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malfutka [58]
3 years ago
8

The radius of a right circular cone is increasing at a rate of 1.4 in/s while its height is decreasing at a rate of 2.1 in/s. At

what rate is the volume of the cone changing when the radius is 120 in. and the height is 175 in.
Mathematics
1 answer:
kow [346]3 years ago
6 0

Given:

\dfrac{dr}{dt}=1.4\text{ in/s}

\dfrac{dh}{dt}=-2.1\text{ in/s}

To find:

The rate of change in volume at r=120\text{ in. and }175\text{ in.}

Solution:

We know that, volume of a cone is

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

Differentiate with respect to t.

\dfrac{dV}{dt}=\dfrac{1}{3}\pi\times \left[(r^2\dfrac{dh}{dt}) + h(2r\dfrac{dr}{dt})\right]

Substitute the given values.

\dfrac{dV}{dt}=\dfrac{1}{3}\times \dfrac{22}{7}\times \left[(120)^2(-2.1) +175(2)(120)(1.4)\right]

\dfrac{dV}{dt}=\dfrac{22}{21}\times \left[-30240+58800\right]

\dfrac{dV}{dt}=\dfrac{22}{21}\times 28560

\dfrac{dV}{dt}=29920

Therefore, the volume of decreased by 29920 cubic inches per second.

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