Differentiation
P1^
5
Tangents and normals
Now that you know how to find the gradient of a curve at any point you can use
this to find the equation of the tangent at any specified point on the curve.
EXAMPLE 5.8 Find the equation of the tangent to the curve y = x^2 + 3 x + 2 at the point (2, 12).
SOLUTION
Calculating d
d
d
d
y
x
y
: x=+ 23 x.
Substituting x = 2 into the expression d
d
y
x
to find the gradient m of the tangent at
that point:
m = 2 × 2 + 3
= 7.
The equation of the tangent is given by
y − y 1 = m(x − x 1 ).
In this case x 1 = 2, y 1 = 12 so
y − 12 = 7(x − 2)
⇒ y = 7 x − 2.
This is the equation of the tangent.
The normal to a curve at a particular point is the straight line which is at
right angles to the tangent at that point (see figure 5.13). Remember that for
perpendicular lines, m 1 m 2 = −1.
–2 O
y = x^2 + 3x + 2
–1 (^2) x
y
y = 7x – 2
(2, 12)
Figure 5.12