#### Explain solution RD Sharma class 12 chapter Derivative as a Rate Measure exercise 12.1 question 4 maths

Hint:

We have to differentiate area of circular disc ($A$) with respect to its circumference ($L$)

Area of circular disc, $A=\pi r^{2}$

Circumference of circular disc, $L=2\pi r$

Given:

Radius of a circular disc, $r=3 cm$

Solution:

Here we have,

Radius of disc, $r=3 cm$

Circumference of disc, $L=2\pi r$

Let’s differentiate with the circumference with respect to radius $r$

$\frac{dL}{dr}=2\pi$

$\therefore \frac{dr}{dL}=\frac{1}{2\pi }\: \; \; \; \; \; \; \; \; .....(i)$            $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

Area of circular disc, $A=\pi r^{2}$.

Let’s differentiate with the area with respect to $L$

$\therefore \frac{d A}{d L}=2 \pi r\frac{dr}{d L}$

$\therefore \frac{d A}{d L}=\frac{2 \pi r}{2 \pi }$                                    {From eq.(i)}

$\therefore \frac{d A}{d L}=r$

Putting value of $r=3cm$$r = 3cm$

Note:

Here we cannot write directly,

$A=\frac{L}{2}r$            It's wrong