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2 Newton’s Method for Solving Equations

2 Newton’s Method for Solving Equations

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24.2 Newton’s Method for Solving Equations



■ Newton’s method is another mathematical

development by the English mathematician and

physicist Isaac Newton (1642–1727).



y



y 5 f(x)



(x 1, f (x1))



r

0



x2 x1

Fig. 24.8



x



715



We have shown that the roots of an equation can be found with great accuracy on a

graphing calculator. In this section, we develop Newton’s method, which uses the derivative to find approximately, but also with great accuracy, the real roots of many

kinds of algebraic and nonalgebraic equations.

Newton’s method is an iterative method, which starts with a reasonable guess for

the root, and then yields a new and better approximation. This, in turn, is used to obtain

an even better approximation, and so on until an approximate answer with the required

degree of accuracy is obtained. Iterative methods are easily programmable for use on a

computer.

Let us consider a section of the curve of y = f1x2 that (a) crosses the x-axis,

(b) always has either a positive slope or a negative slope, and (c) has a slope that either

becomes greater or becomes less as x increases. See Fig. 24.8. (When either (b) or (c) is

not satisfied, Newton’s method may fail to find the root. See Exercises 20 and 23.) The

curve in the figure crosses the x-axis at x = r, which means that x = r is a root of the

equation f1x2 = 0. If x1 is sufficiently close to r, a line tangent to the curve at

3 x1, f1x1 2 4 will cross the x-axis at a point 1x2, 02, with x2 closer to r than x1.

We know that the slope of the tangent line is the value of the derivative at x1, or

mtan = f′ 1x1 2. Therefore, the equation of the tangent line is

y - f1x1 2 = f′1x1 2 1x - x1 2



For the point 1x2, 02 on this line, we have

Solving for x2, we have



-f1x1 2 = f′1x1 2 1x2 - x1 2

x2 = x1 -



f1x1 2

f ′ 1x1 2



Here, x2 is a second approximation to the root. We can repeat the process starting with

x2 in place of x1, obtaining an even better approximation x3. The repetition of this process is what we call Newton’s method, which we now summarize.

/FXUPOT.FUIPE

Let x = r be a root of the equation f1x2 = 0, and let x1 be a first approximation

of r. We obtain approximations x2, x3, . . . by using

xn+1 = xn -



f1xn 2

f′1xn 2



(24.1)



for n = 1, 2, . . . . The number of iterations depends on the required accuracy.

The initial approximation x1 (which must be close to r) may be found by either of

the following procedures:

1. Choose x1 inside an interval 1 a, b 2 where f 1 a 2 and f 1 b 2 have opposite

signs, with x1 closer to the endpoint where the function is closer to zero.

2. Sketch the graph of the function and choose x1 as an estimate of the x-intercept of the function. For some equations, it may be easier to choose x1 as an

estimate of the intersection of two functions.

E X A M P L E 1 6TJOH/FXUPOTNFUIPE



Find the root of x2 - 3x + 1 = 0 between x = 0 and x = 1.

Here, f1x2 = x2 - 3x + 1, f102 = 1, and f112 = -1. This indicates that the

root may be near the middle of the interval. Therefore, we choose x1 = 0.5.

The derivative is

f′1x2 = 2x - 3



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CHAPTER 24 Applications of the Derivative



Therefore, f10.52 = -0.25 and f′10.52 = -2, which gives us

x2 = 0.5 x3



r

x



x 1 5 0.5



This is a second approximation, which is closer to the actual value of the root.

See Fig. 24.9. A second iteration using x2 = 0.375, f10.3752 = 0.015 625, and

f′10.3752 = -2.25 gives us

x3 = 0.375 -



x 2 5 0.375

(0.5, 20.25)



-0.25

= 0.375

-2



0.015 625

= 0.381 944 4

-2.25



We can check this particular result by using the quadratic formula. This tells us the

root is x = 0.381 966 0. Our result using Newton’s method is good to three decimal

places, and additional accuracy may be obtained by using the method again as many

times as needed.



E X A M P L E 2 Newton’s method—application



Fig. 24.9

Practice Exercise



1. In Example 1, let x1 = 0.3, and find x2.



A spherical water-storage tank holds 500.0 m3. If the outside diameter is 10.0000 m,

what is the thickness of the metal of which the tank is made?

Let x = the thickness of the metal. We know that the outside radius of the tank

is 5.0000 m. Therefore, using the formula for the volume of a sphere, we have



■ An explanation and an example of Newton’s



4p

15.0000 - x2 3 = 500.0

3

125.0 - 75.00x + 15.00x2 - x3 = 119.366

x3 - 15.00x2 + 75.00x - 5.634 = 0

f1x2 = x3 - 15.00x2 + 75.00x - 5.634

f′1x2 = 3x2 - 30.00x + 75.00



method due to Thomas Simpson (of Simpson’s

rule) and dating back to 1740 can be found in

the text’s companion website.



