Answer:
The volume of the cylinder with time is increasing approximately at a rate of 16.493 cubic centimeters per minute.
Step-by-step explanation:
The statement is incomplete: The size of a cylinder changes with time. If r increases at the rate of 2 cm/min and h decreases at the rate of 7 cm/min. ¿At what rate is the volume changing at the instant when r = 1 cm and h = 7 cm?
The volume of the cylinder ([tex]V[/tex]), measured in cubic centimeters, is expressed by the following formula:
[tex]V = \frac{\pi}{4}\cdot r^{2}\cdot h[/tex] (1)
Where:
[tex]r[/tex] - Radius, measured in centimeters.
[tex]h[/tex] - Height, measured in centimeters.
The rate of change of the volume ([tex]\frac{dV}{dt}[/tex]), measured in cubic centimeters is obtained by deriving (1) in time:
[tex]\frac{dV}{dt} = \frac{\pi}{2} \cdot r\cdot h\cdot \frac{dr}{dt} + \frac{\pi}{4}\cdot r^{2}\cdot \frac{dh}{dt}[/tex] (2)
Where:
[tex]\frac{dr}{dt}[/tex] - Rate of change of the radius, measured in centimeters per minute.
[tex]\frac{dh}{dt}[/tex] - Rate of change of the height, measured in centimeters per minute.
If we know that [tex]r = 1\,cm[/tex], [tex]h = 7\,cm[/tex], [tex]\frac{dr}{dt} = 2\,\frac{cm}{min}[/tex] and [tex]\frac{dh}{dt} = -7\,\frac{cm}{min}[/tex], then the rate of change of the volume is:
[tex]\frac{dV}{dt} = \frac{\pi}{2}\cdot (1\,cm)\cdot (7\,cm)\cdot \left(2\,\frac{cm}{min} \right) + \frac{\pi}{4}\cdot (1\,cm)^{2}\cdot \left(-7\,\frac{cm}{min} \right)[/tex]
[tex]\frac{dV}{dt} \approx 16.493\,\frac{cm^{3}}{min}[/tex]
The volume of the cylinder with time is increasing approximately at a rate of 16.493 cubic centimeters per minute.
