How do I find the equation of the normal line given a point on the curve?

The normal line to a curve at a particular point is the line through that point and perpendicular to the tangent.

A simple trick to remembering how to find the normal gradient, n, is that the slope of any line perpendicular to a line that has a gradient, m, is just the negative reciprocal, -1/m.


Find the normal gradient to the curve y=2x3 +3x+7 at the point (1,1).

So firstly, let’s recap how to calculate the gradient of the tangent line:

By differentiating y=2x3 +3x+7 , we find

dy/dx = 6x2 +3

Then, by substituting in our point, at x=1 we yield dy/dx=9. This is the tangent gradient of the curve (m=9).

Finally we substitute this into our formula for calculating the normal gradient n=-1/m.

Therefore n=-1/9.


Now, let’s try another example which demonstrates how we use the normal gradient to find an equation for the normal line.

We will use the formula (y-y0) = n(x-x0), where (x0,y0) is a given point.


Consider a curve y=x5+3x2 +2. Find the equation of the normal to the curve at the point (-1,2). Leave your answer in the form y=mx+c.

By differentiating the curve, we have dy/dx = 5x4 +6x.

To find the gradient of the tangent line we substitute in x=-1, which yields

dy/dx = 5(-1)4 +6(-1)

                = 5-6

                =-1 = m

Therefore, we know that the normal gradient is n=-1/m

So n=1


Finally, we substitute this into our formula for the normal line (y-y0) = n(x-x0):

In our example, (x0, y0) = (-1,2)


So   y-2 = 1(x+1)

And my rearranging, we find y = x+3.

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