Since f102 = -5.634 and f10.12 = 1.717, the root may be closer to 0.1 than to

0.0. Therefore, we let x1 = 0.07. Setting up a table, we have these values:

n



y = V4x − 1



1

1



2



−1

Fig. 24.10



f′1xn 2



xn -



f1xn 2

f′ 1xn 2



1



0.07



-0.457 157



72.9147



0.076 269 750 8



2



0.076 269 750 8



-0.000 581 145



72.729 358 7



0.076 277 741 3



E X A M P L E 3 Graphically locating x1



y = x2 − 1



2



−1



f1xn 2



Since x2 = x3 = 0.0763 to four decimal places, the thickness is 0.0763 m. This

means the inside radius of the tank is 4.9237 m, and this value gives an inside volume of 500.0 m3. In using a calculator, the values in the table are more easily found

if the values of xn, f1xn 2, and f′1xn 2 are stored in memory for each step.





y

3



xn



x



Solve the equation x2 - 1 = 14x - 1.

We can see approximately where the root is by sketching the graphs of

y1 = x2 - 1 and y2 = 14x - 1, as shown in Fig. 24.10. We see that they intersect

between x = 1 and x = 2. Therefore, we choose x1 = 1.5. With

f1x2 = x2 - 1 - 14x - 1

2

f′ 1x2 = 2x 14x - 1



we now find the values in the following table:



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24.2 Newton’s Method for Solving Equations



f1xn 2



f′1xn 2



xn -



717



f1xn 2

f′1xn 2



n



xn



1

2



1.5

1.968 313 4



-0.986 067 98 2.105 572 8 1.968 313 4

-0.252 568 59 3.173 759 8 1.888 733 2



3



1.888 733 2



4



1.886 379 4



-0.007 052 69 2.996 295 7 1.886 379 4

-0.000 006 20 2.991 026 5 1.886 377 3



Since x5 = x4 = 1.886 38 to five decimal places, this is the required solution.

(Here, rounded-off values of xn are shown, although additional digits were carried

and used.)





E XE RC IS ES 2 4 .2

In Exercises 1–4, find the indicated roots of the given quadratic equations by finding x3 from Newton’s method. Compare this root with

that obtained by using the quadratic formula.



22. Use Newton’s method on f1x2 = x1>3 with x1 = 1. Calculate

x2, x3, and x4. What is happening as successive approximations

are calculated?



1. In Example 1, change the middle term from -3x to -5x and use

the same x1.



23. To calculate reciprocals without dividing, a computer programmer applied Newton’s method to the equation 1>x - a = 0.

Show that x2 = 2x1 - ax21. From this, determine the expression

for xn.



2. 2x2 - x - 2 = 0

2



(between 1 and 2)



3. 3x - 5x - 1 = 0



(between - 1 and 0)



4. x2 + 4x + 2 = 0



(between - 4 and -3)



In Exercises 5–16, find the indicated roots of the given equations to at

least four decimal places by using Newton’s method.

3



2



5. x - 6x + 10x - 4 = 0



(between 0 and 1)



6. x3 - 3x2 - 2x + 3 = 0



(between 0 and 1)



7. x3 + 6x2 + 9x + 2 = 0



(the smallest root)



8. 2x3 + 2x2 - 11x + 3 = 0

4



3



(the largest root)



2



9. x - x - 3x - x - 4 = 0



(between 2 and 3)



10. 2x4 - 2x3 - 5x2 - x - 3 = 0

4



3



11. x - 2x - 8x - 16 = 0



(between -2 and -1)



(the negative root)



12. 3x4 - 3x3 - 11x2 - x - 4 = 0

13. 2x2 = 12x + 1



14. x3 = 1x + 1



15. x =



1

1x + 2



16. x3>2 =



1

2x + 1



24. The altitude h (in m) of a rocket is given by

h = - 2t 3 + 84t 2 + 480t + 10, where t is the time (in s) of

flight. When does the rocket hit the ground?

25. A solid sphere of specific gravity s sinks in water to a depth h (in

cm) given by 0.009 26h3 - 0.0833h2 + s = 0. Find h for

s = 0.786 (when the diameter of the sphere is 6.00 cm).

26. A dome in the shape of a spherical segment is to be placed over

the top of a sports stadium. If the radius r of the dome is to be

60.0 m and the volume V within the dome is 180 000 m3, find the

height h of the dome. See Fig. 24.11. 1 V = 16ph1h2 + 3r 2 2. 2

h



(the negative root)



(the positive real solution)

(the real solution)



(the real solution)

(the real solution)



In Exercises 17–31, determine the required values to at least 4 decimal

places by using Newton’s method.

