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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the angle of intersection of the curves $y=4-x^2$ and $y=x^2$.

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Toolbox:
  • Angle between the tangents of two curves is given by $ \theta = \tan^{-1} \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$
  • If $y=f(x),$ then $\bigg( \large\frac{dy}{dx} \bigg)_p=$ slope of the tangent to $y=f(x)$ at point $p$
Step 1
$y=4-x^2\: \: \: \: (1)$
$y=x^2\: \: \: \: \: \: \: \:\:\: \: (2)$
On solving the equation we get
$x^2=4-x^2 \Rightarrow 2x^2=4$
$ x^2=2\: x= \pm \sqrt 2$
$ \Rightarrow y= 2$
Consider $y=4-x^2$
differentiating w.r.t $x$ we get,
$ \bigg( \large\frac{dy}{dx} \bigg)=-2x\: \therefore \bigg(\large\frac{dy}{dx} \bigg)_{(\sqrt 2, 2)}=-2\sqrt 2$
Let $m_1=-2\sqrt 2$
Consider $y=x^2$
differentiating w.r.t $x$ we get,
$ \large\frac{dy}{dx}=2x\: \therefore \bigg( \large\frac{dy}{dx} \bigg)_{(\sqrt 2, 2)}=2\sqrt 2$
Let this be $m_2$
The angle between the two curves is given as
$ \tan \: \theta = \bigg| \large\frac{m_1-m_2}{1+m_1m_2} \bigg|$
Substituting for $m_1\: and \: m_2$ we get
$ \tan\: \theta = \bigg| \large\frac{2\sqrt 2-(-2\sqrt 2)}{1+2\sqrt 2 \times -2\sqrt 2} \bigg|$
$ = \bigg| \large\frac{2\sqrt 2+2\sqrt 2}{1-8} \bigg|$
$ \tan\: \theta = \large\frac{4\sqrt 2}{7}$
$ \therefore \theta = \tan^{-1} \bigg| \large\frac{4\sqrt 2}{7} \bigg|$
answered Aug 4, 2013 by thanvigandhi_1
 

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