Question

A spherical balloon is being inflated at the rate of 4 cubic ft/min. The rate, in square feet/min, at which the surface area is increasing when the volume is 32pi/3 ft^3 is...?

Answer 1

When we are inflating a balloon, we are just pumping volume of air into it. .'.dV/dt = 4cu ft/min
V = (4/3)πr^3
.'.dV/dt = (4/3)π[3r^2(dr/dt)]
or, 4 = (4/3)π[3r^2(dr/dt)]
or, (1/π) = r^2(dr/dt).............(1)
When, volume is 32π/3 cu ft
(4/3)πr^3 = 32π/3
r^3 = 8
or, r=2.
.'.dr/dt = 1/(4π)
Now, surface area = S = 4πr^2
.'.dS/dt(at r=2) = 8πr(dr/dt) = 8π*2*[1/(4π)] = 16/4 = 4 ft^2/min
.'.rate of increase of surface area of a sphere = 4ft^2 / min

Answer 2

answer choices:
a. 4pi
b. 2
c. 4
d. 1
e. 2pi
I know this is a related rates problem and I just cant find a good example in my book or notes! I would appreciate any help!

Answer 3

c

Answer 4

Oh wow! Thanks for such a concise and quick reply! I looked through your work and it looks great! I understand this now, and I'll work through it one more time just to solidify it for myself. Again thank you, and I'll award you Best Response!

Answer 5

you welcome I just like helping people lol

Answer 6

Well thanks much then! I'll close this question.