17. Find all the real roots of x3 - 2x2 - 5x + 4 = 0.

18. Find all the real roots of x4 - 2x3 + 3x2 + x - 7 = 0.

3

19. Explain how to find 24 by using Newton’s method.



r

Fig. 24.11



27. The capacitances (in mF) of three capacitors in series are C,

C + 1.00, and C + 2.00. If their combined capacitance is

1.00 mF, their individual values can be found by solving the

equation

1

1

1

+

+

= 1.00

C

C + 1.00

C + 2.00

Find these capacitances.

28. An oil-storage tank has the shape of a right circular cylinder with

a hemisphere at each end. See Fig. 24.12. If the volume of the

tank is 50.0 m3 and the length l is 4.00 m, find the radius r.



20. Explain why Newton’s method does not work for finding the root

of x3 - 3x = 5 if x1 is chosen as 1.

21. Use Newton’s method to find an expression for xn + 1, in terms of

xn and a, for the equation x2 - a = 0. Such an equation can be

used to find 1a.



r



r

Fig. 24.12



l ϭ 4.00 m



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CHAPTER 24 Applications of the Derivative



29. A rectangular block of plastic with edges 2.00 cm, 2.00 cm, and

4.00 cm is heated until its volume doubles. By how much does

each edge increase if each increases by the same amount?



31. The vibration analysis of a three-storey building requires that the

following equation be solved:



30. Water flow in a wide channel approaches a bump. See Fig. 24.13.

The water depth x (in m) over the bump is the largest root of the

equation



Find the root that lies between 0 and 4 * 10-4 to 3 significant

digits.



l3 - 0.0012l2 + 9 * 10-7l - 7.2 * 10-11 = 0.



x3 - 1.015x2 + 0.115 = 0.

Find x if it is between 0.750 m and 1.20 m.



Answer to Practice Exercise



1. x2 = 0.379167

x

Bump

Fig. 24.13



24.3 Curvilinear Motion

7FDUPSTBOE$VSWJMJOFBS.PUJPO t

1BSBNFUFS t 1BSBNFUSJD'PSN t

Acceleration



When velocity was introduced in Section 23.4, the discussion was limited to rectilinear

motion, or motion along a straight line. A more general discussion of velocity is necessary when we discuss the motion of an object in a plane. There are many important applications of motion in a plane, a principal one being the motion of a projectile.

An important concept in developing this topic is that of a vector. The necessary fundamentals related to vectors are taken up in Chapter 9. Although vectors can be used to

represent many physical quantities, we will restrict our attention to their use in describing the velocity and acceleration of an object moving in a plane along a specified path.

Such motion is called curvilinear motion.

In describing an object undergoing curvilinear motion, it is common to express the

x- and y-coordinates of its position separately as functions of time. Equations given in

this form—that is, x and y both given in terms of a third variable (in this case, t)—are

said to be in parametric form, which we encountered in Section 10.6. The third variable, t, is called the parameter. Each value of the parameter t determines a point

(x(t), y(t)), and the set of all such points determines the graph of the curve. The graph

can be sketched by plotting enough points to indicate its shape. In some cases, we can

eliminate the parameter t to obtain an equation in x and y, which we can then graph.

To find the velocity of an object whose coordinates are given in parametric form, we

find its x-component of velocity vx by determining dx>dt and its y-component of velocity vy by determining dy>dt. These are then evaluated, and the resultant velocity is

found from v = 2v 2x + v 2y . The direction in which the object is moving is found from

tan u = vy >vx.

E X A M P L E 1 Parametric form—resultant velocity



If the horizontal distance x that an object has moved is given by x = 3t 2 and the vertical distance y is given by y = 1 - t 2, find the resultant velocity when t = 2.

To find the resultant velocity, we must find v and u, by first finding vx and vy.

After the derivatives are found, they are evaluated for t = 2. Therefore,

dx

vx ͉ t = 2 = 12

= 6t

dt

dy

vy =

= -2t

vy ͉ t = 2 = -4

dt

v = 2122 + 1 -42 2 = 12.6



vx =



tan u =



-4

12



u = -18.4°



find velocity components



magnitude of velocity

direction of motion



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24.3 Curvilinear Motion



To sketch the graph of the path, we can eliminate the parameter t by substituting

t 2 = x>3 (from the equation for x) into the equation for y. Therefore, y = 1 - 3x ,

and the path of the object is a line. The path and velocity vectors are shown in

Fig. 24.14.





y

5

5



10



15

x



0



u



vx

v



vy



−5



719



E X A M P L E 2 Parametric form—resultant velocity



Find the velocity and direction of motion when t = 2 of an object moving such that its

x- and y-coordinates of position are given by x = 1 + 2t and y = t 2 - 3t.



