**Question:**

The volume $V$ and depth $x$ of water in a vessel are connected by the relation $V=5 x-\frac{x^{2}}{6}$ and the volume of water is increasing the rate of $5 \mathrm{~cm}^{3} / \mathrm{sec}$, when $x=2 \mathrm{~cm}$. The rate of which the depth of water is increasing is equal to__________________

**Solution:**

It is given that, $\frac{d V}{d t}=5 \mathrm{~cm}^{3} / \mathrm{sec}$

The volume $V$ and depth $x$ of water in a vessel are connected by the relation $V=5 x-\frac{x^{2}}{6}$.

$V=5 x-\frac{x^{2}}{6}$

Differentiating both sides with respect to $t$, we get

$\frac{d V}{d t}=5 \frac{d x}{d t}-\frac{2 x}{6} \frac{d x}{d t}$

$\Rightarrow \frac{d V}{d t}=\left(5-\frac{x}{3}\right) \frac{d x}{d t}$

When $x=2 \mathrm{~cm}$ and $\frac{d V}{d t}=5 \mathrm{~cm}^{3} / \mathrm{sec}$, we get

$5=\left(5-\frac{2}{3}\right) \frac{d x}{d t}$

$\Rightarrow \frac{d x}{d t}=\frac{15}{13} \mathrm{~cm} / \mathrm{sec}$

Thus, depth of water is increasing at the rate of $\frac{15}{13} \mathrm{~cm} / \mathrm{sec}$.

The volume $V$ and depth $x$ of water in a vessel are connected by the relation $V=5 x-\frac{x^{2}}{6}$ and the volume of water is increasing the rate of $5 \mathrm{~cm}^{3} / \mathrm{sec}$, when $x=2 \mathrm{~cm} .$ The rate of which the depth of water is increasing is equal to $\frac{15}{13} \mathrm{~cm} / \mathrm{sec}$