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 3 inches per se

Question

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 3 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

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Nevaeh 2 weeks 2021-09-13T13:58:31+00:00 1 Answer 0

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    2021-09-13T14:00:07+00:00

    Answer:

    The volume of cone is increasing at a rate 2512 cubic inches per second.                                  

    Step-by-step explanation:

    We are given the following in the question:

    \dfrac{dr}{dt} = 3\text{  inches per second}\\\\\dfrac{dh}{dt} = -3\text{  inches per second}

    Radius = 40 inches    

    Height = 30 inches

    The volume of cone is given by:

    V = \dfrac{1}{3}\pi r^2 h

    Rate of change of volume is given by:

    \dfrac{dV}{dt} = \dfrac{1}{3}\pi (2r\dfrac{dr}{dt}h+\dfrac{dh}{dt}r^2)

    Putting the values, we get,

    \dfrac{dV}{dt} = \dfrac{1}{3}(3.14) \big(2(40)(3)(30)+(-3)(40)^2\big)\\\\\dfrac{dV}{dt} = \dfrac{1}{3}(3.14)(2400)\\\\\dfrac{dV}{dt} = 2512

    Thus, the volume of cone is increasing at a rate 2512 cubic inches per second.

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