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Question
The slope of the tangent to the curve x = 3t2 + 1, y = t3 −1 at x = 1 is ___________ .
Options
1/2
0
`-2`
∞
Solution
`0`
\[\text {Given }: \]
\[x = 3 t^2 + 1 \]
\[y = t^3 - 1\]
\[x = 1\]
\[\text { Now }, \]
\[3 t^2 + 1 = 1\]
\[ \Rightarrow 3 t^2 = 0\]
\[ \Rightarrow t = 0\]
\[\frac{dx}{dt} = 6t \text { and } \frac{dy}{dt} = 3 t^2 \]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{3 t^2}{6t} = \frac{t}{2}\]
\[\text { Thus, we get }\]
\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_{t = 0} =\frac{0}{2}=0\]
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