Fig. 24.14



dx

= 2

dt

dy

vy =

= 2t - 3

dt

v ͉ t = 2 = 222 + 12 = 2.24



vx ͉ t = 2 = 2



vx =



y



vy ͉ t = 2 = 1

magnitude of velocity



1

2



tan u =



10



find velocity components



u = 26.6°



direction of motion



We graph the path by plotting the following points:



5

5



10

x



v



0

−2



t

x

y



0

1

0



1

3

–2



2

5

–2



3

7

0



u



The path and the velocity are shown in Fig. 24.15.



Fig. 24.15



COMMON ERROR







Note that to find the resultant velocity, we first find the necessary derivatives and then

evaluate them. This procedure should always be followed. When a derivative is to be

found, it is incorrect to take the derivative of the evaluated expression (which is a

constant).



Acceleration is the time rate of change of velocity. Therefore, if the velocity, or its

components, is known as a function of time, the acceleration of an object can be found

by taking the derivative of the velocity with respect to time. If the displacement is

known, the acceleration is found by finding the second derivative with respect to time.

Finding the acceleration of an object is illustrated in the following example.

E X A M P L E 3 Parametric form—resultant acceleration



Find the magnitude and direction of the acceleration when t = 2 for an object that is

moving such that its x- and y-coordinates of position are given by the parametric equations x = t 3 and y = 1 - t 2.



y

2

0



vx =

4



8



12



−2

−4



16



x



ay



Fig. 24.16



a



ax =



dvx

d 2x

= 2 = 6t

dt

dt



ax ͉ t = 2 = 12



take second derivatives to find

acceleration components



ax

u



dx

= 3t 2

dt



vy =



dy

= -2t

dt



ay =



dvy

dt



a ͉ t = 2 = 2122 + 1 -22 2 = 12.2

tan u =



ay

ax



= -



2

12



=



u = -9.5°



d 2y

dt 2



= -2



ay ͉ t = 2 = -2



magnitude of acceleration

direction of acceleration



The quadrant in which u lies is determined from the fact that ay is negative and ax is

positive. Thus, u must be a fourth-quadrant angle (see Fig. 24.16).





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CHAPTER 24 Applications of the Derivative



Practice Exercise



1. In Example 3, solve for the acceleration

when t = 2, if x = 0.8t 5>2, instead of

x = t 3.



We now summarize the equations used to find the velocity and acceleration of an

object for which the displacement is a function of time. They indicate how to find the

components, as well as the magnitude and direction, of each.



vx =



dx

dt



vy =



ax =



dvx

d 2x

= 2

dt

dt



ay =



v = 2v 2x + v 2y



tan uv =



COMMON ERROR



■ For reference, Eq. (23.15) rewritten for u as a

function of t is

dun

du

= nun - 1 .

dt

dt



vy

vx



dy

dt

dvy

dt



=



d 2y

dt



2



a = 2a2x + a2y



tan ua =



ay

ax



velocity

components



(24.2)



acceleration

components



(24.3)



magnitude



(24.4)



direction



(24.5)



If the curvilinear path an object follows is given with y as a function of x, the velocity (or

acceleration) is found by taking derivatives of each term of the equation with respect to

time. It is assumed that both x and y are functions of time, although these functions

are not stated. When finding derivatives, we must be careful in using the power rule,

Eq. (23.15), so that the factor du>dt is not neglected.



In the following examples, we illustrate the use of Eqs. (24.2) to (24.5) in applied

situations for which we know the equation of the path of the motion. Again, we must be

careful to find the direction of the vector as well as its magnitude in order to have a

complete solution.

E X A M P L E 4 Velocity at a point along a path



y

5



vy



v



u



vx

x



3



0



−3



vy =



y = 120 − 0.0151x 2



u

vy



v

x



0

Fig. 24.18



dx2

dx

= 2x

dt

dt

using Eq. (24.2)

substituting

magnitude (Eq. 24.4)

direction (Eq. 24.5)







E X A M P L E 5 Velocity and acceleration—application to projectile motion



A helicopter is flying at 18.0 m>s and at an altitude of 120 m when a rescue marker is

released from it. The marker maintains a horizontal velocity and follows a path given

by y = 120 - 0.0151x2, as shown in Fig. 24.18. Find the magnitude and direction of

the velocity and of the acceleration of the marker 3.00 s after release.



18.0 m /s

120 m



dy

1

dx

= a2x b

dt

3

dt



2

xv

3 x

2

vy = 12.002 16.002 = 8.00 cm>s

3

v = 26.002 + 8.002 = 10.0 cm>s

8.00

tan u =

,

u = 53.1°

6.00



Fig. 24.17



y



In a physics experiment, a small sphere is constrained to move along a parabolic path

described by y = 13 x2. If the horizontal velocity vx is constant at 6.00 cm>s, find the

velocity at the point 12.00, 1.332. See Fig. 24.17.

