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

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

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Solution

The equation of the given curve is,

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

On differentiating w.r.t. x, we get $\frac{d y}{d x} = 4 x^{3} - 18 x^{2} + 26 x - 10$

$∴$ Slope of tangent at (1,3) is ${(\frac{d y}{d x})}_{(1, 3)}$ =14-18+26-10=2

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

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

Again, the slope of normal at (1,3) is

$\frac{- 1}{S l o p e o f t a n g e n t a t (1, 3)} = \frac{- 1}{2}$

Hence, the equation of the normal at (1,3) is

$y - 3 = \frac{- 1}{2} (x - 1) \Rightarrow 2 y - 6 = - x + 1 \Rightarrow x + 2 y - 7 = 0$

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

(1, 3)

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

(1, 3)

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

(i) $y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5 a t (0, 5)$

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

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

(iv) $y = x^{2} a t (0, 0)$

(v) $x = c o s t, y = s i n t = \frac{π}{4}$

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

(1, 3)

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

(0, 5)

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

(1, 3)

y = x^{3}

(1, 1)

y = x^{2}

(0, 0)

x = cos t, y = sin t

t = \frac{π}{4}

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