Find the equations of the tangent and normal to the given curves at the indicated points: (i) `y=x^4-6x^3+13 x^2-10 x+5` at `(0," "5)` (ii) `y=x^4-6x^3+13 x^2-10 x+5` at `(1," "3)` (iii) `y=x^3` at `(1," "1)` (iv) `y=x^2` at `(0," "0)` (v) `x" "="

Explanation

(i) Given curve
`y= x^(4)-6x^(3)+13x^(2)-10x+5`
` rArr (dy)/(dx) = 4x^(3) - 18x^(2)+26x-10`
at point (0, 5)
slope of tangent `m=0-0+0-10=-10`
and equation of tangent ` y - 5 =-10 (x-0)`
` rArr 10x + y - 5 = 0`
Equation of normal
`y-5=1/10(x-0)`
` rArr 10y - 50 = x`
` rArr x- 10y + 50 = 0`
(ii) Given curve `y= x^(4)-6x^(3)+13x^(2)-10x+5`
` rArr (dy)/(dx) = 4x^(3) - 18x^(2) + 26x - 10 `
at point (1, 3)
slope of tangent `m=4-18+26y-10=2`
and equation of tangent
`y-3=2(x-1)`
` rArr 2x-y+1=0`
Equation of normal
`y-3=-1/2(x-1)`
` rArr 2y - 6 =- x+1`
` rArr x+2y - 7 =0`
(iii) Given curve
`y = x^(3)`
` rArr (dy)/(dx) = 3x^(2)`
at point (1, 1)
slope of tangent `m = 3 (1)^(2)=3`
and equation of tangent ,brgt ` y-1 = 3 (x-1)`
` rArr 3x-y-2=0`
Equation of normal
` y-1=-1/3 (x-1)`
`rArr 3y-3=-x+1`
` rArr x+ 3y - 4 = 0`
(iv) Equation of curve `y = x^(2)`
`rArr (dy)/(dx) = 2x`
at point (0, 0)
slope of tangent m = 0
and equation of tnagent
`y - 0= 0 (x-0)`
` rArr y= 0`
Equation of normal
`y-0=(-1)/0 (x-0)`
` rArr x= 0`
(v) x= cos t
` rArr (dx)/(dt) = - sin t`
` y = sin t`
`rArr (dy)/(dt) = cos t `
`:. (dy)/(dx) = ((dy)//(dt))/((dx)//(dt)) = (cos t)/(-sin t) = - cot t`
at ` t = pi/4`
x_(1) = cos pi/4 = 1/sqrt2 and y_(1) = sin pi/4 = 1/sqrt2`
Slope of tangent = `-cot pi/4 = - 1`
`:. ` equation of tangent
`y - 1/sqrt2 = - 1 (x- 1/sqrt2)`
` rArr x+ y = 1/sqrt2 rArr x+ y = sqrt2`
Equation of normal ` y - 1/sqrt2 = 1 (x-1/sqrt2)`
` rArr x= y