Question

The slope of the tangent to the curve x = 3t² + 1, y = t³ −1 at x = 1 is

(a) 1/2
(b) 0
(c) −2
(d) ∞

Answer 1

(b) 0

$$\begin{gathered} \mathrm{Given}: \\ x=3t^2+1 \\ y=t^3-1 \\ x=1 \\ \mathrm{Now}, \\ 3t^2+1=1 \\ \Rightarrow 3t^2=0 \\ \Rightarrow t=0 \\ \frac{dx}{dt}=6t\mathrm{and}\frac{dy}{dt}=3t^2 \\ \therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{3t^2}{6t}=\frac{t}{2} \\ \mathrm{Thus},\mathrm{we}\mathrm{get} \\ \text{Slope of the tangent=}{\left(\frac{dy}{dx}\right)}_{t=0}\text{=}\frac{0}{2}\text{=0} \end{gathered}$$