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? recall that the volume of a cylinder is given by v = πr²h. provide your answer below. dtdh = □ m/sec

Explanation:

Step1: Recall the volume formula for a cylinder

The volume of a cylinder is $V = \pi r^{2}h$. Here, the radius $r = 2$m is constant. So $V=4\pi h$.

Step2: Differentiate both sides with respect to time $t$

Using the chain - rule, $\frac{dV}{dt}=4\pi\frac{dh}{dt}$.

Step3: Substitute the given value of $\frac{dV}{dt}$

We know that $\frac{dV}{dt}=- 4$m³/sec (negative because the volume is decreasing due to leakage).
So, $-4 = 4\pi\frac{dh}{dt}$.

Step4: Solve for $\frac{dh}{dt}$

Dividing both sides by $4\pi$, we get $\frac{dh}{dt}=-\frac{1}{\pi}$ m/sec.

Answer:

$-\frac{1}{\pi}$ m/sec