# Spherical balloon related rates problem

1. Mar 17, 2015

### s3a

1. The problem statement, all variables and given/known data
You are blowing air into a balloon at a rate of 4*pi/3 cubic inches per second. (The reason for this strange-looking rate is that it will simplify your algebra a little bit.)

Let r(t), A(t) and V(t) denote the radius, surface area and the volume of your balloon at time t, respectively. (Assume the thickness of the skin is zero.)

Find:
a) r'(t)

b) A'(t)

c) V'(t)

2. Relevant equations
Differentiation.
Chain rule.
Related rates.

3. The attempt at a solution
I know that dV/dt = 4*pi/3 and that dV/dt = 4*pi * r^2 dr/dt, and that 4*pi/3 = 4*pi r^2 * dr/dt, which implies that 1/3 = r^2 * dr/dt.

I also found that dA/dt = 8*pi*r * dr/dt.

My issue is that I now have two equations, 1/3 = r^2 * dr/dt and dA/dt = 8*pi*r * dr/dt, but three unknowns, dr/dt, dA/dt and r.

I'm assuming that I need to find a third relationship/equation, but I cannot figure out what it is.

As always, any help would be very much appreciated!

2. Mar 17, 2015

### andrewkirk

You've derived $\frac{1}{3}=r^2\frac{dr}{dt}$, so you can express $\frac{dt}{dr}$ as a function of $r$. Then you can integrate that with respect to $r$ to get $t$ as a function of $r$. INvert that to get $r$ as a function of $t$ and the rest is just differentiation.

3. Mar 17, 2015

### s3a

Thanks for the response, andrewkirk, but I'm still stuck after having read what you wrote.

Here is what I understood:
dt/dr = 3r^2

d/dr (dt/dr) = d/dr (3r^2)
d^2 t / dr^2 = 6r

d^2 r / dt^2 = 1/(6r)

4. Mar 17, 2015

### LCKurtz

You don't want to differentiate $\frac{dt}{dr}=3r^2$, you want to write it as $3r^2dr = 1dt$ and integrate both sides to get $r$ as a function of $t$. Don't forget you are given $r=0$ when $t=0$.

5. Mar 17, 2015

### s3a

Thanks for the response LCKurtz, but this is for a calculus course that has yet to cover integrals.

6. Mar 17, 2015

### LCKurtz

If you can't integrate then you can't expect to get $r$ in terms of $t$. But you do have $\frac{dr}{dt}=\frac{1}{3r^2}$. You can plug that in to get $\frac{dA}{dt}$ as a function of $r$ just like your $\frac{dr}{dt}$ is. I think that is all you can do in that case.

7. Mar 17, 2015

### LCKurtz

Also, possibly you are expected to write your equation as $3r^2\frac{dr}{dt}= 1$ and notice that looks like the chain rule from differentiating what?

8. Mar 17, 2015

### SteamKing

Staff Emeritus
You know dA/dt = 8πr dr/dt and dV/dt = 4πr2 dr/dt.

Let Q = dr/dt so that dV/dt = 4πr2 * Q = 4π/3. What is Q? What is dA/dt? Now, it's just a problem of algebra.

9. Mar 18, 2015

### s3a

Thanks guys, I'm no longer confused! :)

To keep it concise and free of any material beyond the first calculus course for any potential future readers:
Because the rate of change is constant, the rate of increase is a linear rate, so we can say that dV/dt = ΔV/Δt and since, at t = 0, r = 0, it follows that at t = 0, V = 0 as well.

Therefore, ΔV = $V_{final}$ - $V_{initial}$ = $V_{final}$ - 0 = $V$, and Δt = $t_{final}$ - $t_{initial}$ = $t_{final}$ - 0 = $t$, from which we have the relationship V = 4*pi/3 * t which is equivalent to 4*pi/3 * r^3 = 4*pi/3 * t, and we have t = r^3 as the third equation, and then having 3 equations is enough for solving the 3 unknowns (dA/dt, r and dr/dt).

10. Mar 18, 2015

### LCKurtz

That looks OK. Do you see where $r^3 = t$ might have been deduced from $3r^2\frac{dr}{dt} = 1$? Think of it as a "preview of coming attractions" in integral calc.

11. Mar 22, 2015

### s3a

Yes, I do see how r^3 = t could have been deduced from 3 r^2 * dr/dt = 1, but I honestly prefer noticing that the constant rate of change denotes a linear relationship between change in the volume and time. Having said that, it's nice to see different approaches.

Also, I already took this course, and I was trying to help someone who is taking it now, and it bothered me that I couldn't solve it.