The radius of a cylinder is increasing at a constant rate of 2 inches per minute. The volume remains a constant 440 cubic inches. At the instant when the radius of the cylinder is 44 inches, what is the rate of change of the height? The volume of a cylinder can be found with the equation V= [tex]\pi r^2h[/tex] Round your answer to three decimal places.

Answer:

The rate of change of the height is approximately [tex]-6.577\times 10^{-3}\,\frac{in}{min}[/tex].

Step-by-step explanation:

From Geometry we know that volume of cylinder ([tex]V[/tex]), measured in cubic inches, is expressed by:

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

Where:

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

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

Given that volume remains constant, the expression for the rate of change of the height ([tex]\dot h[/tex]), measured in inches per minute, is:

[tex]2\pi\cdot r \cdot h \cdot \dot r + \pi \cdot r^{2}\cdot \dot h = 0[/tex]

[tex]2\cdot h\cdot \dot r + r\cdot \dot h = 0[/tex]

[tex]r\cdot \dot h = -2\cdot h \cdot \dot r[/tex]

[tex]\dot h = -\frac{2\cdot h\cdot \dot r}{r}[/tex]

[tex]\dot h = - \left(\frac{2\cdot \dot r}{r}\right)\cdot \left(\frac{V}{\pi\cdot r^{2}} \right)[/tex]

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

Where [tex]\dot r[/tex] is the rate of change of the radius, measured in inches per minute.

If we know that [tex]r = 44\,in[/tex], [tex]\dot r = 2\,\frac{in}{min}[/tex] and [tex]V = 440\,in^{3}[/tex], then the rate of change of the height is:

[tex]\dot h = - \frac{2\cdot \left(2\,\frac{in}{min} \right)\cdot \left(440\,in^{3}\right)}{\pi\cdot (44\,in)^{3}}[/tex]

[tex]\dot h \approx -6.577\times 10^{-3}\,\frac{in}{min}[/tex]

The rate of change of the height is approximately [tex]-6.577\times 10^{-3}\,\frac{in}{min}[/tex].

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