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Math Help - help with derivatives (applied problem)

  1. #1
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    help with derivatives (applied problem)

    Can anyone help me figure this problem out?

    Any help is appreciated! Thanks
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  2. #2
    MHF Contributor alexmahone's Avatar
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    What have you tried so far?
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    i answered part a with V=s^3 A=s^2...i dont know what it means to differentiate dV/dt with respect to s...
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    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by colerelm1 View Post
    i answered part a with V=s^3 A=s^2...i dont know what it means to differentiate dV/dt with respect to s...
    Good!

    \displaystyle \frac{dV}{dt}=kA=ks^2

    Do you think k will be positive or negative? (Remember that the volume is decreasing with time.)
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    Did you mean "differentiate V with respect to t? If so, then it's called implicit differentiation:

    \dfrac{dV}{dt} = 3s^2 \dfrac{ds}{dt}
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    Quote Originally Posted by alexmahone View Post
    Good!

    \displaystyle \frac{dV}{dt}=kA=ks^2

    Do you think k will be positive or negative? (Remember that the volume is decreasing with time.)
    ok i know its going to be negative because its decreasing but how do I do part c? I honestly have no clue where to start this one. My thinking is that dV/dt means volume(y axis) with respect to time (x axis) related to some ds/dt where s (y axis) is being differentiated with respect to time (x axis). I know how to read leibniz notation I just can't seem to grasp an understanding of it...
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  7. #7
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by colerelm1 View Post
    ok i know its going to be negative because its decreasing but how do I do part c? I honestly have no clue where to start this one. My thinking is that dV/dt means volume(y axis) with respect to time (x axis) related to some ds/dt where s (y axis) is being differentiated with respect to time (x axis). I know how to read leibniz notation I just can't seem to grasp an understanding of it...
    \displaystyle V=s^3

    Using the Chain Rule,

    \displaystyle \frac{dV}{dt}=\frac{dV}{ds}\frac{ds}{dt}=3s^2\frac  {ds}{dt}
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  8. #8
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    Quote Originally Posted by alexmahone View Post
    \displaystyle V=s^3

    Using the Chain Rule,

    \displaystyle \frac{dV}{dt}=\frac{dV}{ds}\frac{ds}{dt}=3s^2\frac  {ds}{dt}
    ok so for part (d) i got the answer of:

    ks^2 = 3s^2 ds/dt
    which is the same as ds/dt = ks^2 / 3s^2

    Now, for part (e) I don't get how I would use my answer from (d) to write an equation which relates s(0) to s(1) and then use it to find t_melt in terms of the quantity s_1/s_0


    Can anyone help me out?
    Last edited by colerelm1; March 15th 2011 at 02:47 PM.
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