Find the equations of the tangent and normal to the given curve at the indicated point:
$y = x^{3}$ at $(1, 1)$

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Solution

On differentiating with respect to $x$ , we get:
$\frac{d y}{d x} = 3 x^{2}$

${(\frac{d y}{d x})}_{(1, 1)} = 3 (1)^{2} = 3$

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

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

The slope of the normal at $(1, 1)$ is $\frac{- 1}{Slope of the tangent at (1, 1)} = \frac{- 1}{3}$ .

Therefore, the equation of the normal at $(1, 1)$ is given as,
$y - 1 = - \frac{1}{3} (x - 1) \Rightarrow x + 3 y - 4 = 0$

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