51. 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=cosy,y=sint,at,t=π4

(i) we have, y=x4−6x3+13x2−10x+5 slope of tangent, dydx=4x3−18x2+26x−10 dydx|(x,y)=(0,5)=−10. slope of normal =−1−10=110 Hence eqn of tangent at (0, 5) is y−5=−10(x−0)→10x+y−5=0 And eqn of normal at (0, 5) is y−5=110(x−0) ⇒10y−50=x ⇒x−10y+50=0 (ii) We have, y=x4−6x3+13x2−10x+5 Slope of tangent, dydx=4x3−18x2+26x−10. dydx|(x,y)=(1,3)=4(1)3−18(1)2+26(1)−10 =4−18+26−10 = 30 28 = 2 Slope of normal =−12 Hence eqn of tangent at (1, 3) is y−3=2(x−1) ⇒ y−3=2x−2 →2x−y+1=0 And eqn of normal at (1,3) is (y−3)=−12(x−1) ⇒ 2y−6=−x+1 → x+2y−7=0 (iii) We have, y=x3 Slope of tangent, dydx=3x2 dydx|(1,1)=3(1)2=3. And slope of normal =−13 Hence, eqn of tangent at (1, 1) is y−1=3(x−1) ⇒ y−1=3x−3 →3x−y−2=0 And eqn of normal at (1,1) is y−1=−13(x−1) →3y−3=−x+1 ⇒x+3y−4=0. (iv) We have, y=x2 Slope of tangent dydx=2x dydx|(0,0)=0. So, eqn of the tangent at (0,0) is (y−0)=0(x−0) ⇒y=0. ie, x- axis Hence, the eqn of normal is x = 0 ie, y-axis (v) We have, x=costy=sint dxdt=−sint·dydt=cost So, slope of tangent dydx=dy/dtdx/dt=cost−sint=−cott dydx|t=π4=−cotπ4=−1. And slope of normal =−1−1=1

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

$\begin{array}{l} (i) y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5, a t, (0, 5) \\ (i i) y = x^{4} - 6 x^{3} + 13 x^{2} - 10 x + 5, a t, (1, 3) \\ (i i i) y = x^{3}, a t, (1, 1) \\ (i v) y = x^{2}, a t, (0, 0) \\ (v) x = c o s y, y = s i n t, a t, t = \frac{π}{4} \end{array}$

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Vishal Baghel | Contributor-Level 10

9 months ago

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

slope of tangent, $\frac{d y}{d x} = 4 x^{3} - 18 x^{2} + 26 x - 10$

${\frac{d y}{d x} |}_{(x, y) = (0, 5)} = - 10 .$

slope of normal $= \frac{- 1}{- 10} = \frac{1}{10}$

Hence eqⁿ of tangent at (0, 5) is

$y - 5 = - 10 (x - 0) \to 10 x + y - 5 = 0$

And eqⁿ of normal at (0, 5) is

$y - 5 = \frac{1}{10} (x - 0)$

$\Rightarrow 10 y - 50 = x$

$\Rightarrow x - 10 y + 50 = 0$

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

Slope of tangent, $\frac{d y}{d x} = 4 x^{3} - 18 x^{2} + 26 x - 10 .$

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

$= 4 - 18 + 26 - 10$

= 30 28

= 2

Slope of normal $= - \frac{1}{2}$

Hence eqⁿ of tangent at (1, 3) is

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

$\Rightarrow y - 3 = 2 x - 2$

$\to 2 x - y + 1 = 0$

And eqⁿ of normal at (1,3) is

$(y - 3) = - \frac{1}{2} (x - 1)$

$\Rightarrow 2 y - 6 = - x + 1$

$\to x + 2 y - 7 = 0$

(iii) We have, $y = x^{3}$

Slope of tangent, $\frac{d y}{d x} = 3 x^{2}$

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

And slope of normal $= - \frac{1}{3}$

Hence, eqn of tan

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