Assignment 4

Name:

Due June 26, 2007

1. Find the length of each of the following curves:

(a) y = ln (secx), 0 ≤ x ≤ π4.

Solution Notice that ddx

ln sec x = tanx.

∫ π

4

0

√

1 + tan2 xdx =

∫ π

4

0

√sec2 x dx

=

∫ π

4

0

secxdx

= ln (secx + tanx)∣

∣

∣

π

4

0

= ln(√

2 + 1)

− ln (1 + 0)

= ln(√

2 + 1)

(b) y = x3

6+ 1

2x, 1

2≤ x ≤ 1.

Solution Notice that ddx

(

x3

6+ 1

2x

)

= x2

2− 1

2x2 .

∫ 1

1

2

√

1 +

(

x2

2−

1

2x2

)2

dx =

∫ 1

1

2

√

1 +x4

4−

1

2+

1

4x4dx

=

∫ 1

1

2

√

x4

4+

1

2+

1

4x4dx

=

∫ 1

1

2

√

(

x2

2+

1

2x2

)2

dx

=

∫ 1

1

2

(

x2

2+

1

2x2

)

dx

=

(

x3

6−

1

2x

)

∣

∣

∣

1

1

2

=31

48

1

2. Use Simpson’s Rule with n = 6 to estimate the arc length of the curvey = tanx, with 0 ≤ x ≤ π

4.

Solution Notice that ddx

tan x = sec2 x, so that we obtain the integral

∫ π

4

0

√

1 + sec4 x dx.

We obtain ∆x = π24

and Simpson’s Rule yields

π

72

(

√

1 + sec4 0 + 4

√

1 + sec4π

24+ 2

√

1 + sec4π

12+ 4

√

1 + sec4π

8

+ 2

√

1 + sec4π

6+ 4

√

1 + sec45π

24+

√

1 + sec4π

4

)

= 1.278100

3. Find the arc length of the parametric curve x = t3 − 4t, y = t2 − 3t,−2 ≤ t ≤ 2. Approximate numerically if necessary.

Solution Notice that ddt

(

t3 − 4t)

= 3t2 − 4 and ddt

(

t2 − 3t)

= 2t − 3.

∫ 2

−2

√

(3t2 − 4)2+ (2t − 3)

2dt =

∫ 2

−2

√

9t4 − 24t2 + 16 + 4t2 − 12t + 9 dt

=

∫ 2

−2

√

9t4 − 20t2 − 12t + 25 dt

≈ 19.0446

4. Find the area of each of the following surfaces of revolution:

(a) y = x2

4− ln x

2, 1 ≤ x ≤ 4, rotated about the x-axis.

Solution Since we are rotating about the x-axis, y is the radius. Fur-

thermore ddx

(

x2

4− lnx

2

)

= x2− 1

2x. Thus, we have

Area = 2π

∫ 4

1

(

x2

4−

lnx

2

)

√

1 +

(

x

2−

1

2x

)2

dx

= 2π

∫ 4

1

(

x2

4−

lnx

2

)

√

1 +x2

4−

1

2+

1

4x2dx

= 2π

∫ 4

1

(

x2

4−

lnx

2

)

√

x2

4+

1

2+

1

4x2dx

= 2π

∫ 4

1

(

x2

4−

lnx

2

)

√

(

x

2+

1

2x

)2

dx

= 2π

∫ 4

1

(

x2

4−

lnx

2

)(

x

2+

1

2x

)

dx

2

= 2π

∫ 4

1

(

x3

8+

x

8−

x lnx

4−

lnx

4x

)

dx

= 2π

(

x4

32+

x2

16−

x2 lnx

8+

x2

16−

(lnx)2

8

)

∣

∣

∣

4

1

= 2π

(

315

32− 4 ln 2 +

1

2(ln 2)

2

)

= π

(

315

16− 8 ln 2 + (ln 2)

2

)

≈ 1842.129

(b) y = 1 − x2, 0 ≤ x ≤ 1, rotated about the y-axis.

