Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

Find the equation of tangent to the curve \( x-\sin\: 3t, y=\cos\: 2t \: at \: t=\large\frac{\pi}{4}.\)

Can you answer this question?

1 Answer

0 votes
  • Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
Step 1:
$x=\sin 3t$
Differentiating w.r.t $t$ we get,
$\large\frac{dx}{dt}$$=3\cos 3t$
$y=\cos 2t$
Differentiating w.r.t $t$ we get,
$\large\frac{dy}{dt}$$=-2\sin 2t$
$\large\frac{dy}{dx}=\frac{-2\sin 2t}{3\cos 3t}$
$\big(\large\frac{dy}{dx}\big)_{t=\Large\frac{\pi}{4}}=\frac{-2\sin 2(\pi/4)}{3\cos 3(\pi/4)}$
$\qquad\qquad=\large\frac{-2\sin 2(\pi/2)}{3\cos 3(\pi/4)}$
But $\sin\large\frac{\pi}{2}$$=1$
$\cos\large\frac{3\pi}{4}$$=\large\frac{-1}{\sqrt 2}$
$\therefore \large\frac{dy}{dx}$$=m$
$\Rightarrow \large\frac{-2}{3\times -1/\sqrt 2}$
$\Rightarrow \large\frac{2\sqrt 2}{3}$
Step 2:
Equation of the tangent is
Here $x_1=\sin 3t$
$\sin 3(\large\frac{\pi}{4})=\large\frac{-1}{\sqrt 2}$
$y_1=\cos 2\big(\large\frac{\pi}{4}\big)$$=0$
Equation of the tangent is $y-0=\large\frac{2\sqrt 2}{3}$$(x+\large\frac{1}{\sqrt 2}\big)$
$\Rightarrow 3y=2\sqrt 2(\sqrt x+1)$
$\therefore 4x-3y+2\sqrt 2=0$
answered Sep 26, 2013 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App