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Question

Find the equation of the tangent and normal to the given curve at the given points

y=x3 at (1,1)

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Solution

The equation of the given curve is y=x3

On differentiating w.r.t. x, we get dydx=3x2

Slope of tangent at (1,1) is (dydx)(1,1)=3(1)2=3

Thus, the slope of the tangent at (1,1) is 3 and the equation of the tangent is

y1=3(x1)y1=3x3y=3x2

Again, the slope of normal at (1,1)

=1Slope of tangent at (1,1)=13

Hence, the equation of the normal at (1,1) is

y1=13(x1)

3y3=x+1 x+3y4=0


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