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Question
Find the slope of the tangent to the curve x = t2 + 3t − 8, y = 2t2 − 2t − 5 at t = 2 ?
Solution
\[\text { Here, } \]
\[x = t^2 + 3t - 8 \text { and } y = 2 t^2 - 2t - 5\]
\[\frac{dx}{dt} = 2t + 3 \text { and } \frac{dy}{dt} = 4t - 2\]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{4t - 2}{2t + 3}\]
\[\text { Now,} \]
\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_{t = 2} =\frac{8 - 2}{4 + 3}=\frac{6}{7}\]
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