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. $\infty$
• C. -2

Answer 1

$x=3t^2+1,y=t^3-1\therefore \frac{dx}{dt}=6t,\frac{dy}{dt}=3t^2$
Now $\frac{dy}{dx}=\left(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\right)=\frac{3t^2}{6t}=\frac{t}{2}$
For $x=1,3t^2+1=1\Rightarrow t=0$ Slope =$\frac{0}{2}=0$.