Question

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

• A. $0$
• B. $\frac{1}{2}$
• C. $\infty$
• D. $-2$

Answer 1

The correct option is A

$0$

Explanation for the correct answer:

Compute the slope.

$x=3t^2+1$

Differentiating with respect to $t$

$\frac{dx}{dt}=6t$

$y=t^3-1$

Differentiating with respect to $t$

$\frac{dy}{dt}=3t^2$

At $x=1$

$3t^2+1=1$

$\Rightarrow$ $3t^2=0$

$\Rightarrow$ $t=0$

By chain rule,

$\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$

$\Rightarrow$ $\frac{dy}{dx}=\frac{3t^2}{6t}$

$\Rightarrow$ $\frac{dy}{dx}=\frac{1}{2}t$

Substitute $t=0$

$\Rightarrow$ $\frac{dy}{dx}=0$

$\frac{dy}{dx}$ is the slope of the tangent to a curve at a given point.

Hence, the slope of the tangent to the given curve at $x=1$ is $0$.

Hence, option $\left(A\right)$ is the correct answer.