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}=\pi(\frac{21}{8} + 9)\times64\pi\frac{dh}{dt}\\)
step 3
substituting for \\(\frac{dv}{dt}\\) and isolating \\(\frac{dh}{dt}\\), we can conclude that the height of the water is increasing at the following rate (in m/min). (round your answer to four decimal places.)
\\(\frac{dh}{dt}=0.0001\times m/min\\)

Explanation:

Step1: Differentiate volume formula

The volume of a cylinder is $V = \pi r^{2}h$. Differentiating both sides with respect to time $t$ using the chain - rule gives $\frac{dV}{dt}=\pi r^{2}\frac{dh}{dt}$.

Step2: Substitute the value of $r$

Given $r = 8$ m, so $\frac{dV}{dt}=\pi\times(8)^{2}\frac{dh}{dt}=64\pi\frac{dh}{dt}$.

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

We know that $\frac{dV}{dt}=2$ m³/min. Substituting into the equation $\frac{dV}{dt}=64\pi\frac{dh}{dt}$, we get $2 = 64\pi\frac{dh}{dt}$. Then $\frac{dh}{dt}=\frac{2}{64\pi}=\frac{1}{32\pi}\approx0.0099$ m/min.

Answer:

$0.0099$ m/min