A particle moves along the curve $3y = ax^3 + 1$ such that at a point with $x$-coordinate $1$, $y$-coordinate is changing twice as fast as $x$-coordinate. Find the value of $a$.

2

The correct answer is Option (2) → 2 ##

Given: $3y = ax^3 + 1 \dots (i)$

$\frac{dy}{dt} = 2 \left( \frac{dx}{dt} \right)$ at $x = 1 \dots (ii)$

From eq. (i), we have

$\frac{3dy}{dt} = 3x^2 a \frac{dx}{dt} + 0$

$\frac{dy}{dt} = ax^2 \frac{dx}{dt}$

$2 \left( \frac{dx}{dt} \right) = a(1)^2 \frac{dx}{dt} \quad \text{[Using eq. (ii)]}$

$∴a = 2$