Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Sand poured on the ground at the rate of 3.00 m^3/min . . .?

forms a conical pile whose height is one-third the diameter of its base. How fast is the altitude of the pile increasing when the radius of its base is 2.00 m?

This question is an application of derivatives. I really don't know where to start . . thanks for any help!

2 Answers

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  • Dr D
    Lv 7
    1 decade ago
    Favorite Answer

    r = radius at any point t.

    h = 2r/3 = height at any point t.

    V = 1/3 * h * πr^2 = 2/9 * π r^3

    dV/dt = dV/dr * dr/dt

    = 2/9 * πr^2 * dr/dt

    Given that dV/dt = 3 m^3 / min

    dr/dt = 3 / (2/9 * πr^2)

    when r = 2,

    dr/dt = 27 / (8π) = 1.074 m/s

    dh/dt = 2/3 * dr/dt = 0.7162 m/s

  • 1 decade ago

    vol = V = 1/3 pi r^3 h

    h = dia/3 . . . . or . . . . . dia = 3h = 2 r

    r = 3/2 h. . . . . . . substitute to the above equation

    V = 1/3 pi ( 3/2 h )^3 h = 9/8 h^4

    dV/dt = 9/8 pi h^3 dh/dt . . . . differentiating

    dV/dt = 9/8 pi h^3 dh/dt . . . . change h = dia / 3 = 2 r /3

    dV/dt = 9/8 pi ( 2 r /3 )^3 dh/dt

    dV/dt = 1/3 pi r ^3 dh/dt

    dh/dt = 3 dV/dt / (pi r^3) = 3 ( 3 ) / ( pi 2^3 ) = 9 / 8 pi

    dh/dt = 0.358 meter / min

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