Struggling with a rate of change problem.

**KhanDisciple** · February 12th 2013, 09:53 PM

A spherical snowball is melting in such a way that its diameter is decreasing at rate of 3 cm/min. At what rate is the volume of the snowball decreasing when the diameter is 11cm?
When the diameter is 11cm, the volume of the snowball is decreasing at a rate of _______ .

I know that the volume of a sphere is $\frac{4}{3}\pi r^3$ . So I use implicit differentiation and take the derivative of both sides and use the chain rule because I know $r$ is a function of $t$ , and I believe this rate is given as -3, so $\frac{dr}{dt}=-3$ .

$\frac{dV}{dt}=4 \pi r^2 \frac{dr}{dt}$
Since the radius of a circle is one half the diameter... $\frac{dV}{dt}=4 \pi (\frac{11}{2})^2 (-3)$ , however, this answer is incorrect.

I was given a hint that I have to write an equation linking the changing volume V(t) to the changing diameter d(t), where $d(t)=2r(t)$

I don't understand how to figure this out. Also, I don't have the answer so please don't give it to me. Thanks in advance.

**EliteAndoy** · February 12th 2013, 10:58 PM

Hi Khan! Well I'm not really certain of this solution but give it a try

:

1)State the given:

$\frac{dD}{dt}=-3cm/min$ , $D=11cm$ , so we attempt to find $\frac{dV}{dt}$

2)Derive a formula for Volume as a function of diameter:

Volume of a sphere is given as $V=\frac{4}{3}\pi r^3$ , but since $D=2r$ , then $r= \frac{D}{2}$
Then we plug r back in to get: $V=\frac{4}{3}\pi (\frac{D}{2})^3$
So that by simplifying, we get: $V= \frac{1}{6}\pi D^3$

3)Take the derivative of the equation with respect to time:

We get: $\frac{dV}{dt}=\frac{1}{6}\pi 3D^2 \frac{dD}{dt}$

Take it from here dude. Remember that you only have to plug in all the values that is on part 1 of this solution. And $\frac{dV}{dt}=-cm^3/min$ since it is a negative rate of change

**jakncoke** · February 12th 2013, 10:59 PM

--- minut errror gimme a min, nvm the above poster has right answer. cheers

**EliteAndoy** · February 12th 2013, 11:14 PM

Wow did I hehe! Thanks man that really boosted me up on doing this integrals early in the morning

! Anyways dude, I saw your solution, and for some reason you used partial derivatives. Damn, if that worked I would love you to educated me more about it. Cheers!

**jakncoke** · February 12th 2013, 11:18 PM

It did work (As for the partial derivative i dont know how to use latex for normal derivative so i erronously use the partial symbol, its a very bad habit i know!), but i didnt read his question correctly so i made it much more complicated than it had to be, so instead of fixing it, i saw your very correct solution and said, bah humbug.

**EliteAndoy** · February 12th 2013, 11:20 PM

Aww man, what a bummer. I thought someone was finally able to solve it using partial derivatives

. Either way, nice try big man!

**jakncoke** · February 12th 2013, 11:27 PM

I'm not sure where you would use partial derivatives in this case as Volume is a function of one variable D (which itself is a function of variable t) ultimately it is still a function of one variable whether it is r or d or t. If he had said the diamater varied according to time and some other variable, then we could possible use partial derivatives.

**EliteAndoy** · February 12th 2013, 11:30 PM

Exactly my point! So if you are able to solve it using partial derivatives then you are my new hero!

**jakncoke** · February 12th 2013, 11:48 PM

But when there is one variable the partial derivate become the normal derivative no? So i can solve it using partial derivatives!.

**KhanDisciple** · February 13th 2013, 06:32 AM

Thank you for your help, but I don't understand why plugging $(\frac{11}{2})$ (which is what the radius) into the equation $4\pi r^2\frac{dr}{dt}$ does not work.
The equations $4\pi r^2\frac{dr}{dt}$ ,and $\frac{1}{6}\pi 3d^2 \frac{dD}{dt}$ look equivalent to me in that they both should solve the problem. In other words, how is the equation $V=\frac{4}{3}\pi r^3$ different from the equation $V=\frac{4}{3}\pi(\frac{diameter}{2})^3$ before and after they are differentiated. Thanks.

**jakncoke** · February 13th 2013, 06:47 AM

Originally Posted by KhanDisciple

Thank you for your help, but I don't understand why plugging $(\frac{11}{2})$ (which is what the radius) into the equation $4\pi r^2\frac{dr}{dt}$ does not work.
The equations $4\pi r^2\frac{dr}{dt}$ ,and $\frac{1}{6}\pi 3d^2 \frac{dD}{dt}$ look equivalent to me in that they both should solve the problem. In other words, how is the equation $V=\frac{4}{3}\pi r^3$ different from the equation $V=\frac{4}{3}\pi(\frac{diameter}{2})^3$ before and after they are differentiated. Thanks.

you are correct also. They are indeed equivalent. and doing it your way and plugging it $\frac{11}{2}$ for r and $\frac{3}{2}$ for $\frac{dr}{dt}$ will also give you the rate at which volume is changing with time, when diameter is 11.

**KhanDisciple** · February 13th 2013, 06:54 AM

Thank you for your clarification guys, I really appreciate it.

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