Since both y and x change with time, both can be considered to be functions of

time. Therefore, we can take derivatives of y = 13 x2 with respect to time.



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24.3 Curvilinear Motion



721



From the given information, we know that vx = dx>dt = 18.0 m>s. Taking derivatives with respect to time leads to this solution:

= 120 - 0.0151x2

dx

= -0.0302x

dt

= -0.0302xvx

= 13.002 118.02 = 54.0 m

= -0.0302154.02 118.02 = -29.35 m>s

= 218.02 + 1 -29.352 2 = 34.4 m>s

-29.35

tan u =

,

u = -58.5°

18.0

y

dy

dt

vy

x

vy

v



taking derivatives

using Eq. (24.2)

evaluating at t = 3.00 s

magnitude

direction



The velocity is 34.4 m>s and is directed at an angle of 58.5° below the horizontal.

To find the acceleration, we return to the equation vy = -0.0302xvx. Since vx is

constant, we can substitute 18.0 for vx to get

vy = -0.5436x

Again taking derivatives with respect to time, we have

dvy



dx

dt

dt

ay = -0.5436vx

ay = -0.5436118.02 = -9.78 m>s2

= -0.5436



using Eqs. (24.3) and (24.2)

evaluating



We know that vx is constant, which means that ax = 0. Therefore, the acceleration is

9.78 m>s2 and is directed vertically downward.





E XE RC IS ES 2 4 .3

In Exercises 1 and 2, make the given changes in the indicated examples of this section and then solve the resulting problems.

1. In Example 1, change x = 3t 2 to x = 4t 2.

2. In Example 4, change y = x2 >3 to y = x2 >4 and (2.00, 1.33) to

(2.00, 1.00).



In Exercises 3–6, given that the x- and y-coordinates of a moving particle are given by the indicated parametric equations, find the magnitude

and direction of the velocity for the specific value of t to 3 significant

digits. Sketch the curves and show the velocity and its components.

3. x = 3t, y = 1 - t, t = 4

5t

4. x =

, y = 0.11t 2 + t2, t = 2

2t + 1

5. x = t12t + 12 2, y =



6

, t = 0.5

14t + 3

2



6. x = 11 + 2t, y = t - t , t = 4



In Exercises 7–10, use the parametric equations and values of t of

Exercises 3–6 to find the magnitude and direction of the acceleration

in each case.



In Exercises 11–28, find the indicated velocities and accelerations.

11. The water from a fire hose follows a path described by

y = 2.0 + 0.80x - 0.20x2 (units are in metres). If vx is constant

at 5.0 m>s, find the resultant velocity at the point (4.0, 2.0).

12. A roller mechanism follows a path described by y = 14x + 1,

where units are in metres. If vx = 2x, find the resultant velocity

(in m>s) at the point 12.0, 3.02.



13. A float is used to test the flow pattern of a stream. It follows a

path described by x = 0.20t 2, y = - 0.10t 3 (x and y in m, t in

min). Find the acceleration of the float after 2.0 min.



14. A radio-controlled model car is operated in a parking lot. The coordinates (in m) of the car are given by x = 3.5 + 2.0t 2 and

y = 8.5 + 0.25t 3, where t is the time (in s). Find the acceleration

of the car after 2.5 s.

15. An astronaut on Mars drives a golf ball that moves according to

the equations x = 25t and y = 15t - 3.7t 2 (x and y in metres, t

in seconds). Find the resultant velocity and acceleration of the

golf ball for t = 6.0 s.

16. A package of relief supplies is dropped and moves according to

the parametric equations x = 45t and y = - 4.9t 2 (x and y in m, t

in s). Find the velocity and acceleration when t = 3.0 s.



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722



CHAPTER 24 Applications of the Derivative



17. A spacecraft moves along a path described by the parametric

equations x = 10 1 21 + t 4 - 1 2 , y = 40t 3>2 for the first 100 s

after launch. Here, x and y are measured in metres, and t is measured in seconds. Find the magnitude and direction of the velocity

of the spacecraft 10.0 s and 100 s after launch.

18. An electron moves in an electric field according to the equations

x = 8.0> 21 + t 2 and y = 8.0t> 21 + t 2 (x and y in Mm and t

in s). Find the velocity of the electron when t = 0.50 s.

19. In a computer game, an airplane starts at 11.00, 4.002 (in cm) on

the curve y = 3.00 + x -1.50 and moves with a constant horizontal

velocity of 1.20 cm>s. What is the plane’s velocity after 0.500 s?

