What is percentage change in its resistance if radius is increased by $1\%$and area is increased by$1\%$ ?

 Similarly$R=\rho \frac{V}{A^2},\frac{dR}{R}=-2\frac{dA}{A}$

As cross section area $={\mathrm{\pi r}}^2$

$\frac{dA}{A}=2\frac{dr}{r}$

so,$\frac{dR}{R}=-2\frac{dA}{A}=-4\frac{dr}{r}$

If radius is increased by 1%, the resistance is decreased by $4\%.$ If area is increased by $1\%$, the resistance decreased by$2\%.$

The significance of negative sign is that change in resistance is opposite to that of radius and area. We used to calculus in this question because percentage change are small.