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another rates of change question watch

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    a hollow right circular cone whose height is twice the radius is held with the vertex downwards and water is poured in at the rate of 4 cm^3s^-1

    show when the depth of the water = h, V= (pi x h^3) / 12

    dv/dt = 4

    dh/dt = dv/dt x dh/dv

    dh/dv = 2/3 x pi x r x h

    I said r = h/2
    and ended up with (8 x pi x h^2) / 6 so ive obviously gone wrong

    anyone able to show me where i went wrong
    thanks
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    (Original post by hatsuko)
    a hollow right circular cone whose height is twice the radius is held with the vertex downwards and water is poured in at the rate of 4 cm^3s^-1

    show when the depth of the water = h, V= (pi x h^3) / 12

    dv/dt = 4

    dh/dt = dv/dt x dh/dv

    dh/dv = 2/3 x pi x r x h

    I said r = h/2
    and ended up with (8 x pi x h^2) / 6 so ive obviously gone wrong

    anyone able to show me where i went wrong
    thanks
    You are still making exactly the same error as in the previous question

    Ask your teacher
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    (Original post by hatsuko)
    a hollow right circular cone whose height is twice the radius is held with the vertex downwards and water is poured in at the rate of 4 cm^3s^-1

    show when the depth of the water = h, V= (pi x h^3) / 12

    dv/dt = 4

    dh/dt = dv/dt x dh/dv

    dh/dv = 2/3 x pi x r x h

    I said r = h/2
    and ended up with (8 x pi x h^2) / 6 so ive obviously gone wrong

    anyone able to show me where i went wrong
    thanks
    So what is V?

    V= \frac{1}{3} \pi r^2 h

    You need 'h' and no 'r'. So what could you replace 'r' with?
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    (Original post by pleasedtobeatyou)


    So what is V?

    V= \frac{1}{3} \pi r^2 h

    You need 'h' and no 'r'. So what could you replace 'r' with?
    h/2 ?
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    (Original post by hatsuko)
    h/2 ?
    Yes.
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    (Original post by pleasedtobeatyou)
    Yes.
    i subd that in and differentiated and got the answer, but why dont you have to time by 4?
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    (Original post by hatsuko)
    i subd that in and differentiated and got the answer, but why dont you have to time by 4?
    Find (h/2)^2 and substitute it back into the equation for the volume of the cone.

    Remember to square both the numerator and denominator.
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    (Original post by pleasedtobeatyou)
    Find (h/2)^2 and substitute it back into the equation for the volume of the cone.

    Remember to square both the numerator and denominator.
    oh i see what u mean, i thought you had to x4 because water is going in ar 4 cm^3s^-1, but its just asking for the volume at h
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    (Original post by pleasedtobeatyou)
    Find (h/2)^2 and substitute it back into the equation for the volume of the cone.

    Remember to square both the numerator and denominator.
    there is a second part, find in terms of pi, the rate of rise of the water when the depth is 2cm I got

    dh/dt = dv/dt x dh/dv

    dh/dv = pih^3 / 12

    h=2

    answer = 8pi/3, does this look right?
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    (Original post by hatsuko)
    there is a second part, find in terms of pi, the rate of rise of the water when the depth is 2cm I got

    dh/dt = dv/dt x dh/dv

    dh/dv = pih^3 / 12

    h=2

    answer = 8pi/3, does this look right?
    Seriously - ASK YOUR TEACHER
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    (Original post by hatsuko)
    ...
    I have to agree with TenOfThem here.

    There's no point using the chain rule when you don't understand the basic rules of differentiation.

    Ask your teacher for help/clarification ASAP.
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    (Original post by pleasedtobeatyou)
    I have to agree with TenOfThem here.

    There's no point using the chain rule when you don't understand the basic rules of differentiation.

    Ask your teacher for help/clarification ASAP.
    oh, its wrong then?
 
 
 
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