Page 74 - How To Solve Word Problems In Calculus

S = 4πr 2

dS dr
= 8πr
dt dt
When r = 3,

dr
−1 = 24π
dt
dr 1
=−
dt 24π
dr 1
Because is negative, r is shrinking at the rate of in/min.
dt 24π
4 3
(b) The volume of a sphere is V = πr .
3
dS
Given: =−1
dt
dV
Find: when r = 3.
dt
4
V = πr 3
3
dV 2 dr
= 4πr
dt dt
dr 1
From part (a), =− when r = 3.
dt 24π

dV 2 1
= 4π · 3 −
dt 24π
36π
=−
24π

3
=−
2
1
3
The volume is shrinking at the rate of 1 /2 in /min.

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