# Thread: This should be obvious to those of you...

1. ## This should be obvious to those of you...

OK, I'm teaching myself calculus and the book I'm using wasn't particularly clear on this, so I'll give you the problem, the solution, and the part I don't comprehend...

[b]A sponge is in the shape of a right circular cone. As it soaks up the water, it grows in size. At a certain moment, the height equals 6 inches, and is increasing at the rate of .3 inches per second. At that same moment, the radius is 4 inches, and is increasing at the rate of .2 inches per second. How is the volume changing at that time?

So, height = 6 in and is increasing by .3 in/sec
Radius is 4 inches and is increasing by .2 in/sec

The volume of a right circular cone is $V = \frac{1}{3}[pi]r^2h$ (If someone could tell me how to write the symbol for pi in the [ Tex ] tags, that'd be helpful in the future.)

Now is where I get confused, to derive that formula in respect to the variable t which is time. As well, I was unsure of how to apply the product rule when I have three different numbers being multiplied at once...

The book gives this;
$\frac{dV}{dt} = \frac{1}{3}[pi][2r\frac{dr}{dt}h + r^2\frac{dh}{dt}]$

I realize that $\frac{dr}{dt}$ is the rate the radius is increasing and $\frac{dh}{dt}$ is the rate the height is increasing, but knowing when/ where/ why to use them is what I don't get.
I think someone may just need to explain each step in the problem and I'll get it.

2. ## Re: This should be obvious to those of you...

The question asks us to find the rate at which the volume is changing. In other words, we are looking for how fast the volume is changing with respect to time, or $\frac{dV}{dt}$. We have $V = \frac13\pi r^2h$, so we differentiate both sides with respect to $t$,

$\frac{dV}{dt} = \frac d{dt}\!\left[\frac13\pi r^2h\right]$

$=\frac\pi3\cdot\frac d{dt}\!\left[r^2h\right]$ (constant multiple)

$=\frac\pi3\left[r^2\frac{dh}{dt} + h\frac d{dt}\!\left[r^2\right]\right]$ (product rule)

$=\frac\pi3\left(r^2\frac{dh}{dt}+2rh\frac{dr}{dt} \right)$ (power rule with chain rule)

3. ## Re: This should be obvious to those of you...

Why did you use the chain rule on $r^2$?

4. ## Re: This should be obvious to those of you...

Originally Posted by Nervous
Why did you use the chain rule on $r^2$?
Because we are differentiating with respect to $t$, not $r$.

$\frac d{dr}\!\left[r^2\right] = 2r$

but

$\frac d{dt}\!\left[r^2\right] = 2r\frac{dr}{dt}$.

5. ## Re: This should be obvious to those of you...

I see. That's what I was confused about, differentiating in respect to something other than a simple $\frac d{dx}$

6. ## Re: This should be obvious to those of you...

Originally Posted by Nervous
I see. That's what I was confused about, differentiating in respect to something other than a simple $\frac d{dx}$
Well, the important thing isn't which letter we use but that we are differentiating a function of one variable with respect to a different variable. If we were to find $\frac d{dx}\!\left[r^2\right]$, we would still have to use the chain rule: $\frac d{dx}\!\left[r^2\right] = 2r\frac{dr}{dx}$. The power rule doesn't apply since the variables are different.