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

taniska , 10 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 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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