Calculus

An inverted cone has a diameter of 42 in and a height of 15in. If the water flowing out of the container is 35pi in^3/sec , how fast is the depth of the water dropping when the height is 5in?

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  1. let the height of water be h in, let the radius of the water surface be r in
    By similar triangles, r/h = 21/15 = 7/5
    so r = 7h/5

    V = (1/3)pi(r^2)h
    = (1/3)pi(49h^2/25)(h
    = (49/75)pi(h^3)
    dV/dt = (49/25)pi(h^2)dh/dt
    -35pi = (49/25)pi(25)dh/dt

    solve for dh/dt

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  2. -5/7

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