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Math Help - How do I solve the rate of volume change of a cone?

  1. #1
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    How do I solve the rate of volume change of a cone?

    The radius of a right circular cone is increasing at a rate of 3 inches per second and its height is decreasing at a rate of 5 inches per second. At what rate is the volume of the cone changing when the radius is 40 inches and the height is 30 inches?

    ? cubic inches per second


    If someone could help solve this so I can use it to study for my exam that'd be spectacular!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by pseizure2000 View Post
    The radius of a right circular cone is increasing at a rate of 3 inches per second and its height is decreasing at a rate of 5 inches per second. At what rate is the volume of the cone changing when the radius is 40 inches and the height is 30 inches?

    ? cubic inches per second


    If someone could help solve this so I can use it to study for my exam that'd be spectacular!
    first, we must know the formula for the volume of a right-circular cone.

    V = \frac 13 \pi r^2 h

    now we differentiate implicitly with respect to t

    \Rightarrow \frac {dV}{dt} = \frac 23 \pi rh~\frac {dr}{dt} + \frac 13 \pi r^2~\frac {dh}{dt}

    now, we were given all the variables besides \frac {dV}{dt}, which is what we must find. so just plug them in. (Note, when something is decreasing, we treat its rate as negative)
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  3. #3
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    Hello, pseizure2000!

    The radius of a right circular cone is increasing at a rate of 3 inches per second
    and its height is decreasing at a rate of 5 inches per second.
    At what rate is the volume of the cone changing
    when the radius is 40 inches and the height is 30 inches?

    The volume of a cone is: . V \;=\;\frac{\pi}{3}r^2h

    Differentiate with respect to time: . \frac{dV}{dt} \;=\;\frac{\pi}{3}r^2\left(\frac{dh}{dt}\right) + \frac{\pi}{3}2rh\left(\frac{dr}{dt}\right)

    We are given: . \frac{dr}{dt} = 3,\;\frac{dh}{dt} = -5,\;r = 40,\;h = 30

    Plug them in: . \frac{dV}{dt}\;=\;\frac{\pi}{3}(40^2)(-5) + \frac{2\pi}{3}(40)(30)(3) \;=\;-\frac{8000\pi}{3} + \frac{7200\pi}{3}


    Therefore: . \frac{dV}{dt} \;=\;-\frac{800\pi}{3} in³/sec


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    thank you! very well done and easy to understand!
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