A vertical cylinder is leaking water at a rate of 4m3/sec. If the cylinder has a height of 10m and a radius of 2m, at what rate is the height of the water changing when the height is 3m?
Answer:
[tex]\dfrac{1}{\pi}$ m/s[/tex]
Step-by-step explanation:
Volume of a cylinder, [tex]V=\pi r^2h[/tex]
When the water is leaking, the radius of the cylinder remains constant while the height changes with respect to time.
Therefore, the rate of change of the volume,
[tex]\dfrac{dV}{dt}=\pi r^2 \dfrac{dh}{dt}[/tex]
[tex]4=\pi \times 2^2 \times \dfrac{dh}{dt}\\\\\dfrac{dh}{dt}=\dfrac{4}{4\pi}\\\\\dfrac{dh}{dt}=\dfrac{1}{\pi}$ m/s[/tex]
