Question

How do you find the equation of the tangent and normal line to the curve `y=x+cosx` at `x=1`?

Answer 1

`Tangent: y = 0.15852x+1.3818` See Socratic graph for the curve, the point of contact and the tangent. Normal : `y = -6.3084x+7.8487`

Explanation:

`x = 1 radian = 57.296^o`

The point of contact is (1, 1.5403)

y' ar x = 1 is `m = 1-sin 57.2903^o=0.15852`, nearly.

So, the equation of the tangent is

`y-1.5403=0.15852(x-1)` Simplifying,

`y = 0.15852x+1.3818`.

The tangent crosses the curve, elsewhere.

The normal to the curve is given by

`y-1.5403=-1/0.15852(x-1)`, Simplifying,

`y = -6.3084x+7.8487`

graph{(y-0.16x-1.38)(y-x-cos x)(y+6.31x-7.84)=0 [-20, 20, -10, 10]}