Semi-vertical angle $=\frac{60^{\circ}}{2}=30^{\circ}$

$$ \tan 30^{\circ}=\frac{r}{h} $$

TG EAPCET 2024 (Online) 9th May Evening Shift Mathematics - Application of Derivatives Question 101 English Explanation

$$ \begin{array}{ll} \Rightarrow & \frac{1}{\sqrt{3}}=\frac{r}{h} \\\\ \Rightarrow & h=\sqrt{3} r \end{array} $$

When $h=3 m$, then $r=\frac{3}{\sqrt{3}}=\sqrt{3} m$

$$ \begin{aligned} & \text { Volume, } V=\frac{1}{3} \pi r^2 h \\\\ & \Rightarrow \quad 3 V=\pi r^2(\sqrt{3} r) \\\\ & \Rightarrow \quad \sqrt{3} V=\pi r^3 \end{aligned} $$

On differentiating Both sides w.r.t. $t$, we get

$$ \begin{aligned} & \sqrt{3} \frac{d V}{d t}=\pi\left(3 r^2\right) \frac{d r}{d t} \\\\ \Rightarrow & \sqrt{3} \times \frac{1}{\sqrt{3}}=3 r^2 \pi \frac{d r}{d t} \\\\ \Rightarrow & \left.\frac{d r}{d t}\right|_{h=3 m}=\frac{1}{3(\sqrt{3})^2 \pi}=\frac{1}{9 \pi} \mathrm{~m} / \mathrm{min} \end{aligned} $$

Semi-vertical angle $=\frac{60^{\circ}}{2}=30^{\circ}$

$$ \tan 30^{\circ}=\frac{r}{h} $$

$$ \begin{array}{ll} \Rightarrow & \frac{1}{\sqrt{3}}=\frac{r}{h} \\\\ \Rightarrow & h=\sqrt{3} r \end{array} $$

When $h=3 m$, then $r=\frac{3}{\sqrt{3}}=\sqrt{3} m$

$$ \begin{aligned} & \text { Volume, } V=\frac{1}{3} \pi r^2 h \\\\ & \Rightarrow \quad 3 V=\pi r^2(\sqrt{3} r) \\\\ & \Rightarrow \quad \sqrt{3} V=\pi r^3 \end{aligned} $$

On differentiating Both sides w.r.t. $t$, we get

Application Of Derivatives

Continue this paper

Similar Application Of Derivatives PYQs

More practice from this question