Question

Ok, im working with derivatives on this problem. A spherical balloon is inflated with gas at a rate of 20 cubic feet per minute. How fast is the radius of the balloon changing at the instant the radius is (a) 1 foot and (b) 2 feet?

Answer 1

If this is a parametric differential question, you'd have to use:
V = 4/3*pi*r^3
differentiate with respect to r (dv/dr)
use dV/dt is 20 (from the 20 cubic feet per min)
and then use the fact that:

dV/dt = dV/dr * dr/dt

Answer 2

Can u show me some steps?

Answer 3

That's pretty much it. Dubstitute what you know.

dV/dt = 20 ft^3 / min

When the Radius is 1 ft, what is the volume?
When the Radius is 2 ft, what is the volume?

You're looking for dr/dt in each case.

Answer 4

So how would i work in the formula for the volume of a sphere?

Answer 5

You use it to calculate the volume for the two given radii.
You use it to calculate the derivative, DV/dr.

Please show this work.

Answer 6

Well i think the derivative of v is 4pir^2 my teacher isnt the best at explaining this stuff.

Answer 7

Okay, in this case we have V(t) and r(t), not just V and r. In other words, the radius and the volume are functions of time. You must account for this in your derivation.

Your derivtive was good, but better notation would make it look substantially more obvious.

Given \(V = \dfrac{4}{3}\pi r^{3}\), we have \(dV = 4\pi r^{2} dr\).

Recognizing that V and r are functions of time, we have: \(\dfrac{dV}{dt} = 4\pi r^{2} \dfrac{dr}{dt}\)

It turns out we do NOT need to volume related to the individual radii. Just substitute to known values.

Answer 8

Um, the computer is not showing ur equations, can u retype them?

Answer 9

You can produce your own, I would think.

V = 4/3 pi r^3 ==> dV/dt = 4 pi r^2 dr/dt