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Contents
Tangents & Normals
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Equations
of Tangents & Normals
 Say
we have a graph y = f(x), and a point A with coordinates (a, b). The line l1,
which touches the graph at only one point, is the tangent. The line l2,
which is perpendicular to the tangent, is the normal.
The
gradient of the tangent, m1, is given by the derivative of f(x),
i.e. f'(x). To find the gradient of the normal, m2, we use the
property m1m2 = -1. Hence the gradient is the normal is
-1/f'(a).
Hence,
using the general cartesian equations for straight lines, we have:
Equation of tangent:
y - b = f'(a) (x - a), and
Equation of normal:
y - b = -1/f'(a) (x - a)
We
will illustrate this idea with some examples.
Example:
 The
equation of a curve is y = 1/3 (8 - x2)6. Find dy/dx,
and the equation of the normal at the point when x = 3.
Solution:
dy/dx
= 2 (8 - x2)5 (-2x)
= -4x (8 - x2)5
When
x = 3, y = 1/3 (8 - 32)6 = 1/3
dy/dx|x=3
= -4 (3) (8 - 32)5 = 12
Gradient
of normal = -1/12
Equation
of normal:
y - 1/3 = -1/12 (x - 3)
12y - 4 = -x + 3
x + 12y = 7
Example:
Find
the equation of the tangents to the curve x2 - 3xy2
+ y3 = 8, at the point where y = 2.
Solution:
x2
- 3xy2 + y3 = 8
Differentiating
w.r.t x,
When
y = 2,
x2 -
3x(2)2 + (2)3 = 8
x (x - 12) = 0
x = 0 or x = 12
At
(0, 2), dy/dx = 3(4)/3(4) = 1
Equation
of tangent: y - 2 = x
y = x + 2
At
(12, 2), dy/dx = [3(4)-2(12)]/[3(4)-6(12)(2)] = 1/11
Equation
of tangent: y - 2 = 1/11 (x - 12)
11y = x + 10
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