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Math Help - Related Rates with a cone

  1. #1
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    Related Rates with a cone

    The volume of a cone of radius r and height h is given by v=1/3*pi*r^2*h. If the radius and height both increase at a constant rate of 2 cm per second, at which rate in cubic centimeters per second is the volume increasing when the height is 8 cm and the radius is 6 cm.

    I'm kinda puzzled by this problem. Should I do the product rule with the 1/3*pi*r^2*h part? Thanks a lot.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Spoiler:
    Write down what you are given, what you are required to find, etc.

    Given:

    \dfrac{dr}{dt} = 2

    \dfrac{dh}{dt} = 2

    V = \dfrac13 \pi r^2h

    Required:

    \dfrac{dV}{dt} at h = 8, and r = 6.

    You will need to express either r or h in terms of h and r respetively. To find this relation, make a quick sketch of the cone, and you'll see that at the instant asked for:

    \dfrac{h}{r} = \dfrac86

    h = \dfrac{4r}{3}

    This way, you can use the chain rule:

    \dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}

    Substitute h in the formula for V, find the derivative and used the chain rule with what you already know


    EDIT: Sorry, I just realised that this is a type of question I never encountered as both r and h are changing and I see that the chain rule doesn't apply here. Sorry for the misleading.
    Last edited by Unknown008; January 12th 2011 at 09:35 PM.
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  3. #3
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    Quote Originally Posted by zhangxupage2 View Post
    The volume of a cone of radius r and height h is given by v=1/3*pi*r^2*h. If the radius and height both increase at a constant rate of 2 cm per second, at which rate in cubic centimeters per second is the volume increasing when the height is 8 cm and the radius is 6 cm.

    I'm kinda puzzled by this problem. Should I do the product rule with the 1/3*pi*r^2*h part? Thanks a lot.
    sounds like a good plan ... carry it out.
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  4. #4
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    Is Unknown008's plan workable? Why do I get different answers with the product rule and the substitution method?
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  5. #5
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    Quote Originally Posted by zhangxupage2 View Post
    Is Unknown008's plan workable? Why do I get different answers with the product rule and the substitution method?
    How can we know when you haven't shown any working?
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  6. #6
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    Sorry for not posting my work.

    This is what I did:

    dV/dt=pi/3*(2r*dr/dt*h+dh/dt*r^2)

    Plug in the numbers: pi/3 (2*6*2*8+2*6*6)

    Simplify: pi/3 * 264 = 88pi

    Is this correct?

    Thanks a lot!
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  7. #7
    MHF Contributor Unknown008's Avatar
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    I see how my method was wrong now. It involved considering the rate of change of one parameter only (ie either r or h, not both at the same time). I now realise my mistake and I'll not make the same mistake again.
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  8. #8
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    Quote Originally Posted by zhangxupage2 View Post
    Sorry for not posting my work.

    This is what I did:

    dV/dt=pi/3*(2r*dr/dt*h+dh/dt*r^2) Mr F says: This is correct.

    Plug in the numbers: pi/3 (2*6*2*8+2*6*6)

    Simplify: pi/3 * 264 = 88pi

    Is this correct?

    Thanks a lot!
    Assuming your arithmetic is correct (I don't check arithmetic) you will have the correct answer. But you should include the unit too ....
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