Find the equations of the tangent and normal to the given curve at the indicated point:
$y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5$ at $(1, 3)$

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Solution

We have
$y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5$

$\Rightarrow \frac{d y}{d x} = 4 x^{3} - 18 x^{2} + 26 x - 10$

$∴ {(\frac{d y}{d x})}_{(1, 3)} = 4 - 18 + 26 - 10 = 2$

Thus, the slope of the tangent at $(1, 3)$ is $2$ .

So, equation of the tangent is given as,

$(y - 3) = 2 (x - 1) \Rightarrow 2 x - y + 1 = 0$

The slope of the normal at $(1, 3)$ is $\frac{- 1}{Slope of the tangent at (1, 3)} = - \frac{1}{2}$ .

Therefore, the equation of the normal at $(1, 3)$ is given as,
$(y - 3) = - \frac{1}{2} (x - 1) \Rightarrow x + 2 y - 7 = 0$

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

$y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5 a t (1, 3)$

Find the equations of the tangent and normal to the given curves at the indicated points:

(i) y = x⁴ − 6x³ + 13x² − 10x + 5 at (0, 5)

(ii) y = x⁴ − 6x³ + 13x² − 10x + 5 at (1, 3)

(iii) y = x³ at (1, 1)

(iv) y = x² at (0, 0)

(v) x = cos t, y = sin t at

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