20. In an aerobic exercise machine, weights are lifted and a person’s

hands are constrained to move along arcs of the ellipse

16x2 + 9y2 = 9 (in m). If the person’s hands move upward at

0.100 m>s, and start at y = 0, at what velocity is each moving

after 1.50 s?

21. Find the resultant acceleration of the spacecraft in Exercise 17 for

the specified times.

22. A ski jump is designed to follow the path given by the equations

x = 3.50t 2 and y = 20.0 + 0.120t 4 - 3.002t 4 + 1

10 … t … 4.00 s 2 (x and y in m, t in s). Find the velocity and

acceleration of a skier when t = 4.00 s. See Fig. 24.19.



v



1 3

23. A rocket follows a path given by y = x - 90

x (distances in km).

If the horizontal velocity is given by vx = x, find the magnitude

and direction of the velocity when the rocket hits the ground (assume level terrain) if time is in minutes.



24. A ship is moving around an island on a route described by

y = 3.0x2 - 0.20x3. If vx = 1.2 km>h, find the velocity of the

ship where x = 3.5 km.

25. A computer’s hard disk is 88.9 mm in diameter and rotates at

3600 r>min. With the centre of the disk at the origin, find the velocity components of a point on the rim for x = 30.5 mm, if

y 7 0 and vx 7 0.

26. A robot arm joint moves in an elliptical path (horizontal major

axis 8.0 cm, minor axis 4.0 cm, centre at origin). For y 7 0 and

-2 cm 6 x 6 2 cm, the joint moves such that vx = 2.5 cm>s.

Find its velocity for x = -1.5 cm.

27. An airplane ascends such that its gain h in altitude is proportional

to the square root of the change x in horizontal distance travelled.

If h = 280 m for x = 400 m and vx is constant at 350 m>s, find

the velocity at this point.

28. A meteor travelling toward the earth has a velocity inversely proportional to the square root of the distance from the earth’s centre.

State how its acceleration is related to its distance from the centre

of the earth.



Answer to Practice Exercise



1. a = 4.70, u = - 25.2°

Fig. 24.19



24.4 Related Rates

%FSJWBUJWFTXJUI3FTQFDUUP5JNF t

Rates of Change are Related



COMMON ERROR



Often, variables vary with respect to time, and are therefore implicitly functions of

time. If a relation is known to exist relating them, the time rate of change of one can be

expressed in terms of the time rate of change of the other(s). This is done by taking the

derivative with respect to time of the expression relating the variables, even if t does

not appear in the expression, as in Examples 4 and 5 of Section 24.3. Since the time

rates of change are related, this is referred to as a related-rates problem. Consider the

following examples.

Remember that all variables in related rates problems are functions of time, even if t

does not appear explicitly in the expression. Once again, a common error is not to include the derivative with respect to t as a factor when applying the power rule or the

chain rule.



E X A M P L E 1 Related rates—voltage and temperature



The voltage E of a certain thermocouple as a function of the temperature T (in °C) is

given by E = 2.800T + 0.006T 2. If the temperature is increasing at the rate of

1.00 °C/min, how fast is the voltage increasing when T = 100°C?

Since we are asked to find the time rate of change of voltage, we first take derivatives with respect to time. This gives us

dE

dT

dT

= 2.800

+ 0.012T

dt

dt

dt



d

dT

10.006T 2 2 = 0.006 a2T

b

dt

dt



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24.4 Related Rates



723



again being careful to include the factor dT>dt. From the given information, we know

that dT>dt = 1.00°C>min and that we wish to know dE>dt when T = 100°C. Thus,

dE 2

= 2.80011.002 + 0.01211002 11.002 = 4.00 V>min

dt T = 100

COMMON ERROR







When working with related rates, the derivative must be taken before given values are

substituted. A common error is to substitute before taking derivatives, so that the functions appear constant when they are not. For instance, in Example 1 we are finding the

time rate of change of voltage for a specified value of T. Different values of T will lead

to different values of dE >dT.



E X A M P L E 2 Related rates—distances



The distance q that an image is from a certain lens in terms of p, the distance of the

object from the lens, is given by

q =

Object



10p

p - 10



If the object distance is increasing at the rate of 0.200 cm>s, how fast is the image

distance changing when p = 15.0 cm? See Fig. 24.20.

Taking derivatives with respect to time, we have

dp



q

p



don’t forget the



Image

Fig. 24.20



dq

=

dt



1p - 102 a10



dp

dp

b - 10p a b

dt

dt



1p - 102 2



-100

=



Now, substituting p = 15.0 and dp>dt = 0.200, we have



dt



dp

dt



1p - 102 2



-10010.2002

dq

2

=

dt p = 15

115.0 - 102 2

= -0.800 cm>s



Practice Exercise



1. In Example 2, change each 10 to 12 and

then solve.



