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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives

Find the equations of the tangent and normal to the given curves at the indicated points: $ (iii) \: y = x^3\; at \;(1, 1)$

This is third part of the multi-part question q14

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1 Answer

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  • 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}$$(x-x_1)$
Step 1:
Given : $y=x^3$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}_{(1,1)}$$=3(1)$
$m=3$
Step 2:
Therefore equation of the tangent at $(1,1)$ is $(y-1)=3(x-1)$
$\Rightarrow 3x-y-2=0$
Step 3:
Equation of the normal at $(1,1)$ is $(y-1)=\large\frac{-1}{3}$$(x-1)$
$\Rightarrow 3y-3=-x+1$
$\Rightarrow x+3y-4=0$
answered Jul 11, 2013 by sreemathi.v
 

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