Leibniz notation for derivatives,

Example 1:

Find the length of the arc of the semicubical parabola y2 = x3 between the point (1,1) and (4,8).

Solution:

,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. When x = 1, u = 13/4; when x = 4, u = 10.

Therefore 4

1

4

1

2

4

91)(1 dxxdx

dx

dyL

.27

13131080

][27

8)

4

13(10

23

23

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

If a curve has the equation x = g(y), , then by the interchanging the roles of x and y in the formula, we obtain

.11 )()]('[2

2

d

c

d

c dydyLdy

dxyg

dyc

Example 2:

Find the length of the arc of the parabola y2 = x from (0,0) to (1,1).

Solution:

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

.11 1

0

21

0

2

4)( dydyL ydy

dx

Make the trigonometric substitution tan21y

ddy 2

21 sec sectan141 22 yand

When y = 0, = 0, so ; when , sotan 0

2 tan

say. ,2arctan

Thus

, which gives

aa ddL 032

0 sec2

1sec

2

1sec

.|tansec|lntansec4

12

1

2

1|tansec|lntansec 0

a

.4

)25ln(

2

5 L

Since , we have 2tan and 5sec so ,5tan1sec 22

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 difficult or even impossible to evaluate explicitly.

Thus we sometimes have to be content with finding an approximation to the length of a curve as in the following example.

Example 3

(a) Set up an integral for the length of the arc of the hyperbola xy = 1 from the point (1,1) to (2,1/2).

(b) Use Simpson’s 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

11 2

1 2

4

2

1 4

2

1

2

)( dxdxdxL

x

x

xdx

dy

(b) Using Simpson’s Rule with a=1, b=2, n=10, x=0.1,

We have

./11)( 4

xxf

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

•The arc length function

If a smooth curve C has the equation y = f(x), , let s(x) be the distance along C from the initial point P0 (a , f(a)) to the point Q(x , f(x)). Then s 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

a dtxs tf

.11 )()]('[2

2

dx

dyxf

dx

ds

bxa

The above equation shows that the rate of change of s with respect to x is always at least 1 and is equal to 1 when f ' (x), the slope of the curve, is 0.

The differential of arc length is

This equation is sometimes written in the symmetric form

.1 )(2

dxdsdx

dy

.)()()(222

dydxds The geometric interpretation of it is shown in the figure.

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 dxLdx

dy

Or we can solve to get

which gives

dydy

dxds )(1

2

.1 )(2

d

c dyLdy

dx

Example

Find the arc length function for the curve y = x2 – (lnx)/8 taking P0 (1 ,1) as the starting point.

Solution

Thus the arc length function is given by

.8

1211

8

12

8

12)]('[

22

2

xx

xx

xxxf

xxxf

8

12)('

xx dtt

tdttfxs 11

2 )8

12()]('[1)(

1ln8

1ln

8

1 2

1

2

xxtt

x

8.3 Area of a Surface of Revolution

A surface of revolution is formed when a curve is rotated 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.

•Some simple surfaces

(a) The lateral surface area of a circular cylinder with radius r and height h is taken to be A = 2rh because we can imagine cutting the cylinder and unrolling it to obtain a rectangle with dimensions 2r and h.

(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 l and central angle = 2r/l.

In general, the area of a sector of a circle with radius l and angle is

l 2

2

1

Therefore, we define the lateral surface area of a cone to be A = rl

rll

rllA

2

2

1

2

1 22

In this case, it is

(c) The area of the band (or frustum of a cone) with 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 orPutting them together, we get

or

where r = (½)(r1 + r2) is the average radius of the band.

lrlrrlrllrA 21121112 )(

r

ll

r

l

2

1

1

1

lrlrlr 11112 lrrr 112 )(

)( 21 lrlrA

rl A2

•The definition of surface area

Consider the surface obtained by rotating the curve y = f(x), , about the x-axis, where f is positive and has a continuous derivative.

Take a partition P of [a,b] determined by points xi with a = x0< x1<…< xn= b. Let yi = f(xi), then the point Pi (xi , yi) lies on the curve. The part of the surface between xi-1 and xi is approximated by taking the line segment Pi-1Pi and rotating it about the x-axis. The result is a band (a frustum of a cone) with slant height l = |Pi-1Pi | and average radius

r = (½)(yi-1 + yi), so its surface area is

As in section 8.2, we have

where xi* [xi-1, xi].

ii

ii PPyy

1

1

22

xxfPP iii i *' 2

1 1

bxa

When xi is small, we have yi = f(xi) f(xi*) and yi-1 = f(xi-1) f(xi

*), since f is 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

1 122

2

xxfxf i

n

ii i

1

2* *'12

dxxfxfxxfxf b

aii

n

ini )(1)(212lim '*' 22*

1

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 curve y = f(x), , about the x-axis as

With the Leibniz notation for derivatives,

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

dxxfxfS b

a )(1)(2 ' 2

dxdx

dyyS b

a

2

12

dyxS d

c dy

dx

2

12

bxa

dyc

Using the notation for arc length given in Section 8.2,

where

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

where

ydsS 2

xdsS 2

dxdx

dyds

2

1

dydy

dxds

2

1

Example 1:The curve is an arc of the circle .

