The slope of the tangent to the curve $x = 3t^{2} + 1,\ y = t^{3} - 1$, at $x = 1$is

Question

Dr. Abubacker · Accepted Answer

(a)
Given curve is $x = 3t^{2} + 1,\ y = t^{3} - 1$
For $x = 1,\ \ 3t^{2} + 1 = 1\ \ \  \Rightarrow \ \ t = 0$
$	herefore\ \ \frac{dx}{dt} = 6t,\ \frac{dy}{dx} = 3t^{2}$
Now, $\frac{dy}{dx} = \left( \frac{\frac{dy}{dt}}{\frac{dx}{dt}} ight) = \frac{3t^{2}}{6t} = \frac{t}{2}$
$	herefore\ \ \left( \frac{dy}{dx} ight)_{(t = 0)} = \frac{0}{2} = 0$

Q. The slope of the tangent to the curve $x = 3t^{2} + 1,\ y = t^{3} - 1$, at $x = 1$is

Explanation

Reviewed by Dr. Abubacker

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