If the radius of a circle increases from $5\mathrm{ }\mathrm{c}\mathrm{m}$ to $5.1\mathrm{ }\mathrm{c}\mathrm{m}$, find the increase in it's area.

 Let $r_1$ be the radius of the circle, $r_2$ be the radius of new circle with an increased radius and $∆r$ be the change in radius.

Given $r_1=5cm$and $r_2=5.1cm$.

Change in radius =$∆r=r_2-r_1=(5.1-5)cm=0.1cm$.

Let $A$ be the area of the circle with radius $5cm$ and change in area is $∆A$.

We know $∆r=dr$and $∆A=dA$.

The formula of area of a circle is $\mathrm{\pi }{\left(\mathrm{radius}\right)}^2 {\mathrm{units}}^2$.

Now, rate of change of area $A$ with respect to radius is $\frac{dA}{dr}$.

$\Rightarrow \frac{dA}{dr}=\frac{d\left({\mathrm{\pi r}}^2\right)}{dr}=\mathrm{\pi }\frac{\mathrm{d}\left({\mathrm{r}}^2\right)}{\mathrm{dr}}=2\mathrm{\pi r}.$

As the formula of differentiation of $x^n=n.x^{n-1}$.

Now, Change in area= $∆A=dA=\frac{dA}{dr}\left(dr\right)$

By substituting the values of $\frac{dA}{dr},dr$ in above equation 

$$\begin{gathered} \Rightarrow ∆A=\left(\frac{dA}{dr}\right)dr=\left(2\mathrm{\pi r}\right)0.1 \\ =2\times \mathrm{\pi }\times 5\times 0.1=10\times 0.1\times \mathrm{\pi }=\mathrm{\pi } {\mathrm{cm}}^{2.} \end{gathered}$$

 Therefore, Increase in area $=\pi {\mathrm{cm}}^2$.