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].
