Related Rates air in a balloon

[AmiraM]

Homework Statement

Air is being pumped into a spherical balloon so that its volume increases at a rate of 70 cm³/s. How fast is the surface area of the balloon increasing when its radius is 5 cm³?

Homework Equations

A ball with radius r has the following volume (V) and surface area (S):
V = (4/3)(πr³)
S = 4πr²

The Attempt at a Solution

The problem seems to be asking for dS/dt at r = 5. dV/dt appears to be equivalent to S.

dS/dt = (d/dt)(4πr²)
dS/dt = 8πr(dV/dt) <--- This step is the one that's really bugging me, especially that dV/dt tagged onto the end.
dS/dt = 8π5(70)
dS/dt = 2800π
dS/dt = 8796.459

Which is apparently incorrect. I'm having a lot of trouble grasping these and other related rate problems, mainly because I am not sure how to properly set up the equations and every tutorial I've found on the subject (including my textbook) simply breeze over that step by saying "Find an equation that describes the given relationship."

Any help is much appreciated.

 

[Char. Limit]

Well, firstly, what you'll want to do is find S as a function of V. This can be done by solving your function V = 4 pi r^3 / 3 for r, and then plugging that into the r in your function for S.

So basically, solve your volume function for r, and then plug that into the r in your surface area function.

 

[AmiraM]

Thanks!

Alright, so I tried what you said and came out with:

S = 4π(sqrt3(3v/4π))²

Taking the derivatives using the chain rule leaves me with:

dS/dt = 8π(3v'/4π)^1/3(3v'/4π)^-2/3, or
dS/dt = 8(3v'/4)^-1/3

When I plug v' = 70 into the equation, it returns 2.137, which is unfortunately also incorrect. I can show my deriving process step-by-step if it helps anyone spot where I'm making my error.

 

[Char. Limit]

Thanks!

Alright, so I tried what you said and came out with:

S = 4π(sqrt3(3v/4π))²

Taking the derivatives using the chain rule leaves me with:

dS/dt = 8π(3v'/4π)^1/3(3v'/4π)^-2/3, or
dS/dt = 8(3v'/4)^-1/3

When I plug v' = 70 into the equation, it returns 2.137, which is unfortunately also incorrect. I can show my deriving process step-by-step if it helps anyone spot where I'm making my error.

You did something wrong here. Taking the derivative with the chain rule should yield a (-1/3) term and then v' on the outside.

 

[jhae2.718]

You are applying the chain rule incorrectly. [tex]\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx}[/tex]

 

[AmiraM]

Yes, there was definitely something wrong with the way I derived. I tried deriving it again, step-by-step this time, and ended up with the correct answer. Thanks so much guys, you were really helpful :)