Find the equation of tangent to the curve x = sin 3t, y = cos 2t, at t = $\frac{π}{4}$

Solution

$X = \sin 3 t \Rightarrow \frac{dx}{dt} = 3 \cos 3 t ∴ x_{(t = \frac{π}{4})} = \sin 3 (\frac{π}{4}) = \frac{1}{\sqrt{2}} y = coc 2 t \Rightarrow \frac{dy}{dt} = - 2 \sin 2 t ∴ y_{(t = \frac{π}{4})} = \cos 2 t = \cos 2 (\frac{π}{4}) = 0 \Rightarrow \frac{dy}{dx} = \frac{dy}{dt} . \frac{dt}{dx} = - 2 \sin 2 t \frac{1}{3 \cos 3 t} = - \frac{2}{3} (\frac{\sin 2 t}{\cos 3 t})$

$∴ {\frac{dy}{dx}}_{(t = \frac{π}{4})} = \frac{- 2}{3} \frac{\sin (2 x \frac{π}{4})}{\cos (3 x \frac{π}{4})} = \frac{- 2}{3} \frac{\sin (\frac{π}{2})}{\cos (\frac{3 π}{4})} = \frac{2 \sqrt{2}}{3} Therefore, the equation of the tangent at the point (\frac{1}{\sqrt{2}}, 0) is y - 0 = \frac{2 \sqrt{2}}{3} (x - \frac{1}{\sqrt{2}}) y = \frac{2 \sqrt{2}}{3} x - \frac{2}{3} 3 y - 2 \sqrt{2 x} + 2 = 0$

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Application Of Derivatives

Find the equation of tangent to the curve x = sin 3t, y = cos 2t, at t = $\frac{π}{4}$

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

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For Daily Free Study Material Join wiredfaculty

Application Of Derivatives

Find the equation of tangent to the curve x = sin 3t, y = cos 2t, at t = π4

Some More Questions From Application of Derivatives Chapter

The radius of a circle is increasing uniformly at the rate of 4 cm per second Find the rate at which the area of the circle is increasing when the radius is 8 cm.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Find the equation of tangent to the curve x = sin 3t, y = cos 2t, at t = $\frac{π}{4}$