Rate problem

**daigo** · July 10th 2012, 03:45 AM

A snowball melts in volume @ 1 cm³/min. At what rate is the diameter decreasing when the diameter is 10 cm?

Volume of a sphere: $\frac{4}{3}\pi r^{3}$
Volume of a sphere in terms of diameter: $\frac{\pi d^{3}}{6}$

Rate of volume: $\frac{d}{d(time)}volume = -1 cm^{3}/min.$

Rate of diameter: $\frac{d}{d(time)}10 cm = ?$

I'm not sure what to do at this point

**emakarov** · July 10th 2012, 04:03 AM

Let's denote the diameter by D. Then $\frac{dV}{dt}=\frac{dV}{dD}\frac{dD}{dt}$ (d-d-d-t

). You know $\frac{dV}{dt}$ and $\frac{dV}{dD}$ , so you can find $\frac{dD}{dt}$ .

**daigo** · July 10th 2012, 04:07 AM

Originally Posted by emakarov

Let's denote the diameter by D. Then $\frac{dV}{dt}=\frac{dV}{dD}\frac{dD}{dt}$ (d-d-d-t

).

I don't see how $\frac{dV}{dt}=\frac{dV}{dD}\frac{dD}{dt}$ ?

But I found:
$<br /> \frac{dV}{dD} = \frac{\pi}{6} \cdot [D^{3}]' \\<br /> = \frac{\pi}{6} \cdot 3D^{2} \\<br /> = \frac{\pi D^{2}}{2}$

**tom@ballooncalculus** · July 10th 2012, 04:12 AM

Just in case a picture helps...

... where (key in spoiler) ...

Spoiler:

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Balloon Calculus Drawing with LaTeX and Asymptote!

**emakarov** · July 10th 2012, 04:12 AM

Don't you see that we can cancel dD? (Half kidding.) Seriously, this is the chain rule.

**daigo** · July 10th 2012, 04:25 AM

The one thing I am not understanding is where $\frac{dV}{dD}$ came from. How did you know that you need to take the derivative of the volume with respects to the diameter for the problem?

**tom@ballooncalculus** · July 10th 2012, 04:31 AM

By seeing that you are trying to differentiate a composite function of t with respect to t. And the chain rule says this will involve differentiating with respect to the whole of the inner function of t. (D is the inner function of t.)

**HallsofIvy** · July 10th 2012, 04:33 AM

$\frac{dV}{dt}= \frac{dV}{dD}\frac{dD}{dt}$ is the chain rule. Check your Calculus text.

The problem asks you to find $\frac{dD}{dt}$ and you are told $\frac{dV}{dt}$ . " $dV/dD$ ", which you can easily get from the formula for V in terms of D, is the "missing link" in the chain rule.

**emakarov** · July 10th 2012, 04:36 AM

We choose to represent the function V(t) as V(D(t)). The chain rule says that $\frac{dV}{dt}=\frac{dV}{dD}\frac{dD}{dt}$ .

Properly speaking, where something came from is not a mathematical question. The solution to this question is a sequence of statements, each of which is either true or false (hopefully, true). The only proper question is why a certain statement is true.

Edit: Yes, it's legitimate to ask metamathematical questions, such as what reasoning one used to come up with a certain solution. Hopefully, post #8 answered this question.

**daigo** · July 10th 2012, 05:04 AM

It makes more sense to use the Chain rule now that I see there is a function V(D(t)). I did not know this was the function that I was supposed to use to solve the problem. The issue I am having is how to formulate the problem so I can solve it. I did not know to express the volume as the diameter when the time changes the diameter...I am still a bit confused as to how I can get this, but I think I will skip this one for now and come back to it later because I have other problems I need to finish as well as other classes.

**daigo** · July 10th 2012, 05:38 AM

Oh, I see now. Taking the derivative of the Volume with respect to time:

$\frac{dV}{dt} = [\frac{\pi d^{3}}{6}]'$

However, taking the derivative of the diameter results in using the Chain rule which I get.

I just didn't know how to come up with the formula: $\frac{dV}{dt} = [\frac{\pi d^{3}}{6}]'$ to solve this. All I needed was to know how to arrive at $\frac{dV}{dt} = [\frac{\pi d^{3}}{6}]'$ which I didn't know how to in the first place.

**HallsofIvy** · July 10th 2012, 05:53 AM

Originally Posted by daigo

It makes more sense to use the Chain rule now that I see there is a function V(D(t)). I did not know this was the function that I was supposed to use to solve the problem.

There is never a function that your are supposed to use to solve a problem- look at all functions and formulas that have anything to do with the problem.
In this case you are told that "a snowball melts in volume @ 1 cc/min" so the rate of change of V with respect to t is surely relevant.

The issue I am having is how to formulate the problem so I can solve it. I did not know to express the volume as the diameter when the time changes the diameter...I am still a bit confused as to how I can get this, but I think I will skip this one for now and come back to it later because I have other problems I need to finish as well as other classes.

**tom@ballooncalculus** · July 10th 2012, 05:58 AM

Originally Posted by daigo

All I needed was to know how to arrive at $\frac{dV}{dt} = [\frac{\pi d^{3}}{6}]'$ which I didn't know how to in the first place.

Think of it as balancing.

V = f(d)

would imply

3*V = 3*f(d)

and

2 + V = 2 + f(d)

etc... ?

So wouldn't it imply...

$\frac{d}{dt} V = \frac{d}{dt} f(d)$

... ?

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