Thus, the image distance is decreasing (the significance of the minus sign) at the

rate of 0.800 cm>s when p = 15.0 cm.



In many related-rate problems, the function is not given but must be set up according

to the statement of the problem. The following examples illustrate this type of problem.

E X A M P L E 3 Related rates—volume and radius



A spherical balloon is being blown up such that its volume increases at the constant rate of

2.00 m3 >min. Find the rate at which the radius is increasing when it is 3.00 m. See Fig. 24.21.

We are asked to find the relation between the rate of change of the volume of a

sphere with respect to time and the corresponding rate of change of the radius with

respect to time. Therefore, we are to take derivatives of the expression for the volume of a sphere with respect to time:

V =



dV/ dt ϭ 2.00 m3/min



4 3

pr

3



dV

dr

= 4 pr 2 a b

dt

dt



r



2.00 = 4p13.002 2 a



Pump

Fig. 24.21



volume of sphere

take derivatives with respect to time



dr

b

dt



dr 2

1

=

dt r = 3

18.0p

= 0.0177 m>min



substitute

solve for



dV

= 2.00 m3 >min and r = 3.00 m

dt



dr

dt







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724



CHAPTER 24 Applications of the Derivative

E X A M P L E 4 Related rates—force and distance



The force F of gravity of Earth on a spacecraft varies inversely as the square of the distance r of the spacecraft from the centre of Earth. A particular spacecraft weighs 4500 N

on the launchpad (F = 4500 N for r = 6370 km). Find the rate at which F changes later

as the spacecraft moves away from the Earth at the rate of 12 km>s, where r = 8500 km.

We begin by setting up the equation relating F and r. Note that we substitute

known values of F and r in order to obtain the constant of proportionality. As we

said earlier, the values from the related rates problem are only substituted after taking derivatives.

F =

4500 =



k



inverse variation



r2

k



63702

k = 1.83 * 1011 N # km2

1.83 * 1011

F =

r2

dF

dr

= 11.83 * 1011 2 1 -22 1r -3 2

dt

dt

11

-3.66 * 10 dr

=

dt

r3

dF 2

-3.66 * 1011

=

1122

dt r = 8500 km

85003

= -7.2 N>s



substitute F = 4500 N, r = 6370 km

solve for k

substitute for k in equation

take derivatives with respect

to time



evaluate derivative for

r = 8500 km, dr>dt = 12 km>s

gravitational force is decreasing







These examples show the following method of solving a related-rates problem.

Steps for Solving Related-Rates Problems

1. Identify the variables and rates in the problem.

2. If possible, make a sketch showing the variables.

3. Determine the equation relating the variables.

4. Differentiate with respect to time.

5. Solve for the required rate.

6. Evaluate the required rate.

E X A M P L E 5 Related rates—distances

To Alaska

A



x



W



Vancouver



z



y



B

To Seattle

S

Fig. 24.22



Two cruise ships leave Vancouver, British Columbia, at noon. Ship A travels west at

12.0 km>h (before leaving toward Alaska), and ship B travels south at 16.0 km>h

(toward Seattle). How fast are they separating at 2 p.m.?

In Fig. 24.22, we let x = the distance travelled by A and y = the distance

travelled by B. We can find the distance between them, z, from the Pythagorean

theorem. Therefore, we are to find dz > dt for t = 2.00 h. Even though there are three

variables, each is a function of time. This means we can find dz > dt by taking derivatives of each term with respect to time. This gives us

z2 = x 2 + y 2

dy

dz

dx

2z

= 2x

+ 2y

dt

dt

dt



x1dx>dt2 + y1dy>dt2

dz

=

z

dt



using the Pythagorean theorem



(step 3)



taking derivatives with respect to time

solve for



dz

dt



(step 5)



(step 4)



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24.4 Related Rates



At 2 p.m., we have the values

x = 24.0 km, y = 32.0 km, z = 40.0 km

dx>dt = 12.0 km>h, dy>dt = 16.0 km>h



725



(step 6)

d = rt and Pythagorean theorem



124.02 112.02 + 132.02 116.02

dz 2

=

= 20.0 km>h

dt z = 40

40.0



from statement of problem

substitute values







E XE RC IS ES 2 4 .4

In Exercises 1 and 2, make the given changes in the indicated examples of this section and then solve the resulting problems.

1. In Example 1, change 0.006 to 0.012.

2. In Example 3, change “volume” to “surface area” and change m3 >min

to m2 >min.



In Exercises 3–6, assume that all variables are implicit functions of

time t. Find the indicated rates.

3. y = 5x2 - 4x; dx>dt = 0.5 when x = 5; find dy>dt.

4. x2 + 2y2 = 9; dx>dt = 3 when x = 1 and y = 2; find dy>dt.

