Cambridge Additional Mathematics

(4 4),

y

x

normal

tangent gradientmN

gradientmT

O

370 Applications of differential calculus (Chapter 14)

NORMALS

Anormalto a curve is a line which is perpendicular to the tangent at the point of contact.

The gradients of perpendicular lines are negative reciprocals of each other, so:

The gradient of the normal to the curve at x=a is mN=¡

1

f^0 (a)

.

The equation of the normal to the curve at x=a is y=f(a)¡

1

f^0 (a)

(x¡a).

Reminder: If a line has gradient^45 and passes through (2,¡3), another quick way to write down its

equation is 4 x¡ 5 y= 4(2)¡5(¡3) or 4 x¡ 5 y=23.

If the gradient was¡^45 , we would have:

4 x+5y= 4(2) + 5(¡3) or 4 x+5y=¡ 7.

Example 2 Self Tutor

Find the equation of the normal to y=

8

p

x

at the point where x=4.

When x=4, y=

8

p

4

=

8

2

=4. So, the point of contact is (4,4).

Now as y=8x

¡^12

,

dy

dx

=¡ 4 x

¡^32

) when x=4, mT=¡ 4 £ 4

¡^32

=¡^12

) the normal at (4,4) has gradient mN=^21.

) the equation of the normal is

2 x¡ 1 y= 2(4)¡1(4)

or 2 x¡y=4

EXERCISE 14A

1 Find the equation of the tangent to:

a y=x¡ 2 x^2 +3 at x=2 b y=

p

x+1 at x=4

c y=x^3 ¡ 5 x at x=1 d y=p^4

x

at (1,4)

e y=

3

x

¡

1

x^2

at (¡ 1 ,¡4) f y=3x^2 ¡

1

x

at x=¡ 1.

2 Find the equation of the normal to:

a y=x^2 at the point (3,9) b y=x^3 ¡ 5 x+2 at x=¡ 2

c y=

5

p

x

¡

p

x at the point (1,4) d y=8

p

x¡

1

x^2

at x=1.

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Additional Mathematics
Y:\HAESE\CAM4037\CamAdd_14\370CamAdd_14.cdr Wednesday, 8 January 2014 12:03:39 PM BRIAN