The radius of a right circular cone is increasing at a rate of 1.5 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 107 in.

Respuesta :

Answer:

The volume of the cone is changing at a rate of approximately 8670.796 cubic inches per second.

Step-by-step explanation:

Geometrically speaking, the volume of the right circular cone ([tex]V[/tex]), in cubic inches, is defined by the following formula:

[tex]V = \frac{1}{3}\cdot \pi \cdot r^{2}\cdot h[/tex] (1)

Where:

[tex]r[/tex] - Radius, in inches.

[tex]h[/tex] - Height, in inches.

Then, we derive an expression for the rate of change of the volume ([tex]\dot V[/tex]), in cubic inches per second, by derivatives:

[tex]\dot V = \frac{1}{3}\cdot \pi \cdot (2\cdot r\cdot h \cdot \dot r + r^{2}\cdot \dot h)[/tex] (2)

Where:

[tex]\dot r[/tex] - Rate of change of the radius, in inches per second.

[tex]\dot h[/tex] - Rate of change of the height, in inches per second.

If we know that [tex]r = 120\,in[/tex], [tex]\dot r = 1.5\,\frac{in}{s}[/tex], [tex]h = 107\,in[/tex] and [tex]\dot h = -2.1\,\frac{in}{s}[/tex], then the rate of change of the volume is:

[tex]\dot V \approx 8670.796\,\frac{in^{3}}{s}[/tex]

The volume of the cone is changing at a rate of approximately 8670.796 cubic inches per second.

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