Application of Derivatives 2 Question 3
3. A water tank has the shape of an inverted right circular cone, whose semi-vertical angle is $\tan ^{-1} \frac{1}{2}$. Water is poured into it at a constant rate of $5 cu m / min$. Then, the rate (in $m / min$ ) at which the level of water is rising at the instant when the depth of water in the tank is $10 m$ is
(2019 Main, 9 April II)
(a) $\frac{2}{\pi}$
(b) $\frac{1}{5 \pi}$
(c) $\frac{1}{15 \pi}$
(d) $\frac{1}{10 \pi}$
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Solution:
Key Idea Use formula: Volume of cone $=\frac{1}{3} \pi r^{2} h$, where $r=$ radius and $h=$ height of the cone.
Given, semi-vertical angle of right circular cone
$$ =\tan ^{-1} \frac{1}{2} $$
Let $\quad \alpha=\tan ^{-1} \frac{1}{2}$
$\Rightarrow \quad \tan \alpha=\frac{1}{2}$
$\Rightarrow \quad \frac{r}{h}=\frac{1}{2} \quad$ [from fig. $\tan \alpha=\frac{r}{h}$ ]
$$ \Rightarrow \quad r=\frac{1}{2} h $$
$\because$ Volume of cone is $(V)=\frac{1}{3} \pi r^{2} h$
$\therefore \quad V=\frac{1}{3} \pi \frac{1}{2} h \quad(h)=\frac{1}{12} \pi h^{3} \quad$ [from Eq. (i)]
On differentiating both sides w.r.t. ’ $t$ ‘, we get
$$ \begin{array}{rlrl} & \frac{d V}{d t} & =\frac{1}{12} \pi\left(3 h^{2}\right) \frac{d h}{d t} \\ \Rightarrow & \frac{d h}{d t} & =\frac{4}{\pi h^{2}} \frac{d V}{d t} \\ \Rightarrow & \frac{d h}{d t} & =\frac{4}{\pi h^{2}} \times 5 & {\left[\because \text { given } \frac{d V}{d t}=5 m^{3} / min\right]} \end{array} $$
Now, at $h=10 m$, the rate at which height of water level is rising $=\left.\frac{d h}{d t}\right| _{h=10}$
$$ =\frac{4}{\pi(10)^{2}} \times 5=\frac{1}{5 \pi} m / min $$
