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Math Help - Struggling with a rate of change problem.

  1. #1
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    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 \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.
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  2. #2
    Junior Member EliteAndoy's Avatar
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    Re: Struggling with a rate of change problem.

    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
    Last edited by EliteAndoy; February 12th 2013 at 11:04 PM. Reason: Typo on variables
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    Re: Struggling with a rate of change problem.

    --- minut errror gimme a min, nvm the above poster has right answer. cheers
    Last edited by jakncoke; February 12th 2013 at 11:11 PM.
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    Junior Member EliteAndoy's Avatar
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    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! 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!
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    Senior Member jakncoke's Avatar
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    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.
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    Junior Member EliteAndoy's Avatar
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    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. Either way, nice try big man!
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  7. #7
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    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.
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    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!
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  9. #9
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    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!.
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    Re: Struggling with a rate of change problem.

    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.
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    Senior Member jakncoke's Avatar
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    Re: Struggling with a rate of change problem.

    Quote Originally Posted by KhanDisciple View Post
    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.
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  12. #12
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    Re: Struggling with a rate of change problem.

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