Solution Since we are rotating about the y-axis, x is the radius. Fur-

ther dy

dx= −2x. We obtain the integral

∫ 1

02πx

√1 + 4x2 dx. Setting

u = 1 + 4x2, du = 8xdx, u (0) = 1, u (1) = 5, we obtain

π

4

∫ 5

1

√u du =

π

6u

3

2

∣

∣

∣

5

1

=π

6

(

5√

5 − 1)

≈ 5.33041

(c) 8y3 = 3x2, 1 ≤ x ≤ 8, rotated about the x-axis.

Solution The equation of the curve can be solved for either x ory. If we solve for y, we obtain an integral which must be solvednumerically. If we solve for x, the integral can be solved analytically,but it involves many radicals and is very confusing. We solve for

y =3√

3

2x

2

3

Area = 2π

∫ 8

1

3√

3

2x

2

3

√

√

√

√1 +

(

3√

3

3 3√

x

)2

dx

≈ 87.8551

5. An airplane travelling at 150 ft/s at a height of 512 ft wishes to drop asupply of food to a target on the ground.

(a) How long will it take the supplies to reach the ground from the air-plane?

Solution We have y′′ (t) = −32, y′ (0) = 0, and y (0) = 512 andwe need to find the positive zero of y (t). y′ (t) = −32t, so thaty (t) = −16t2 + 512.

512 − 16t2 = 0

512 = 16t2

32 = t2

t = 4√

2

3

(b) At what distance from the target should the supplies be released?

Solution We have x′′ (t) = 0, x′ (0) = 150, x(

4√

2)

= 0. Thus,x′ (t) = 150, x (t) = 150t + c.

150(

4√

2)

+ c = 0

c = −600√

2

x (0) = −600√

2

So the supplies should be released at a distance of −600√

2 feet (about848.53 feet before the target).

6. In order to successfully juggle balls, the balls must be tossed a consistenthorizontal distance to be caught. If a ball is tossed with initial horizon-tal velocity v0x

and initial vertical velocity v0y, show that the horizontal

distance traveled is ω =v0x

v0y

16feet.

Solution We have y′′ (t) = −32 and y′ (0) = v0y, so that y′ (t) = −32t+v0y

.With y (0) = 0, we obtain y (t) = −16t2 + v0y

t. The ball will land wheny (t) = 0.

0 = −16t2 + v0yt

16t = v0y

t =v0y

16

Thus, the ball lands afterv0y

16seconds. Since x′′ (t) = 0 and x′ (0) = v0x

,x′ (t) = v0x

. Since x (0) = 0, we obtain x (t) = v0xt. The horizontal

distance travelled is then x( v0y

16

)

=v0x

v0y

16as desired.

7. A team of movers moves a total of 7,000 pounds up twenty feet from amoving truck to a third-story apartment. How much work is done by thesemovers?

Solution Work is given by∫

F dy.

W =

∫ 20

0

7000 dy

= 7000y∣

∣

∣

20

0

= 140, 000ft · lbs

8. Compute the weight in ounces of a baseball bat extending from x = 0 to

x = 32 with density ρ (x) =(

1

45+ x+3

700

)2slugs/in.

Solution Mass is simply the integral of the density between the appropriateendpoints.

m =

∫ 32

0

(

1

45+

x + 3

700

)2

dx

4

=700

3

(

1

45+

x + 3

700

)

∣

∣

∣

32

0

=700

3

[

(

1

45+

35

700

)3

−(

1

45+

3

700

)3]

=207266

2480625

This is the mass in slugs. The weight in pounds on earth is 32m, so thatthe weight in ounces is thus 512 207266

2480625≈ 42.7796oz.

9. Compute the center of mass of the baseball bat in the previous problem.

Solution The center of mass is determined by dividing the moment∫

xρ (x) dx

by the mass. That is,

x̄ =

∫ 32

0x(

1

45+ x+3

700

)2dx

(

207266

2480625

)

=4271648

207266≈ 20.6095

so that the center of the bat is 20.6095 inches from the end of the bat(presumably the handle end).

5