MATH 1B—SOLUTION SET FOR CHAPTERS 8.1, 8

MATH 1BSOLUTION SET FOR CHAPTERS 8.1, 8.2

Problem 8.1.1. Use the arc length formula to find the length of the curve y =

2 ? 3x, ?2 x 1. Check your answer by noting that the curve is a line segment

and calculating its length by the distance formula.

Solution. First, note:

y 0 = ?3

q



2

1 + (y 0 ) = 10

(Note that this is a constant, which is as it should bethe curve is a line, and a

line should have the same amount of arc length per unit horizontal distance. In

fact, it should be the secant of the angle the line makes with the x-axis!)

So, using the arc length formula, the length of the curve on ?2 x 1 is

Z x=1

Z 1

ds =

10dx

x=?2

?2



1

= 10 [x]?2



= 3 10

Of course, since this curve is a line, using the arc length formula is like using a

flamethrower to kill ants. Since the line has endpoints at (?2, 8) and (1, ?1), its

length must be:

p





32 + (?9)2 = 90 = 3 10

as desired.

Problem 8.1.9. Find the length of the curve given by x =

1

3 y(y

? 3), 1 y 9.

Solution. In this case, were probably (almost certainly) better off integrating up

the y-axis. Taking the derivative, we have:





dx

1 y?3

=

+ y

dy

3 2 y

1

= (3y ? 3)

6 y

y?1

=

2 y

Thus,

s

ds =

1+

s

=

(y ? 1)2

4y

(y + 1)2

4y

Typeset by AMS-TEX

1

2

MATH 1BSOLUTION SET FOR CHAPTERS 8.1, 8.2

On the range that were interested in, y + 1 is positive. Thus, the arc length is:

Z y=9

Z 9

y+1

ds =

dy

y=1

1 2 y

Substituting u =



y, so du =

1



2 y,

we now have

Z

=

3

(u2 + 1)du

1

u3

=

+u

3

32

=

3



So the curve has arc length

3

1

32

3 .

Problem 8.1.13. Find the arc length of the curve given by y = cosh x, 0 x 1.

Solution. As long as you remember how cosh is defined and what its derivative is,

this ones easy. Recall:

y 0 = sinh x

so

p

p

1 + (y 0 )2 = 1 + sinh2 x

p

= cosh2 x

Z

x=1

x=0

= cosh x

Z 1

ds =

cosh xdx

0

1

= [sinh x]0

1

1

= e?

2

2e

Problem 8.1.30.

(a) Sketch the curve y 3 = x2

(b) Set up two integrals for the arc length from (0, 0) to (1, 1), one along x and

one along y.

(c) Find the length of the arc of this curve from (?1, 1) to (8.4).

Proof. (a) Its clear that this curve is single-valued, since f (x) = x3 is invertible

(so for any given x, theres only one value of y that satisfies the equation y 3 = x2 ).

2

Thus, the curve is the same as y = x 3 . This function is even, and has first derivative

1

2 ?3

. This is positive on x > 0, negative on x < 0, and undefined at zero itself.

3x

4

The second derivative is ? 92 x? 3 , which is negative everywhere (except at 0, where

it too is undefined). Thus the curve

is concave down everywhere. Such a curve

p

looks something like the plot of |x|,

2

1

(b) Solving for y, we have y = x 3 . Then y 0 = 23 x? 3 , and so

Z

Z 1r

4 2

ds =

1 + x? 3 dx

9

0

MATH 1BSOLUTION SET FOR CHAPTERS 8.1, 8.2

3

Because the integrand is undefined at x = 0, this integral is improper. We thus

write:

Z 1r

4 2

= lim

1 + x? 3 dx

+

9

s0

s

r

Z 1

1

4

2

x? 3 x 3 + dx

= lim+

9

s0

s

Z 13

9



3

udu

=

lim

2

+

2 s0 s 3 + 94

3

2 h 3 i 13

9

u2 2 4

=

lim+

2 s0 3

s2 +9

  32   32

4

13

?

