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

$y = x^{3} a t (1, 1)$

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Solution

The equation of the given curve is $y = x^{3}$

On differentiating w.r.t. x, we get $\frac{d y}{d x} = 3 x^{2}$

$∴$ Slope of tangent at (1,1) is ${(\frac{d y}{d x})}_{(1, 1)} = 3 (1)^{2} = 3$

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

$y - 1 = 3 (x - 1) \Rightarrow y - 1 = 3 x - 3 \Rightarrow y = 3 x - 2$

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

= $\frac{- 1}{S l o p e o f t a n g e n t a t (1, 1)} = \frac{- 1}{3}$

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

$y - 1 = \frac{- 1}{3} (x - 1)$

$\Rightarrow 3 y - 3 = - x + 1 \Rightarrow x + 3 y - 4 = 0$

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