A cylindrical tank with radius $4 \mathrm{~m}$ is being filled with water at a rate of $3 \mathrm{~m}^{3} / \mathrm{min}$. How fast is the height of the water increasing?
Step 1
If $h$ is the water's height, the volume of the water is $V=\pi r^{2} h$. We must find $d V / d t$. Differentiating both sides of the equation gives the following.
\[
\frac{d V}{d t}=\pi r^{2} \pi r^{2} \frac{d h}{d t}
\]
Step 2
Substituting for $r$, this becomes $\frac{d V}{d t}=$ $\frac{d h}{d t}$
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Answer
Thus, the height of the water is increasing at a rate of $\boxed{\frac{3}{16\pi}}$ meters per minute.
Step 1 ：Given that the volume of the water is increasing at a rate of $3 m^3/min$, we need to find how fast the height of the water is increasing. The volume of a cylinder is given by $V = \pi r^2 h$. Differentiating both sides with respect to time $t$, we get $\frac{dV}{dt} = \pi r^2 \frac{dh}{dt}$.
Step 2 ：Substituting the given values into the equation, we get $3 = \pi (4)^2 \frac{dh}{dt}$. Solving for $\frac{dh}{dt}$, we get $\frac{dh}{dt} = \frac{3}{16\pi}$.
Step 3 ：Thus, the height of the water is increasing at a rate of $\boxed{\frac{3}{16\pi}}$ meters per minute.
