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tangents and normals find the equation of the tangent to the curve x= sin 3t , y=cos 2t at t=pi/4

tangents and normals
find the equation of the tangent to the curve x= sin 3t , y=cos 2t at t=pi/4

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
8 years ago
Ans:
Hello Student,
Please find answer to your question,

x = sin3t
dx = 3cos3t.dt…..........(1)
y = cos2t
dy = -2sin2t.dt…..........(2)
Divide (2) by (1)
\frac{dy}{dx} = \frac{-2sin2t}{3cos3t}
t = \frac{\pi }{4}
\frac{dy}{dx} = \frac{-2sin2\frac{\pi }{4}}{3cos3\frac{\pi }{4}}
\frac{dy}{dx} = \frac{2\sqrt{2}}{3}
This is slope of tangent.
Point is
(\frac{1}{\sqrt{2}}, 0)
Equation is:
(y-0) = \frac{2\sqrt{2}}{3}(x - \frac{1}{\sqrt{2}})
y = \frac{2\sqrt{2}}{3}(x - \frac{1}{\sqrt{2}})


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