Question

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

Options

• 1/2

• 0

• `-2`

• ∞

Answer 1

`0`

 

\[\text {Given }: \]

\[x = 3 t^2 + 1 \]

\[y = t^3 - 1\]

\[x = 1\]

\[\text { Now }, \]

\[3 t^2 + 1 = 1\]

\[ \Rightarrow 3 t^2 = 0\]

\[ \Rightarrow t = 0\]

\[\frac{dx}{dt} = 6t \text { and } \frac{dy}{dt} = 3 t^2 \]

\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3 t^2}{6t} = \frac{t}{2}\]

\[\text { Thus, we get }\]

\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_{t = 0} =\frac{0}{2}=0\]