Struggling with a rate of change problem.

• Feb 12th 2013, 09:53 PM
KhanDisciple
Struggling with a rate of change problem.
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 $\displaystyle \frac{4}{3}\pi r^3$. So I use implicit differentiation and take the derivative of both sides and use the chain rule because I know $\displaystyle r$ is a function of $\displaystyle t$, and I believe this rate is given as -3, so $\displaystyle \frac{dr}{dt}=-3$.

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

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.
• Feb 12th 2013, 10:58 PM
EliteAndoy
Re: Struggling with a rate of change problem.
Hi Khan! Well I'm not really certain of this solution but give it a try(Wink):

1)State the given:

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

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

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

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

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

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 $\displaystyle \frac{dV}{dt}=-cm^3/min$ since it is a negative rate of change :D
• Feb 12th 2013, 10:59 PM
jakncoke
Re: Struggling with a rate of change problem.
--- minut errror gimme a min, nvm the above poster has right answer. cheers
• Feb 12th 2013, 11:14 PM
EliteAndoy
Re: Struggling with a rate of change problem.
Wow did I hehe! Thanks man that really boosted me up on doing this integrals early in the morning:D! 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!
• Feb 12th 2013, 11:18 PM
jakncoke
Re: Struggling with a rate of change problem.
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.
• Feb 12th 2013, 11:20 PM
EliteAndoy
Re: Struggling with a rate of change problem.
Aww man, what a bummer. I thought someone was finally able to solve it using partial derivatives:D. Either way, nice try big man!
• Feb 12th 2013, 11:27 PM
jakncoke
Re: Struggling with a rate of change problem.
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.
• Feb 12th 2013, 11:30 PM
EliteAndoy
Re: Struggling with a rate of change problem.
Exactly my point! So if you are able to solve it using partial derivatives then you are my new hero! :D
• Feb 12th 2013, 11:48 PM
jakncoke
Re: Struggling with a rate of change problem.
But when there is one variable the partial derivate become the normal derivative no? So i can solve it using partial derivatives!.
• Feb 13th 2013, 06:32 AM
KhanDisciple
Re: Struggling with a rate of change problem.
Thank you for your help, but I don't understand why plugging $\displaystyle (\frac{11}{2})$ (which is what the radius) into the equation $\displaystyle 4\pi r^2\frac{dr}{dt}$ does not work.
The equations $\displaystyle 4\pi r^2\frac{dr}{dt}$ ,and $\displaystyle \frac{1}{6}\pi 3d^2 \frac{dD}{dt}$ look equivalent to me in that they both should solve the problem. In other words, how is the equation $\displaystyle V=\frac{4}{3}\pi r^3$ different from the equation $\displaystyle V=\frac{4}{3}\pi(\frac{diameter}{2})^3$ before and after they are differentiated. Thanks.
• Feb 13th 2013, 06:47 AM
jakncoke
Re: Struggling with a rate of change problem.
Quote:

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

you are correct also. They are indeed equivalent. and doing it your way and plugging it $\displaystyle \frac{11}{2}$ for r and $\displaystyle \frac{3}{2}$ for $\displaystyle \frac{dr}{dt}$ will also give you the rate at which volume is changing with time, when diameter is 11.
• Feb 13th 2013, 06:54 AM
KhanDisciple
Re: Struggling with a rate of change problem.
Thank you for your clarification guys, I really appreciate it.