Question

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 waters height, the volume of the water is v = πr²h. we must find dv/dt. differentiating both sides of the equation gives the following.
\\(\frac{dv}{dt}=\frac{1}{16}\times\pi r^{2}\frac{dh}{dt}\\)
step 2
substituting for r, this becomes \\(\frac{dv}{dt}=\square\times\frac{dh}{dt}\\).
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Explanation:

Step1: Differentiate volume formula

Given $V = \pi r^{2}h$, using the product - rule (since $r$ is constant here, $\frac{dV}{dt}=\pi r^{2}\frac{dh}{dt}$).

Step2: Substitute the value of $r$

We know $r = 8$ m. Substituting $r = 8$ into $\frac{dV}{dt}=\pi r^{2}\frac{dh}{dt}$, we get $\frac{dV}{dt}=\pi\times8^{2}\frac{dh}{dt}=64\pi\frac{dh}{dt}$.

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

We are given that $\frac{dV}{dt}=2$ m³/min. So, $2 = 64\pi\frac{dh}{dt}$. Then $\frac{dh}{dt}=\frac{2}{64\pi}=\frac{1}{32\pi}$ m/min.

Answer:

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