Find the area of the surface obtained by rotating this arc about 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

xx

dx

dy

x

xxx

dx

dy2

2/1

4)2()4(

2

1 2

11,4 2 xxy 422 yx

Example 2

The arc of the parabola y = 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 2

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 3Find the area of the surface generated by rotating the curve

y =ex, , about the x-axis.

Solution y = ex so dy/dx = ex, we have

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

)12ln(2)tanln(sectansec

2

12

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

)1 2 ln( 2 ) 1 ln( 12 2

e e e e S

10 x

8.4 Application to Economics(1) Consumer surplus

The demand function p(x) is the price that a company has to charge in order to sell x units of a commodity.

•The demand function and demand curve

Usually, selling larger quantities requires lowering prices, so the demand function is a decreasing function. The graph of a typical demand function, called a demand curve is shown in the figure. If X is the amount of the commodity that is currently available, then P = p(X) is the current selling price.

• The consumer surplusPartition the interval [0, X] into n subintervals, each of length x = X/n, with xi be the right endpoint of the ith subinterval.

For the consumers between xi-1 and xi, the price they are willing to pay is about p(xi), the price they actually pay is P. Therefore, we can consider they have saved an amount of

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

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

The consumer surplus represents the amount of money saved by consumers by purchasing the commodity at price P, corresponding to an amount demanded of X.

The figure below shows the interpretation of the consumer surplus as the area under the demand curve and above the line p = P.

Example

The demand for a product, in dollars, is p = 1200 – 0.2x – 0.0001x2. Find the consumer surplus when the sales level is 500.

Solution

The corresponding price for X = 500 is

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

Therefore, the consumer surplus is

33.333,33$3

0001.01.0)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• Continuously compounded interest rate

With continuously compounded interest rate r, the value of a savings account y(t) increases at a rate proportional to that value, i.e.

Then at time t, the value of y is

(see Example of section 8.1).

• Present value of a future amount

If A0 is the amount that will grow to A in t years, then A0 ert = A and so A0 = Ae-rt . A0 is called the present value of A.

eyty rt)0()(

.rydtdy

• Present value of an income stream

Suppose that income will be received over a period of time from t = a to t = b at a rate of f(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 income received 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 the present value of the income stream f(t).

tttf i )(

ttfe itir )(

n

i

tir ttfe i1)(

n

ba

rt dttfe )(

Example:

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 is f(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

15

5

)1.0(15

5)1.0( ]

1.08000[)8000(

ee

tt dt

8.5 Curves Defined by Parametric Equations

Suppose that x and y are both given as continuous functions of a third variable t (called a parameter) by the parametric equations

x = f(t) y = g(t) .

Each value of t determines a point (x, y), which we can plot in a coordinate plane. As t varies, the point (x, y) = ( f(t), g(t) ) varies and traces out a curve C.

If we interpret t as 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:

Identify the curve defined by the parametric equations x = t2 – 2t and y = t +1.

equations moves along the curve in the direction of the arrows as t increases.

Eliminate the parameter t as follows:

From y = t +1 we obtain t = y –1. Substitute it into x = t2 – 2t, it gives

x = (y-1)2 – 2(y-1) = y2 – 4y +3

and so the curve represented by the given parametric equations is a parabola.

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

Solution:

Example 2:

What curve is represented by the parametric equations x = cos t and y = sin t , 0 t 2 ?

Solution:

Eliminate t by noting that x2 + y2 = cos2t + sin2t = 1Thus the point (x, y) moves on the unit circle x2 + y2 = 1. Notice that the parameter t can be interpreted as the angle shown in the figure.

As t increases from 0 to 2 , the point (x, y) = (cos t, sin t) moves once around the circle in the counterclockwise direction starting from the point (1,0) .

Example 3:

What curve is represented by the parametric equations x = sin2t and y = cos2t , 0 t 2 ?

Thus again the point (x, y) moves on the unit circle x2 + y2 = 1.

But as t increases 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 t by noting that x2 + y2 = sin22t + cos22t = 1

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 -1x1, so the parametric equations represent only the part of the parabola for which -1 x 1.

Since sin t is periodic, the point (x, y)=(sin t, sin2t ) moves back and forth infinitely often along the parabola from (-1,1) to (1,1).

Example 4

Sketch the curve represented by the parametric equations x = sint and y = sin2t.

