Question

What is the equation of the line normal to `f(x)=e^(sqrtx)` at `x=4`?

Answer 1

It is `y=-4/e^2x+9.55`.

Explanation:

The line passing from `x=4` and tangent to `e^sqrt(x)` has as slope

`d/dxe^sqrt(x)` evaluated when `x=4`. The derivative can be calculated with the chain's rule

`d/dxe^sqrt(x)=e^sqrt(x)*d/dxsqrt(x)=e^sqrt(x)/(2sqrt(x))`.

The slope is

`m=e^sqrt(4)/(2sqrt(4))=e^2/4`.

So the tangent has equation

`y=e^2/4x+q`

The `q` can be found imposing the passage of the line from the point `(4, e^sqrt(4))=(4, e^2)`

`e^2=e^2/4*4+q` then `q=0` and the equation of the tangent is

`y=e^2/4x`

The orthogonal has the negative inverse slope, so the equation is

`y=-4/e^2x+q`

Again, to find `q` we have to impose the passage from `(4, e^2)`

`e^2=-4/e^2*4+q`

`q=e^2+16/e^2=(e^4+16)/e^2\approx9.55` and finally the equation of the orthogonal is

`y=-4/e^2x+9.55` or

`y=-0.54x+9.55`.