Q. The slope of the tangent to the curve $x = 3t^{2} + 1,\ y = t^{3} - 1$ at $x = 1$ is
A) $\frac{1}{2}$
B) 0
C) -2
D) $\infty$
Explanation
(b)
Given, $x = 3t^{2} + 1$ and $y = t^{3} - 1$
$\therefore\ \ \ \frac{dx}{dt} = 6t$ and $\frac{dy}{dt} = 3t^{2}$
$\therefore\ \ \ \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx} = \frac{t}{2}$
Since, $x = 1\ \ \ \Rightarrow \ \ \ 1 - 3t^{2} + 1$
$\Rightarrow \ \ \ \ \ t = 0$
$\therefore\ \ \frac{dy}{dx} = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.