Solution

Example 5:The curve traced out by a point P on the circumference of a circle as the circle rolls along a straight line is called a cycloid. If the circle has radius r and rolls along the x axis and if one of the positions of P is the origin, find parametric equations for the cycloid.

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

Let the coordinates of P be (x, y), then

x = |OT| - |PQ| = r - r sin = r( - sin)

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

This is also valid for other values of ( try it by yourself ! ).

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

parameter ( =0 when P is at the origin). For 0< < /2, the distance it

So the parametric equations of the cycloid are

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 of cycloid:

(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.

By eliminating the parameter, some curves defined by parametric equations x = f(t) and y = g(t) can be expressed in the form y = F(x). If we substitute x = f(t) and y = g(t) in the equation y = F(x), we get

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

If g, F and f are differentiable, the Chain Rule gives

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

If f ' (t)≠ 0, we can solve for F ' (x):

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

Since the slope of the tangent to the curve y = F(x) at (x, F(x)) is F ' (x), the above equation enables us to find tangent to parametric curves without having to eliminate the parameter.

8.6 Tangents and Areas

(1) Tangents

Using Leibniz notation, we can rewrite the above equation as

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). This information is useful when sketching parametric curves.

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

dt

dxdt

dy

dx

dy 0dt

dx

dt

dxdx

dy

dt

d

dx

dy

dx

d

xd

yd

2

2

Example 1(a) Find dy/dx and d2y/dx2 for the cycloid x = 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

r

r

d

dxd

dy

dx

dy

cos1

11cossinsin)cos1(cos

cos1

sin

)cos1()cos1(22

d

d

dx

dy

d

d

Solution

(a)

)cos1(

1

)cos1(cos1

1

22

2

rr

d

dxdx

dy

d

d

xd

yd

(b) When = / 3, we have

and

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

or

2

3

33sin

3

rrx

23cos1

rry

/

/

)/cos(

)/sin(

dx

dy

2

3

33

2

rrx

ry

2

33

ryx

(c) The tangent is horizontal when dy/dx = 0, which occurs when sin = 0 and 1-cos0, that is, = (2n-1), where n is an integer. The corresponding point on the cycloid is ((2n-1)r, 2r).

When = 2n, both dy/d and dx/d are 0. There are vertical tangents at these points. We can verify this by using l’Hospital’s Rule:

sin

coslim

cos1

sinlimlim

222 nnn dx

dy

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

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

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

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

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

Example 2 A curve C is defined by x = t2 and y = t3-3t.

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

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

(c) Determine where the curve rises or falls and where it is concave upward or downward.

(d) Sketch the curve.

If the curve is given by parametric equations x = f(t) and y = g(t), t , then we can adapt the earlier formula by using

the Substitution Rule for Definite Integrals as follows:

b

a dxxFA )(

b

a dttftgydxA )(')(

dttftgor )(')(

))()((or

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

dttftg

dttftgdttftfFdxxFAtfx

b

a

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

where F(x) 0.

Example 1

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

.20

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

Solution

One arch of the cycloid is given by

rr22 32

23

r ydxA 2

0

2

0 )cos1()cos1( drr

2

0

22 )cos1( dr

2

0

2 )2coscos21( dr

2

0

2 )]2cos1(21cos21[ dr

4sinsin2 41

23

2

0

2

r

Example 2

Find the area of the region enclosed by the loop of the curve defined by x = t2 and y = 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 (3, 0), the corresponding parameter values are .

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

Solution

3t

35

24

5

44 33

2/52/3

tdtttdttA tt 2)3(2)3( 3

0

33

0

3

tdttt 2)3(3

3

3

dttt23

3

4 62

tt 3552 2

3

3

8.7 Polar CoordinatesA coordinate system represents a point in the plane by an ordered pair of numbers called coordinates. So far we have 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 positive x-axis in Cartesian coordinates.

If P is any other point in the plane, let r be the distance from O to P and let be the angle (usually measured in radians) between the polar axis and the line OP as in the figure. Then the point P is represented by the ordered pair (r, ) and r, are called polar coordinates of P.

We use the convention that an angle is positive if measured in the counterclockwise direction from the polar axis and negative in the clockwise direction.

If P = O, then r = 0 and we agree that (0, ) represents the pole for any value of .

Ox

We extend the meaning of polar coordinates (r, ) to the case in which r is 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 of O.

If r > 0, the point (r, ) lies in the same quadrant as ; if r < 0, the point (r, ) lies in the quadrant on the opposite side of the pole.

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

(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 fourth quadrant 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 to the origin and the polar axis coincides with the positive x-axis. If the point P has Cartesian coordinates (x, y) and polar 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 when the polar coordinates are known. We can also use the below equations to find r and if the Cartesian coordinates of a point are known:

r2 = x2 + y2 tan= y/x

Notice that the above equations do not uniquely determine when x and y are given, because as increases through

the interval 0<2, each value of tan occurs twice.