=

9

9



13 13 ? 8

=

27

We could instead have solved for x (on 0 x 1, the curve is single-valued in

3

either x or y). In this case, we have x = y 2 , so

dx

3 1

= y2.

dy

2

Our arc length is thus

Z

Z

ds =

0

1

r

9

1 + ydy

4

Z 13

4

4

udu

9 1

8 h 3 i 13

4

=

u2

27 "

1

#



8 13 13 ? 8

=

27

8



13 13 ? 8

=

27

=

In either case, we get the same answer, as we shouldthis is, after all, the arc

length of a curve!

(c) Now we have to be careful. On the range ?1 x 8, the curve is a function

in y, but is not invertible. Probably the laziest (and therefore best) way to proceed

is as follows: First, note that we already know the arc length between (0, 0) and

(1, 1). Next, realize that since the function is odd, the length of the curve between

(?1, 1) and (0, 0) must be the same as the length between (0, 0) and (1, 1). This

leaves only the curve between (1, 1) and (8, 4). On this range, the curve is invertible,

4

MATH 1BSOLUTION SET FOR CHAPTERS 8.1, 8.2

so we can just use the second method above, to get

Z

Z 4r

9

dx =

1 + ydy

4

1

Z

4 10

udu

=

9 13

4

"

#



8

13 13

=

10 10 ?

27

8





80 10 ? 13 13

=

27

So, our total arc length is 2 13



13?8

27

+





80 10?13 13

,

27

or





80 10+13 13?8

.

27

3

Problem 8.1.31. Find the arc length function for the curve y = 2x 2 , starting

with the point P0 (1, 2).

Solution. The arc length function is defined by:

Z xq

2

1 + (y 0 ) dt

s(x) =

1

1

2

Since y 0 = 3x , this is

Z

s(x) =

x



1 + 9tdt

1

Z



1

01+9x udu

9 1

i

3

2 h

=

(1 + 9x) 2 ? 10 10

27

h

i

3

2

So the arc length function is s(x) = 27

(1 + 9x) 2 ? 10 10 .

=

Problem 8.1.34. A steady wind blows a kite due west. The kites height above

ground from horizontal position x = 0 to x = 80f t is given by

y = 150 ?

1

(x ? 50)2

40

Find the distance traveled by the kite.

Solution. It should be clear that the distance traveled by the kite is precisely the

arc length of its path, as it travels along its parabolic path. (That the path above

describes a downward-opening parabola isnt important to the problem, but is worth

noting. Its always nice to see old friends like parabolae).

1

(x ? 50), so the arc length is:

In this case, y 0 = ? 20

Z

Z 80 r

1

ds =

1+

(x ? 50)2 dx

400

0

Z 30 r

1 2

=

1+

u du

400

?50

Z arctan( 32 )

= 20

sec3 d

5

arctan(? 2 )

MATH 1BSOLUTION SET FOR CHAPTERS 8.1, 8.2

To find

R

5

sec3 d, we use the usual trick:

Z

sec3 d = sec tan ?

Z

Z

= sec tan ?

Z

2

Z

sec tan2 d

sec (sec2 ? 1)d

sec3 = sec tan + ln | sec + tan |

sec3 =

1

1

sec tan + ln | sec + tan |

2

2

Thus, returning to our arc length problem, the distance traveled by the kite in

feet is:



d=

arctan( 32 )

1

1

sec tan + ln | sec + tan |

2

2

arctan(? 5 )

2

 p

arctan( 32 )

p

1

1

=

1 + tan2 tan + ln | 1 + tan2 + tan |

2

2

arctan(? 25 )

#

"  r

r

r

r 



13 1

13 3

1 29 ?5

29 5

1 3

1

? ln

+

?

?

+ ln

=

2 2

4

2

4

2

2 4

2

2

4

2







3 13 + 5 29 1

3 + 13

=

+ ln

8

2

29 ? 5

Problem 8.2.1. Set up, but do not evaluate, an integral for the area of the surface

obtained by rotating

y = ln x, 1 x 3

about the x-axis.

Solution. This ones easy (since we dont have to evaluate the integral!): y 0 =

so

r

Z 3

1

A=

2 ln x 1 + 2 dx

x

1

1

x,

Problem 8.2.3. Set up, but do not evaluate, an integral for the area of the surface

obtained by rotating

y = sec x, 0 x /4

about the y-axis.

Solution. First, note that y 0 = sec x tan x. Thus,

Z

A=

0

/4

p

2x 1 + sec2 x tan2 xdx

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