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Question
Find the equation of the tangent to the curve x = sin 3t, y = cos 2t at
\[t = \frac{\pi}{4}\] ?
Solution
\[x = \sin 3t \text { and } y = \cos 2t\]
\[\frac{dx}{dt} = 3 \cos 3t \text { and } \frac{dy}{dt} = - 2 \sin 2t\]
\[ \therefore \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{- 2 \sin 2t}{3 \cos 3t}\]
\[\text { Slope of tangent },m= \left( \frac{dy}{dx} \right)_{t = \frac{\pi}{4}} =-\frac{- 2 \sin \left( \frac{\pi}{2} \right)}{3 \cos \left( \frac{3\pi}{4} \right)}=\frac{- 2}{\frac{- 3}{\sqrt{2}}}=\frac{2\sqrt{2}}{3}\]
\[ x_1 = \sin \left( 3 \times \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \text { and }y_1 = \cos \left( 2 \times \frac{\pi}{4} \right) = 0\]
\[\text { So }, \left( x_1 , y_1 \right) = \left( \frac{1}{\sqrt{2}}, 0 \right)\]
\[\text { Equation of tangent is },\]
\[y - y_1 = m \left( x - x_1 \right)\]
\[ \Rightarrow y - 0 = \frac{2\sqrt{2}}{3}\left( x - \frac{1}{\sqrt{2}} \right)\]
\[ \Rightarrow 3y = 2\sqrt{2}x - 2\]
\[ \Rightarrow 2\sqrt{2}x - 3y - 2 = 0\]
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