question a vertical cylinder is leaking water at a rate of 6 m³/sec. if the cylinder has a height of 10 m…

question a vertical cylinder is leaking water at a rate of 6 m³/sec. if the cylinder has a height of 10 m and a radius of 4 m, at what rate is the height of the water changing when the height is 5 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 6 m³/sec. if the cylinder has a height of 10 m and a radius of 4 m, at what rate is the height of the water changing when the height is 5 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 $r = 4$m 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 $\frac{dV}{dt}=- 6$m³/sec (negative because the volume is decreasing) and $r = 4$m. Substituting these values into $\frac{dV}{dt}=\pi r^{2}\frac{dh}{dt}$, we have $-6=\pi\times(4)^{2}\times\frac{dh}{dt}$.

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

First, simplify the right - hand side: $\pi\times(4)^{2}\times\frac{dh}{dt}=16\pi\frac{dh}{dt}$. Then, solve the equation $-6 = 16\pi\frac{dh}{dt}$ for $\frac{dh}{dt}$. We get $\frac{dh}{dt}=-\frac{6}{16\pi}=-\frac{3}{8\pi}$m/sec.

Answer:

$-\frac{3}{8\pi}$ m/sec