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jonny [76]
2 years ago
15

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 2.3 in/s. At

what rate is the volume of the cone changing when the radius is 110 in. and the height is 151 in.
Mathematics
1 answer:
Angelina_Jolie [31]2 years ago
8 0

Answer:

The volume is changing at a rate given by:

\frac{dV}{dt} =19559.56\,\,\frac{in^3}{s}

Step-by-step explanation:

Let's recall the formula for the volume of acone, since it is the rate of the cone changing what we need to answer:

Volume of cone = \frac{1}{3} B\,*\,H

where B is the area of the base (a circle of radius R) which equals = \pi\,R^2

and where H stands for the cone's height.

We apply the derivative over time operator (\frac{d}{dt}) on both sides of the volume equation, making sure that we apply the rule for the derivative of a product:

V=\frac{1}{3} B\,*\,H\\\\V= \frac{1}{3} \pi\,R^2\,H\\\frac{dV}{dt} =\frac{\pi}{3}\,( \frac{d(R^2)}{dt} H+R^2\,\frac{dH}{dt} )\\\frac{dV}{dt} =\frac{\pi}{3}\,( 2\,R\,\frac{dR}{dt}\, H+R^2\,\frac{dH}{dt} )\\\frac{dV}{dt} =\frac{\pi}{3}\,( 2\,(110\,in)(1.4\,\frac{in}{s} )\,(151\,in)+(110\.in)^2\,(-2.3)\frac{in}{s} )\\\frac{dV}{dt} =19559.56\,\,\frac{in^3}{s}

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