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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.
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)
=-1 = m
Therefore, we know that the normal gradient is n=-1/m
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.see more