Page 165 - Calculus Workbook For Dummies

149
Chapter 8: Using Differentiation to Solve Your Problems

h(18)

r
dv dh
2. List the rates: = 200 cubic ft/min = ?
dt dt
1 2
3. a. The formula thing: V cone = π r h
3
b. Write an equation relating r and h so that you can get rid of r: r h
=
1 2 1 3
What could be simpler? Now get rid of r: V = π h h = π h
3 3
dV 2 dh
4. Differentiate: = π h
dt dt
dh
5. Substitute and solve for .
dt
200 = $ 2 dh
π 18
dt
dh . .196 ft/min . 2 1 inches/min
dt 3
6. Check whether this answer makes sense.
Calculate the increase in the height of the cone from the critical moment (h = 18) to ⁄200 minute
1
1 3
after the critical moment. When h = 18, V = π 18h , or about 6107.256 cubic feet. ⁄200 minute
1
^
3
later, the volume (which grows at a rate of 200 cubic feet per minute) will increase by 1 cubic
foot to about 6108.256 cubic feet. Now solve for h:
1
6108 .256 = π h 3
3
6108 .256
h =
3 1 π
3
. 18 .000982
1
Thus, in ⁄200 minute, the height would grow from 18 feet to 18.000982 feet. That’s a change of
$
.000982 feet. Multiply that by 200 to get the change in 1 minute: .000982 200 . .196
It checks.
i s t = t 5 + 4
2
^ h
a. At t = 2, the platypus’s position is s t = 24 feet from the back of your boat.
^ h
b. v t = l ^h s t = 10 t, so at t = 2, the platypus’s velocity is s 2 = 20 feet/second (20 is positive
h
^
l ^ h
so that’s toward the front of the boat).
c. Speed is the absolute value of velocity, so the speed is also 20 ft/sec.
d. Acceleration, a t ^ h, equals v t = m ^h s t = 10. That’s a constant, so the platypus’s acceleration
l ^
h
feet / second
is 10 at all times.
second