A cylindrical water container has developed a leak at the bottom. The water is leaking at the rate of 5 cm^3 / s from the leak. If the radius of the…

Let $\displaystyle \mathrm{V}, r, h$ be the volume, radius and height of cylindrical container. Given $\displaystyle \frac{d \mathrm{V}}{d t}=-5 \mathrm{~cm}^{3} / \mathrm{s}$ \[\begin{aligned} & \mathrm{V}=\pi r^{2} h=\pi(15)^{2} h=225 \pi h \\ & \therefore \frac{d \mathrm{V}}{d t}=225 \pi \frac{d h}{d t} \Rightarrow-5=225 \pi \frac{d h}{d t} \\ & \Rightarrow \frac{d h}{d t}=-\frac{5}{225 \pi}=-\frac{1}{45 \pi} \end{aligned} \] ∴ Height of the water is decreasing at the rate of $\displaystyle \frac{1}{45 \pi} \mathrm{~cm} / \mathrm{s}$