Application of Derivatives

Let x be the length of edge,v be the value and s be the surface area of the cube then,

y = x3.

and S = 6x2, where x is a fxn of time.

Now, $\frac{dY}{dF}=8c{m}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$

$\Rightarrow \frac{d}{dt}$ (x3) = 8

$\frac{d{x}^3}{dx}·\frac{dx}{dt}=8$ (by chain rule)

$\Rightarrow$ 3x2 $\frac{dx}{dx}=8.$

$\frac{dx}{dt}=\frac{8}{3x^2.}$

Now, $\frac{dS}{dt}=\frac{d}{dt}$ (bx2) = $\frac{d\left(6x^2\right)}{dx}-\frac{dx}{dx}$ = 12x $\frac{8}{3x^2}=\frac{32}{x}$ $c{m}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$.

When x = 12 cm,

$\frac{dS}{dt}=\frac{32}{12}=\frac{8}{3}$ $c{m}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$.

(a) r = 3cm

When r = 3 cm,

$\frac{d}{dr}$ = 2 * π * 3 cm = 6π.

Thus, the area of the circle is changing at the rate of 6π cm $c{m}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$.

(b) r = 4cm

whenr = 4cm,

$\frac{d}{dr}$ = 2 * π * 4cm = 8π.

Thus, the area of the circle is changing at the rate of 8 $c{m}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}$.