Question

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. $\infty$

Answer 1

Answer: (B)

$0$

Given, $x=3t^2+1,y=t^3-1$

Slope of the tangent to the given curve is $\frac{dy}{dx}=\frac{dy}{dt}\times \frac{dt}{dx}$

$=3t^2\times \frac{1}{6t}$

$=\frac{t}{2}$

Since the slope has to be calculated at $x=1,$ i.e. at $3t^2+1=1$, we get $t=0$

Thus, the required slope is $0.$