4 The normal line to a curve

We have already noted that, at any point ( x 0 , y 0 ) on a curve y = f ( x ) , the tangent and normal lines are perpendicular. Thus if the equations of the tangent and normal lines are, respectively

y = m x + c y = n x + d

then m = − 1 n or, equivalently n = − 1 m .

We have also noted, for the tangent line

m = d f d x x 0

so n can easily be obtained. To find d , we again use the fact that the normal line y = n x + d and the curve have a point in common:

y 0 = n x 0 + d and y 0 = f ( x 0 )

so n x 0 + d = f ( x 0 ) leading to d = f ( x 0 ) − n x 0 .

Task!

Find the equation of the normal line to curve y = sin 3 x at the point x = π 4 .

[The equation of the tangent line was found in the previous Task.]

First find the value of m :

m = − 3 2

Hence find the value of n :

n m = − 1 ∴ n = 2 3

The equation of the normal line is y = 2 3 x + d . Now find the value of d to 2 d.p.. (Remember the normal line must pass through the curve at the point x = π 4 .)

2 3 π 4 + d = sin π 4 ∴ d = 1 2 − 2 3 π 4 ≃ 0.34

Now obtain the equation of the normal line to 2 d.p.:

y = 0.47 x + 0.34 . The curve and the normal line are shown in the following figure:

The image shows a wave function plotted on a coordinate system with a straight line labeled "normal line" passing through the origin.

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Task!

Find the equation of the normal line to the curve y = x 3 at x = 1 .

First find f ( x ) , x 0 , d f d x x 0 , m , n :

f ( x ) = x 3 , x 0 = 1 , d f d x 1 = 3 x 2 1 = 3 ∴ m = 3 and n = − 1 3 Now use the property that the normal line y = n x + d and the curve y = x 3 pass through the point ( 1 , 1 ) to find d and so obtain the equation of the normal line:

1 = n + d ∴ d = 1 − n = 1 + 1 3 = 4 3 . Thus the equation of the normal line is y = − 1 3 x + 4 3 . The curve and the normal line through ( 1 , 1 ) are shown below:

A graph showing a curve, a tangent line, and a normal line passing through a specific point on the curve.

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Exercises
  1. Find the equations of the tangent and normal lines to the following curves at the points indicated
    1. y = x 4 + 2 x 2 , ( 1 , 3 )
    2. y = 1 − x 2 , ( 2 2 , 2 2 ) What would be obtained if the point was ( 1 , 0 ) ?
    3. y = x 1 ∕ 2 , ( 1 , 1 )
  2. Find the value of a if the two curves y = e − x and y = e a x are to intersect at right-angles.
    1. f ( x ) = x 4 + 2 x 2 d f d x = 4 x 3 + 4 x , d f d x x = 1 = 8

      tangent line y = 8 x + c . This passes through ( 1 , 3 ) so    y = 8 x − 5

      normal line y = − 1 8 x + . ̣ This passes through ( 1 , 3 ) so y = − 1 8 x + 25 8 .

    2. f ( x ) = 1 − x 2 d f d x = − x 1 − x 2 d f d x x = 2 2 = − 1

      tangent line y = − x + c . This passes through 2 2 , 2 2 so y = − x + 2

      normal line y = x + d . This passes through 2 2 , 2 2 so y = x .

      At ( 1 , 0 ) the tangent line is x = 1 and the gradient is infinite (the line is vertical), and the normal line is y = 0 .
    3. f ( x ) = x 1 2 d f d x = 1 2 x − 1 2 d f d x x = 1 = 1 2

      tangent line: y = 1 2 x + c . This passes through ( 1 , 1 ) so y = 1 2 x + 1 2

      normal line: y = − 2 x + d . This passes through ( 1 , 1 ) so y = − 2 x + 3 .
  2. The curves will intersect at right-angles if their tangent lines, at the point of intersection, are perpendicular.

    Point of intersection: e − x = e a x i.e. − x = a x ∴ x = 0 ( a = − 1 not sensible)

    The tangent line to y = e a x is  y = m x + c  where m = a e a x x = 0 = a

    The tangent line to y = e − x is  y = k x + g  where k = − e − x x = 0 = − 1

    These two lines are perpendicular if a ( − 1 ) = − 1 i.e. a = 1.

    A graph showing two exponential functions, \(y = e^{-x}\) and \(y = e^{x}\), plotted on a coordinate system.

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