this question has several parts that must be completed sequentially. if you skip a part of the question, you…

this question has several parts that must be completed sequentially. if you skip a part of the question, you will not receive any points for the skipped part, and you back to the skipped part. tutorial exercise a cylindrical tank with radius 8 m is being filled with water at a rate of 2 m³/min. how fast is the height of the water increasing? step 1 if h is the water’s height, the volume of the water is v = πr²h. we must find dv/dt. differentiating both sides of the equation gives the following. dv/dt = dh/dt

this question has several parts that must be completed sequentially. if you skip a part of the question, you will not receive any points for the skipped part, and you back to the skipped part. tutorial exercise a cylindrical tank with radius 8 m is being filled with water at a rate of 2 m³/min. how fast is the height of the water increasing? step 1 if h is the water’s height, the volume of the water is v = πr²h. we must find dv/dt. differentiating both sides of the equation gives the following. dv/dt = dh/dt

Answer

Explanation:

Step1: Differentiate volume formula

Given $V = \pi r^{2}h$, where $r$ is a constant ($r = 8$ m). Using the constant - multiple rule of differentiation, $\frac{dV}{dt}=\pi r^{2}\frac{dh}{dt}$. Since $r = 8$, we have $\frac{dV}{dt}=\pi\times8^{2}\frac{dh}{dt}=64\pi\frac{dh}{dt}$.

Step2: Substitute known values

We know that $\frac{dV}{dt}=2$ m³/min. Substituting into the equation $\frac{dV}{dt}=64\pi\frac{dh}{dt}$, we get $2 = 64\pi\frac{dh}{dt}$.

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

To find $\frac{dh}{dt}$, we can rewrite the equation as $\frac{dh}{dt}=\frac{2}{64\pi}=\frac{1}{32\pi}$ m/min.

Answer:

$\frac{1}{32\pi}$ m/min