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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the equations of the tangent and normal to the given curves at the indicated points: $ (v) \: x = \cos t,y=\sin t\; at \;(t=\large\frac{\pi}{4})$

This is fifth part of the multi-part question q14

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Toolbox:
  • Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
  • Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1)$
Step 1:
Given : $x=\cos t,y=\sin t$
Consider $x=\cos t$
Differentiating w.r.t $t$ we get,
$\large\frac{dx}{dt}$$=-\sin t$
Step 2:
Consider $y=\sin t$
Differentiating w.r.t $t$ we get,
$\large\frac{dy}{dt}$$=\cos t$
Step 3:
Therefore $\large\frac{dy}{dx}=\frac{dy}{dt}$$\times\large\frac{dt}{dx}$
$\qquad\qquad\quad\;=\cos t\times \large\frac{-1}{\sin t}$
$\qquad\qquad\quad\;=\cot t$
$\large\frac{dy}{dx}$ at $(t=\large\frac{\pi}{4})$ is
$\large\frac{dy}{dx}_{(t=\Large\frac{\pi}{4})}$$=-\cot(\large\frac{\pi}{4})$
But $\cot \large\frac{\pi}{4}$$=1$
Therefore slope of the tangent at $(t=\large\frac{\pi}{4})$ is $-1$
Step 4:
Equation of the tangent is $(y-\sin t)=1(x-\cos t)$
But $\cos t=\sin t=\large\frac{1}{\sqrt 2}$
$(y-\large\frac{1}{\sqrt 2})$$=-1(x-\large\frac{1}{\sqrt 2})$
$\Rightarrow \sqrt 2y-1=-\sqrt 2 x+1$
$\Rightarrow \sqrt 2x+\sqrt 2y-2=0$
$\Rightarrow x+y-\sqrt 2=0$
Step 5:
Equation of the normal is $(y-\sin t)=1(x-\cos t)$
$\Rightarrow (y-\large\frac{1}{\sqrt 2})=$$x-\large\frac{1}{\sqrt 2}$
$\Rightarrow x-y=0$
$\Rightarrow x=y$
answered Jul 11, 2013 by sreemathi.v
 

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