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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Show that the normal at any point \(\theta\) to the curve $ x = a \cos\theta + a \: \theta \sin\: \theta, y = a\sin\theta - a\theta \cos\theta$ is at a constant distance from the origin.

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Toolbox:
  • $\large\frac{d}{dx}$$(\sin\theta)=\cos\theta$
  • $\large\frac{d}{dx}$$(\cos\theta)=-\sin\theta$
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
We have $x=a\cos\theta+a\theta\sin\theta$
$\qquad\quad\; y=a\sin\theta-a\theta\sin\theta$
Differentiating with respect to $x$
$\large\frac{dx}{d\theta}$$=-a\sin\theta+a\theta\cos\theta+a\sin\theta$
$\quad\;=a\theta\cos\theta$
$\large\frac{dy}{d\theta}$$=a\cos\theta-a\cos\theta+a\theta\sin\theta$
$\quad\;=a\theta\sin\theta$
Step 2:
Slope of the tangent =$\large\frac{dy}{dx}=\frac{dy}{d\theta}\div\frac{dx}{d\theta}$
$\qquad\qquad\qquad\qquad\quad=\large\frac{a\theta\sin\theta}{a\theta\cos\theta}$
$\qquad\qquad\qquad\qquad\quad=\large\frac{\sin\theta}{\cos\theta}$
$\qquad\qquad\qquad\qquad\quad=\tan\theta$
Step 3:
Slope of the normal at $\theta=\large\frac{-1}{\Large\frac{dy}{dx}}$
$\Rightarrow \large\frac{-1}{\tan\theta}$
$\Rightarrow -\cot \theta$
The equation of the normal at the point $'\theta'$ is
$[y-(a\sin\theta-a\theta\cos\theta)]=-\cot\theta[x-a\cos\theta+a\theta\sin\theta]$
On simplifying,we get
$x\cos\theta+y\sin\theta=a$ which is the equation of the normal.
answered Aug 9, 2013 by sreemathi.v
 

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