Provide solution for RD Sharma maths class 12 chapter 12 Derivative as a Rate Measurer exercise multiple choise question 10

$A.\; \; 54\pi \; \; cm^{2}/min$

Hint:

The lateral surface area of cone, with radius r and height h is defined as

\begin{aligned} &L(r, h)=\pi r \sqrt{r^{2}+h^{2}} \quad \ldots \ldots \text { (i) }\\ \end{aligned}

Given:

\begin{aligned} &r=7 \mathrm{~cm}, h=24 \mathrm{~cm} \text { and } \frac{d r}{d t}=3 \mathrm{~cm} / \mathrm{min}, \frac{d h}{d t}=-4 \mathrm{~cm} / \mathrm{min}\\ \end{aligned}

Solution:

→ Differentiating (i) with respect to t,we get

\begin{aligned} &\frac{d l}{d t}=\pi\left(\sqrt{r^{2}+h^{2}} \frac{d r}{d t}+r \times \frac{1}{2 \sqrt{r^{2}+h^{2}}} \times\left(2 r \frac{d r}{d t}+2 h \frac{d h}{d t}\right)\right)\\ &\text { Substituting values }\\ &\frac{d l}{d t}=\pi\left(\sqrt{7^{2}+24^{2}} \times 3+7 \times \frac{1}{2 \sqrt{7^{2}+24^{2}}} \times(2 \times 7 \times 3+2 \times 24 \times-4)\right)\\ &\frac{d l}{d t}=54 \pi \mathrm{cm}^{2} / \mathrm{min} \end{aligned}