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Math Help - Rates of change - surface area vs. volume

  1. #1
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    Rates of change - surface area vs. volume

    At first this question seemed easy, however I still can't get the answer. I've been using the formula:

    dx/dt = dx/dy * dy/dt

    The surface area of a cube is changing at the rate of 8cm^2/second. How fast is the volume changing when the surface area is 60 cm?

    Thanks for your help
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  2. #2
    Super Member malaygoel's Avatar
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    Quote Originally Posted by freswood
    At first this question seemed easy, however I still can't get the answer. I've been using the formula:

    dx/dt = dx/dy * dy/dt

    The surface area of a cube is changing at the rate of 8cm^2/second. How fast is the volume changing when the surface area is 60 cm?

    Thanks for your help
    S=6a^2=60cm^2
    \frac{dS}{dt}=12a\frac{da}{dt}=8cm^2/s
    \frac{dV}{dt}=3a^2\frac{da}{dt}
    Get the value of a from first equation
    Divide equation 3 by equation 2
    You will get the answer

    KeepSmiling
    Malay
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  3. #3
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    I'm quite confused

    What is dx/dy? I'm not really sure how to relate surface area and volume. And what's the final answer? I want to see if you get the same answer as what's in the book.
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  4. #4
    Super Member malaygoel's Avatar
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    Quote Originally Posted by freswood
    I'm quite confused

    What is dx/dy? I'm not really sure how to relate surface area and volume. And what's the final answer? I want to see if you get the same answer as what's in the book.
    I think the answer is 2\sqrt{10}

    KeepSmiling
    Malay
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  5. #5
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    I'm still not sure where you got those values from.

    dx/dt = 8, but what is dx/dy and how did you get it? Is it a formula for surface area with respect to volume?
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  6. #6
    Super Member malaygoel's Avatar
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    Quote Originally Posted by freswood
    I'm still not sure where you got those values from.

    dx/dt = 8, but what is dx/dy and how did you get it? Is it a formula for surface area with respect to volume?
    I am sorry but I don't want to give you a solution straightaway.
    Answer these simple questions and solve the problem yourself.
    1)What do you intend to find?Write down an expression for it.
    2)I am listing a few things. Tell whether you can find their value, if yes, what is their value:
    S
    a
    V
    dS/dt
    da/dt
    dS/da
    dV/da
    dS/dV
    dV/dS
    3)Write an expression for dV/dt. What values do you need for it?Can you find those values in ques.2?
    I hope that you can solve the problem now. I still you have any problem, post it and i will try to tell you, straightaway this time.
    To find dy/dx . There are two cases(for explicit functions):you want to find dy/dx in terms or yuo want to find dy/dx in terms of y.
    To find in terms of x, express y as a function of x and differentiate.
    to find in terms of y, express x as a function of y and differentiate.

    Keep Smiling
    Malay
    Last edited by malaygoel; June 27th 2006 at 11:26 PM.
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  7. #7
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    Quote Originally Posted by malaygoel
    I am sorry but I don't want to give you a solution straightaway.
    Answer these simple questions and solve the problem yourself.
    1)What do you intend to find?Write down an expression for it.
    2)I am listing a few things. Tell whether you can find their value, if yes, what is their value:
    S
    a
    V
    dS/dt
    da/dt
    dS/da
    dV/da
    dS/dV
    dV/dS
    3)Write an expression for dV/dt. What values do you need for it?Can you find those values in ques.2?
    I hope that you can solve the problem now. I still you have any problem, post it and i will try to tell you, straightaway this time.
    To find dy/dx . There are two cases(for explicit functions):you want to find dy/dx in terms or yuo want to find dy/dx in terms of y.
    To find in terms of x, express y as a function of x and differentiate.
    to find in terms of y, express x as a function of y and differentiate.

    Keep Smiling
    Malay
    1) dV/dt
    2)
    S = 6a^2 where a is side length
    a = 10^.5
    V = a^3
    dS/dt = 8
    da/dt
    dS/da = 12a
    dV/da = 3a^2
    dS/dV
    dV/dS

    3) For dV/dt you need dx/dt and dS/dV

    But that's the whole problem - I don't know how to find dS/dV. I know you think you're doing the right thing, and I appreciate your help, but I like to be able to work backwards. This question is causing me a lot of grief. I'm doing maths a year ahead, and I find it hard to keep up with the older girls.
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  8. #8
    Super Member malaygoel's Avatar
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    Quote Originally Posted by freswood
    1) dV/dt
    2)
    S = 6a^2 where a is side length
    a = 10^.5
    V = a^3
    dS/dt = 8
    da/dt
    dS/da = 12a
    dV/da = 3a^2
    dS/dV
    dV/dS

    3) For dV/dt you need dx/dt and dS/dV
    No you don't
    Now see.
    dV/dt = \frac{dV}{da}*\frac{da}{dt}
    dV/da=3a^2(you know the value of a)
    You couldn't find da/dt
    You know dS/dt = 8
    \frac{dS}{da}*\frac{da}{dt}=8
    Find the value of da/dt from here and the question is solved.

    KeepSmiling
    Malay
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  9. #9
    Grand Panjandrum
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    Quote Originally Posted by freswood
    At first this question seemed easy, however I still can't get the answer. I've been using the formula:

    dx/dt = dx/dy * dy/dt

    The surface area of a cube is changing at the rate of 8cm^2/second. How fast is the volume changing when the surface area is 60 cm?

    Thanks for your help
    I think (but am probably wrong) the problem here is that x,t,and y
    in dx/dt = dx/dy * dy/dt, are serving as place holders for whatever is
    relevant to the problem at hand.

    For this problem you replace x by V (representing the volume of the
    cube) and y by S (representing the surface area of the cube) and let
    t represent time. Then your ralation between derivatives becomes:

    dV/dt=dV/dS * dS/dt.

    You are told dS/dt=8 (cm^2/s), so you need to find dV/dS.

    The volume V of a cube of side a is a^3, and the surface area of
    such a cube is 6a^2, so expressing V in terms of S gives:

    V=(S/6)^(3/2),

    and the rest should be plain sailing.

    RonL
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  10. #10
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    That's brilliant! Thanks so much! It was just that I couldn't understand how to relate surface area to volume, so couldn't come up with the equation dS/dV.
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