3679 發問於 科學及數學數學 · 7 年前

M2 rate of change help plz!!

Water is pumped out at a rate of 3π cm3 per second from a cup that is in the shape of an inverted right circular cone. The height of the cup is 15 cm and its base radius is 5 cm. How fast is the water level falling when the depth of the water is 3 cm?

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  • 7 年前
    最愛解答

    Answer: 3 cm/s

    Solutions:

    let h cm and r cm be the depth and the base radius of water respectively

    by similar triangle, 15 : 5 = h : r => r = h / 3

    the volume of water = π r^2 h / 3 = π h^3 / 27

    by differentiate both sides w.r.t. t,

    d(the volume of water) / dt = (π h^2 / 9)(dh/dt)

    - 3π = (π h^2 / 9)(dh/dt)

    dh/dt = -27 / h^2

    when the depth of the water is 3 cm, dh/dt = -27 / (3)^2 = -3

    thus, the water level level is falling at a rate of 3 cm / second

    資料來源: knowledge
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