See the next page

Therefore, in converting from Cartesian to polar coordinates, it is not good enough just to find r and that satisfy the equations, we must choose so that the point (r, ) lies in the correct quadrant.

2o

x

yxy tan

Example 2

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

32

32

3sin2sin

12

12

3cos2cos

ry

rx

Solution

Since r = 2 and = /3,

s.coordinateCartesian in )3(1, ispoint theTherefore,

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

1tan

2112222

x

y

yxr

Example 3

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

Solution If we choose r to be positive, then

•The Graph of a Polar Equation

The graph of a polar equation r = f(), or more generally F(r,)=0, consists of all points P that have at least one polar representation (r, ) whose coordinates satisfy the equation.

Example 1

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

Solution

The curve consists of all points (r, ) with r = 2. Since r represents the distance from the point to the pole, the curve r = 2 represents the circle with center O and radius 2. In general, the equation r = a represents a circle with center O radius |a|.

Solution This curve consists of all points (r, ) such that the polar angle is 1 radian. It is the straight line that passes through O and makes an angle of 1 radian with the polar axis.

Example 2 Sketch the polar curve = 1.

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 equation

r = 2cos. (b) Find a Cartesian equation for this curve.

Solution

(a) We find the values of r for some convenient values of and plot the corresponding points (r, ). Then we join these 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 , we obtain the same points again.

(b) Multiply r to both sides of the equation r = 2cos :

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 the circle has the equation r =2cos . The angle OPQ is a right angle and so

r/2 = cos.

Example 4 Sketch the curve r = 1 + sin .

Solution

First sketch the graph of r = 1 + sin in Cartesian coordinates by shifting the sine curve up one unit. This enables us to read at a glance the values of r that correspond to increasing values of . We see that as increases from 0 to /2, r increase from 1 to 2; as increases from /2 to , r decrease from 2 to 1; as increases from to 3/2, r decrease from 1 to 0; as increases from 3/2 to 2, r increase from 0 to 1. If we let increases beyond 2 or decrease beyond 0, we would simply retrace our path. Then we sketch out the complete curve as in the figure. It is called a cardioid because it is shaped like a heart.

.

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

Solution

We first sketch r = cos2 , 0 < 2, in Cartesian coordinates. As increases from 0 to /4, r decrease from 1 to 0, and so we draw the corresponding portion of the polar curve. As increases from /4 to /2, r decrease from 0 to –1. This means that the distance from O increases 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 a four-leaved rose.

When sketching polar curves it is sometimes helpful to take advantage of symmetry. The following are three rules. (a) If a polar equation is

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

(b) If a polar equation is unchanged when r is replaced by -r. The curve is symmetric about the pole.

(c) If a polar equation is unchanged when is replaced by -. The curve is symmetric about the vertical line =/2.

The curve sketched in Examples 3 and 5 are symmetric about the polar axis. The curves in Example 4 and 5 are symmetric about = π/2. The four-leaved rose is also symmetric about the pole.

These symmetry properties could be used in sketching curves. 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 as a parameter and write its parametric equations

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

Then using the method for finding slopes of parametric curves we have

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

We locate vertical tangents at the points when dx/d= 0 (provided that dy/d0).

sincos

cossin

rd

dr

rd

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

if

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

tandx

dy 0d

dx

Example

(a) For the cardioid r = 1 + sin, find the slope of the tangent line when = /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

rd

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 method used to derive the formula:

2cossin

2sincos

2cossin

cossin2cossinsinsin)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

131

31

)31)(32(

31

)31)(231(

)31(2

1

))3sin(21))(3sin(1(

))3sin(21)(3cos(

3

dx

dy

(b) Observe that

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 l’Hospital’s Rule, we have

By symmetry,

Thus there is a vertical tangent line at the pole.

6

5,

6,

2

3,0)sin21)(sin1(

6

11,

6

7,

2

3,

2,0)sin21(cos

whend

dx

whend

dy

cos

sinlim

3

1

sin1

coslim

3

1lim

)23(

)23()23( dx

dy

dx

dylim

)23(

d

dy

d

dx

r = 1 + sin

Page558 173

2

3 1

3 ,

1

3

t

ty

t

tx

Solution

).,22(at tangent verticala and ),,22( and (0,0)

at tangent horizontal has curve thetherefore,

,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

3

1

31

2 ,0

2 ,0

)21(3

)1(2

)1(

)21(3)21(

)1(2

33

43

23

3

23

23

2

2

t

t

t

t

t

tt

t

dt

dxdx

dy

dt

d

dx

yd

x

y

)23,

23(

)2,2( 21

23

)2,2( 23

21

0

1t

t

1t

t t

+ + – +

– – – –

x

y

– + + –

curve

dtdx

dtdy

2

2

dxyd

)1,( )2,1( 32

),2( 32

),1(