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Math Help - Frustrum rate of change

  1. #1
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    Frustrum rate of change

    A shock wave is advancing in the shape of a frustrum of a right circular cone. The upper radius, x of the frustrum is increasing at the rate of 2m per sec, the lower radius, y is increasing at the rate of 3m per sec, and the height, z is increasing at the rate of 1.5m per sec. Using the chain rule find the rate of increase of the volume, V of the frustrum at the moment when x = 10m, y = 16m and z = 8m.
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  2. #2
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    Hello, Shaun!

    A shock wave is advancing in the shape of a frustrum of a right circular cone.
    The upper radius, x of the frustrum is increasing at the rate of 2m/sec,
    the lower radius, y is increasing at the rate of 3m/sec,
    and the height, z is increasing at the rate of 1.5m/sec.

    Using the chain rule find the rate of increase of the volume, V of the frustrum
    at the moment when x = 10m, y = 16m and z = 8m.
    The volume of a frustum of a cone (or pyramid) is given by:
    . . . . . . . . . . . . . . . _____
    . . V .= .(h/3)(B1 + √B1B2 + B2)

    where h is the height and B1, B2 are the areas of the two bases.


    Our problem has: .B1 = πx, B2 = πy, h = z.
    . . . . . . . . . . . . . . . . . . .______
    Then: .V .= .(z/3)(πx + √πxπy + πy)

    . .or: . V .= .(π/3)z(x + xy + y)


    Can you finish it now?
    Differentiate with respect to time (product rule and chain rule)
    . . then plug in the given values.

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  3. #3
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    Quote Originally Posted by Shaun Gill View Post
    A shock wave is advancing in the shape of a frustrum of a right circular cone. The upper radius, x of the frustrum is increasing at the rate of 2m per sec, the lower radius, y is increasing at the rate of 3m per sec, and the height, z is increasing at the rate of 1.5m per sec. Using the chain rule find the rate of increase of the volume, V of the frustrum at the moment when x = 10m, y = 16m and z = 8m.
    The volume V of the frustrum is
    V = (pi/3)z[x^2 +y^2 +xy]
    Differentiate both sides with respect to time t,
    dV/dt = (pi/3){(dz/dt)[x^2 +y^2 +xy] +z[2x(dx/dt) +2y(dy/dt) +x(dy/dt) +y(dx/dt)}
    Substituting all the givens,
    dV/dt = (pi/3){(1.5)[10^2 +16^2 +10*16] +(8)[2(10)(2) +2(16)(3) +10(3) +16(2)]}
    dV/dt = (pi/3){774 +1584}
    dV/dt = 786pi

    Therefore, at that instant, the volume is increasing at the rate of 786pi cubic meters per second. -------------answer.
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