# Thread: is 4πr² the rate of change of the volume of a sphere?

1. ## is 4πr² the rate of change of the volume of a sphere?

I need to find the rate of change of the volume of a sphere with respect to the radius r .

I know the volume of a sphere is (4πr³) / 3

And that dV/dr = (4π/3)(3r²) = 4πr²

If r was defined I would plug in the value for it and go through the rest of the problem, but since its not defined is that the final answer?

2. Originally Posted by Brazuca
I need to find the rate of change of the volume of a sphere with respect to the radius r .

I know the volume of a sphere is (4πr³) / 3

And that dV/dr = (4π/3)(3r²) = 4πr²

If r was defined I would plug in the value for it and go through the rest of the problem, but since its not defined is that the final answer?
The rate of change in y with respect to x is $\frac{dy}{dx}$
So similarly, the rate of change in Volume, V, with respect to radius, r is $\frac{dV}{dr}$

Sorry to confuse you, in most related rate questions you'll be giving the rate at which r changes with respect to time, you're answer is correct for this question

in context though, normally volume changes with respect to time, and the radius changes with respect to time,

so $\frac{dV}{dt}=\frac{dV}{dr}\ \frac{dr}{dt}$

3. Originally Posted by Brazuca
I need to find the rate of change of the volume of a sphere with respect to the radius r .

I know the volume of a sphere is (4πr³) / 3

And that dV/dr = (4π/3)(3r²) = 4πr²

If r was defined I would plug in the value for it and go through the rest of the problem, but since its not defined is that the final answer?

You're not a millionaire just yet though.

4. Oh so how do I get it to be dV / dt?

So does that mean I have the wrong answer? Cause I don't know what I can do after I got dV / dr

5. Originally Posted by Brazuca
Oh so how do I get it to be dV / dt?

So does that mean I have the wrong answer? Cause I don't know what I can do after I got dV / dr

Rate of change is synonymous with derivative. So, you need to find the derivative the function V(r) with respect to r, where V(r)= $\frac{4}{3}\pi{r^3}$.

Or $\frac{dV}{dr}=\frac{d}{dr}(\frac{4}{3}\pi{r^3})$

So, taking the derivative here is easy

Since you are differentiating with respect to r and there are no other variables involved in the function, the chain rule is unnecessary. The only rules that apply here are the constant multiple rule, which says that a constant can be "factored" out of the differentiation process

So $\frac{d}{dr}(\frac{4}{3}\pi{r^3})$ becomes $\frac{4}{3}\pi*\frac{d}{dr}(r^3)$.

and the other rule that is applicable is the power rule which states that the derivative of a varible raised to the n power is

$nx^{n-1}$.

So using this rule

$\frac{d}{dr}(r^3)=3r^{3-1}=3r^2$

Such that $\frac{dV}{dr}=\frac{d}{dr}(\frac{4}{3}\pi{r^3})=\f rac{4}{3}\pi*3r^2=4\pi{r^2}$

Your question was that you needed to find the rate of change with respect to r. Well, you now have an expression that will give you that rate of change given any r. I suspect that you were given more information about this problem. If r is not explicitly given, can you think of anywayh that you can find it from the information that was given? are you sure that you were not asked to find the rate of change of the shere with respect to time?

6. Thanks everyone.

That's all the info I was given VonNemo. Which is why I though it was too simple.

I was just worried I had it wrong, there were more steps, or that it was suppose to end up being unsolvable or something like that.

Since before this question I did 3 problems just like that , but with the r being defined, so I though it not being defined would require more stuff to be done to the 4πr² in order to get a final value or something.

7. Originally Posted by Brazuca
Thanks everyone.

That's all the info I was given VonNemo. Which is why I though it was too simple.

I was just worried I had it wrong, there were more steps, or that it was suppose to end up being unsolvable or something like that.

Since before this question I did 3 problems just like that , but with the r being defined, so I though it not being defined would require more stuff to be done to the 4πr² in order to get a final value or something.
No problem.