Question

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A spherical balloon is expanding at a rate of 60 pi cubic inches per second. How fast is the surface area of the balloon expanding when the radius of the balloon is 4 inches?

Answer 1

Am I solving for dr/dt?

Answer 2

Solving for dA/dt it looks like.
Where A is the surface area of a sphere.

Answer 3

Volume of a sphere,
\[\Large\rm V=\frac{4}{3}\pi r^3\]Surface Area of a sphere,\[\Large\rm A=4\pi r^2\]

Answer 4

ok I was not using both of those

Answer 5

the cubic inches per second threw me off

Answer 6

So we're given \(\Large\rm r=4\), and \(\Large\rm dV/dt=60\), yes?
Or does expanding mean the rate at which it's moving outward? radius change?
Oh no no, the units are cubic, so they're talking about volume change, ok ok ok.

Answer 7

yes 60 pi for dv/dt

Answer 8

\[60 \pi\]

Answer 9

I am confused on this one

Answer 10

I was using surface area of a sphere

Answer 11

So ummm let's see if we can figure this out.
Differentiating our Volume function, with respect to time,\[\Large\rm \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}\]Does that look ok?

Answer 12

yes that part looks ok

Answer 13

So in order to find dA/dt, it looks like we'll need to deal with this dr/dt first.
So our first step, yes, is to solve for dr/dt using the information they gave us.

Answer 14

ok let me do that

Answer 15

15/16

Answer 16

or .9375

Answer 17

Ok great.\[\Large\rm \frac{dr}{dt}=\frac{15}{16}\]

Answer 18

Let's jump over to our surface area function now, we're done using the volume formula, it gave us what we needed.

Answer 19

30 pi

Answer 20

inches^2 per second

Answer 21

Ooo yay good job \c:/

Answer 22

Thank you