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3. The radius r of a sphere is increasing at a constant rate of 0.04 centimeters per second. (Note: The volume of a sphere with radius r is v=4/3pir^3 ).
a. At the time when the radius of the sphere is 10 cm, what is the rate of increase of its volume?
b. At the time when the volume of the sphere is 36pi cubic centimeters, what is the rate of increase of the area of a cross section through the center of the sphere?
c. At the time when the volume and the radius of the sphere are increasing at the same numerical rate, what is the radius?

  • calculus -

    dV/dt = 4πr^2 dr/dt

    a) when r=10 and dr/dt = .04
    dV/dt = 4π(100)(.04) = ...

    b) when V = 36π
    36π = (4/3)πr^3
    27 = r^3
    r = 3
    cross section is a circle, A = πr^2
    dA/dt = 2πr dr/dt
    = 2π(3)(.04)
    = ..

    c) dV/dt = 4πr^2 dr/dt, but dr/dt = dV/dt = .04
    .04 = 4πr^2 (.04)
    4πr^2 = 1
    r^2 = 1/(4π)
    r = 1/(2√π) = ...

  • application of derivatives -

    The radius of an air bubble is increasing at what rate is the volume of the bubble increasing when the radius is 1 centimetre?

  • calculus -

    The volume of sphere is increasing at the rate of the rate increase in its surface area when the radius is 10cm is

  • calculus -

    So confusing

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