Question

How do you find the equation of a line normal to the function `y=sqrtx(1-x)^2` at x=4?

Answer 1

`y-18=-4/57(x-4)`

Explanation:

Take the derivative of `y`. You will need quotient rule and chain rule:

`dy/dx=sqrtx (-2(1-x)) + (1-x)^2 ((1/2)(x^(-1/2)))`

Plug in `4`:

`dy/dx = sqrt4 (-2(1-4)) + (1-4)^2 ((1/2)(4^(-1/2)))=57/4`

That is the slope of the tangent line but we need the slope of the normal line. So we need to take the negative reciprocal of `57/4` which gives us `-4/57`

Now we need a point. Plug in `x=4` into the initial equation:

`y=sqrt(4)(1-4)^2=18`

So now we have the point `(4,18)` and we have the slope `-4/57`. Plug into point-slope form:

`y-18=-4/57(x-4)`