Application of Derivatives — Class 12 Maths Solution

We have the equation of the curve,$6y = {x^3} + 2$ ...(i)
and $\cfrac{{dy}}{{dt}} = 8\cfrac{{dx}}{{dt}}$

Differentiating (i) w.r.t. $t$, we obtain

$6\cfrac{{dy}}{{dt}} = 3{x^2}\cfrac{{dx}}{{dt}} \Rightarrow 6 \times \cfrac{{8dx}}{{dt}} = 3{x^2}\cfrac{{dx}}{{dt}} \Rightarrow \cfrac{{dx}}{{dt}}\left( {48 - 3{x^2}} \right) = 0$

$\cfrac{{dx}}{{dt}}$ cannot be equal to zero

Therefore, $48 - 3{x^2} = 0 \Rightarrow x = \pm 4$

$\Rightarrow$ ${x^2} = 16 \Rightarrow x = \pm 4$

When $x = 4 \Rightarrow 6y = {4^3} + 2 \Rightarrow y = \cfrac{{66}}{6} = 11$

When $x = - 4 \Rightarrow 6y = {\left( { - 4} \right)^3} + 2 \Rightarrow y = \cfrac{{ - 62}}{6} = \cfrac{{ - 31}}{3}$

Hence, the required points are $\left( {4,11} \right)$ and $\left( { - 4,\cfrac{{ - 31}}{3}} \right)$.