Normal from differentiation

Finding the normal to a curve at a given point is very similar to finding the tangent. We just need to remember one thing:

A normal is perpendicular to the tangent and curve at that point. That means that its gradient is the negative reciprocal of the tangent’s gradient.

For exmaple:

• If the gradient of the tangent is 2, the gradient of the normal is -\frac12.
• If the gradient of the tangent is -\frac34, the gradient of the normal is \frac43.
• If the gradient of the tangent is 5, the gradient of the normal is -\frac15.

Finding the gradient of a normal

To find the gradient of a normal to a curve at a given point, we first need to find the gradient of the curve at that point (i.e. the gradient of the tangent).

We can do this by differentiating the function to get its derivative (gradient function), and then substituting in the x-coordinate of the point we want the normal at.

Then, remember to take the negative reciprocal of that gradient to get the gradient of the normal!

Find the gradient of the normal to the curve f(x) = x^3 - 2x + 1 at the point where x = 2.

• First, we find the first derivative:
  • f'(x) = 3x^{3-1} - 2\times1x^{1-1} + 0
  • = 3x^2 - 2
• Next, we substitute in x = 2 to find the gradient of the curve at that point:
  • f'(2) = 3(2)^2 - 2
  • = 3\times4 - 2
  • = 12 - 2
  • = 10
• Finally, we take the negative reciprocal to find the gradient of the normal:
  • Gradient of normal = -\frac{1}{10}
• Answer: The gradient of the normal at the point where x = 2 is -\frac{1}{10}.

Finding the full equation of the normal

Once we have the gradient of the normal, we can find its full equation in the same way as we would for a tangent.

Find a point that the normal passes through - we can do that by substituting the x-coordinate of the intersection point into the original function to get the y-coordinate.

Then, substitute the gradient and the point into the equation of a straight line to find the y-intercept, and finally write the full equation of the normal.

Find the full equation of the normal to the curve f(x) = x^3 - 2x + 1 at the point where x = 2.

(continuing from where we left off above)

• We already know that the normal passes through the point where x = 2. So we substitute x = 2 into the original function to find the corresponding y-coordinate:
  • f(2) = (2)^3 - 2(2) + 1
  • = 8 - 4 + 1
  • = 5
• So the normal passes through the point (2, 5).
• Now we can substitute all the information we have into the equation of a straight line to find the y-intercept, c:
  • y = mx + c
  • 5 = -\frac{1}{10}(2) + c
  • 5 = -\frac{2}{10} + c
  • 5 = -\frac{1}{5} + c
  • c = 5 + \frac{1}{5}
  • c = \frac{25}{5} + \frac{1}{5}
  • c = \frac{26}{5}
• Now we have our gradient and y-intercept, we can write the full equation of the normal:
• y = -\frac{1}{10}x + \frac{26}{5}

Find the equation of the normal to the curve f(x) = 2x^2 + 3x + 1 at the point where x = -1.

• First, we find the first derivative:
  • f'(x) = 2\times2x^{2-1} + 1\times3x^{1-1} + 0
  • = 4x + 3
• Next, we substitute in x = -1 to find the gradient of the curve at that point:
  • f'(-1) = 4(-1) + 3
  • = -4 + 3
  • = -1
• Now, we take the negative reciprocal to find the gradient of the normal:
  • Gradient of normal = 1
• Now, we find the corresponding y-coordinate by substituting x = -1 into the original function:
  • f(-1) = 2(-1)^2 + 3(-1) + 1
  • = 2(1) - 3 + 1
  • = 2 - 3 + 1
  • = 0
• So the normal passes through the point (-1, 0).
• Now we can substitute all the information we have into the equation of a straight line to find the y-intercept, c:
  • y = mx + c
  • 0 = 1(-1) + c
  • 0 = -1 + c
  • c = 0 + 1
  • c = 1
• Now we have our gradient and y-intercept, we can write the full equation of the normal:
  • y = 1x + 1
  • y = x + 1
• Answer: y = x + 1