Application of Derivatives — Class 12 Maths Solution

Let internal radius $= r$ and external radius $= R$

Therefore,Volume of hollow spherical shell, $V = \frac{4}{3}\pi \left( {{R^3} - {r^3}} \right)$

$\Rightarrow$ $V = \frac{4}{3}\pi \left[ {{{(3.0005)}^3} - {{(3)}^3}} \right]$ ……..(i)

Now, we shall use differentiation to get approximate value of ${(3.0005)^3}$.

Let ${(3.0005)^3} = y + \Delta y$

and $x = 3,\Delta x = 0.0005$

Also, let $y = {x^3}$

On differentiating both sides w.r.t. $x$, we get
$\frac{{dy}}{{dx}} = 3{x^2}$

Therefore,$\Delta y = \frac{{dy}}{{dx}} \times \Delta x = 3{x^2} \times 0.0005$
$= 3 \times {3^2} \times 0.0005$

$= 27 \times 0.0005 = 0.0135$

Also, ${(3.0005)^3} = y + \Delta y$
$= {3^3} + 0.0135 = 27.0135$

Therefore,$V = \frac{4}{3}\pi [27.0135 - 27.000]$ [By using Eq. (i)]

$= \frac{4}{3}\pi [0.0135] = 4\pi \times (0.0045)$

$= 0.0180\pi {\rm{c}}{{\rm{m}}^3}$