5. x2 + 3y2 + 2y = 10; dx>dt = 2 when x = 3 and y = -1; find

dy>dt.

6. z = 2x2 - 3xy; dx>dt = - 2 and dy>dt = 3 when x = 1 and

y = 4; find dz>dt.

In Exercises 7–44, solve the problems in related rates.

7. How fast is the slope of a tangent to the curve y = 2> 1x + 12

changing where x = 3.0 if dx>dt = 0.50 unit>s?



8. How fast is the slope of a tangent to the curve y = 2(1 - 2x)2

changing where x = 1.5 if dx>dt = 0.75 unit>s?

9. The velocity v (in m/s) of a pulse travelling in a certain string is a

function of the tension T (in N) in the string given by v = 181T.

Find dv>dt if dT>dt = 0.20 N/s when T = 25 N.

10. The force F (in N) on the blade of a certain wind generator as a

function of the wind velocity v (in m/s) is given by F = 0.024v 2.

Find dF>dt if dv>dt = 0.75 m>s2 when v = 28 m>s.

11. The electric resistance R (in Ω) of a certain resistor as a function

of the temperature T (in °C) is R = 4.000 + 0.003T 2. If the temperature is increasing at the rate of 0.100°C>s, find how fast the

resistance changes when T = 150°C.

12. The kinetic energy K (in J) of an object is given by K = 12 mv 2,

where m is the mass (in kg) of the object and v is its velocity. If a

250-kg wrecking ball accelerates at 5.00 m>s2, how fast is the

kinetic energy changing when v = 30.0 m>s?

13. The length L (in cm) of a pendulum is slowly decreasing at the

rate of 0.100 cm>s. What is the time rate of change of the period

T (in s) of the pendulum when L = 16.0 cm, if the equation relating the period and length is T = p1L>245?

14. The voltage V that produces a current I (in A) in a wire of radius r

(in mm) is V = 0.030I>r 2. If the current increases at 0.020 A>s

in a wire of 0.040 mm radius, find the rate at which the voltage is

increasing.

15. A plane flying at an altitude of 2.0 km is at a direct distance

D = 24.0 + x2 from an airport control tower, where x is the

horizontal distance to the tower. If the plane’s speed is 350 km>h,

how fast is D changing when x = 6.2 km?



16. A variable resistor R and an 8@Ω resistor in parallel have a

8R

. If R is changing

combined resistance RT given by RT =

8 + R

at 0.30 Ω >min, find the rate at which RT is changing when

R = 6.0 Ω.



17. The radius r of a ring of a certain holograph (an image produced

without using a lens) is given by r = 10.40l, where l is the

wavelength of the light being used. If l is changing at the rate of

0.10 * 10-7 m>s when l = 6.0 * 10-7 m, find the rate at

which r is changing.

18. An earth satellite moves in a path that can be described by

y2

x2

+

= 1, where x and y are in thousands of kilometres.

72.5

71.5

If dx>dt = 12 900 km>h for x = 3200 km and y 7 0, find

dy>dt.

19. The magnetic field B due to a magnet of length l at a distance r is

k

, where k is a constant for a

given by B =

3r 2 + 1l>22 2 4 3>2

given magnet. Find the expression for the time rate of change of B

in terms of the time rate of change of r.

20. An approximate relationship between the pressure p and volume

V of the vapour in a diesel engine cylinder is pV 1.4 = k, where k

is a constant. At a certain instant, p = 4200 kPa, V = 75 cm3,

and the volume is increasing at the rate of 850 cm3 >s. What is the

time rate of change of the pressure at this instant?



21. A swimming pool with a rectangular surface 18.0 m long and

12.0 m wide is being filled at the rate of 0.80 m3 >min. At one end

it is 1.0 m deep, and at the other end it is 2.5 m deep, with a constant slope between ends. How fast is the height of water rising

when the depth of water at the deep end is 1.0 m?



22. An engine cylinder 15.0 cm deep is being bored such that the radius increases by 0.100 mm>min. How fast is the volume V of the

cylinder changing when the diameter is 9.50 cm?

23. Fatty deposits have decreased the circular cross-sectional opening

of a person’s artery. A test drug reduces these deposits such that

the radius of the opening increases at the rate of 0.020 mm>month.

Find the rate at which the area of the opening increases when

r = 1.2 mm.

24. A computer program increases the side of a square image on the

screen at the rate of 0.25 cm>s. Find the rate at which the area of

the image increases when the edge is 6.50 cm.

25. A metal cube dissolves in acid such that an edge of the cube decreases by 0.500 mm>min. How fast is the volume of the cube

changing when the edge is 8.20 mm?

26. A metal sphere is placed in seawater to study the corrosive effect

of seawater. If the surface area decreases at 35 cm2 >year due to

corrosion, how fast is the radius changing when it is 12 cm?



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