question a vertical cylinder is leaking water at a rate of 4 m³/sec. if the cylinder has a height of 10 m…

question a vertical cylinder is leaking water at a rate of 4 m³/sec. if the cylinder has a height of 10 m and a radius of 2 m, at what rate is the height of the water changing when the height is 3 m? submit an exact answer in terms of π. recall that the volume of a cylinder is given by v = πr²h. provide your answer below: dh/dt = □ m/sec

question a vertical cylinder is leaking water at a rate of 4 m³/sec. if the cylinder has a height of 10 m and a radius of 2 m, at what rate is the height of the water changing when the height is 3 m? submit an exact answer in terms of π. recall that the volume of a cylinder is given by v = πr²h. provide your answer below: dh/dt = □ m/sec

Answer

Explanation:

Step1: Differentiate volume formula

The volume of a cylinder is $V=\pi r^{2}h$. Since the radius $r = 2$ is constant, differentiating with respect to time $t$ using the constant - multiple rule, we get $\frac{dV}{dt}=\pi r^{2}\frac{dh}{dt}$.

Step2: Substitute known values

We know that $\frac{dV}{dt}=- 4$ (negative because the volume is decreasing) and $r = 2$. Substituting these values into $\frac{dV}{dt}=\pi r^{2}\frac{dh}{dt}$, we have $-4=\pi\times(2)^{2}\times\frac{dh}{dt}$.

Step3: Solve for $\frac{dh}{dt}$

First, simplify the right - hand side: $\pi\times(2)^{2}\times\frac{dh}{dt}=4\pi\frac{dh}{dt}$. Then, solve the equation $-4 = 4\pi\frac{dh}{dt}$ for $\frac{dh}{dt}$. Divide both sides by $4\pi$: $\frac{dh}{dt}=-\frac{1}{\pi}$.

Answer:

$-\frac{1}{\pi}$