Question

What is the equation of the normal line of `f(x)=x^2` at `x=5`?

Answer 1

`y = -1/10x + 51/2`

Explanation:

The normal line is the line perpendicular to the tangent line. Then, the steps are as follows:

• Find the slope `m_"tangent"` of tangent line
  -Find the first derivative of the function
  -Evaluate the first derivative at the desired point

• Find the slope `m_"norm"` of the normal line
  -Let `m_"norm" = -1/m_"tangent"`

• Find the line with slope `m_"norm"` passing through the given point
  -Using point-slope form: `(y-y_1) = m_"norm"(x-x_1)`

Let's go through the process.

`f'(x) = d/dxx^2 = 2x`

`m_"tangent" = f'(5) = 2(5) = 10`

`m_"norm" = -1/m_"tangent" = -1/10`

`(x_1, y_1) = (5, f(5)) = (5, 25)`

Substituting, we get the equation of the normal line.

`y - 25 = -1/10(x - 5)`

`=> y = -1/10x + 51/2`