Higher Engineering Mathematics, Sixth Edition

312 Higher Engineering Mathematics

Problem 22. Find the equation of the normal to

the curvey=x^2 −x−2 at the point (1,−2).

m=1 from Problem 21, hence the equation of the

normal is

y−y 1 =−

1

m

(x−x 1 )

i.e. y−(− 2 )=−

1

1

(x− 1 )

i.e. y+ 2 =−x+ 1

or y=−x− 1

Thus the lineCD in Fig. 28.12 has the equation

y=−x−1.

Problem 23. Determine the equations of the

tangent and normal to the curvey=

x^3

5

at the point

(

− 1 ,−

1

5

)

Gradientmof curvey=

x^3

5

is given by

m=

dy

dx

=

3 x^2

5

At the point

(

− 1 ,−^15

)

,x=−1andm=

3 (− 1 )^2

5

=

3

5

Equation of the tangentis:

y−y 1 =m(x−x 1 )

i.e. y−

(

−

1

5

)

=

3

5

(x−(− 1 ))

i.e. y+

1

5

=

3

5

(x+ 1 )

or 5 y+ 1 = 3 x+ 3

or 5 y− 3 x= 2

Equation of the normal is:

y−y 1 =−

1

m

(x−x 1 )

i.e. y−

(

−

1

5

)

=

− 1

( 3 / 5 )

(x−(− 1 ))

i.e. y+

1

5

=−

5

3

(x+ 1 )

i.e. y+

1

5

=−

5

3

x−

5

3

Multiplyingeach term by 15 gives:

15 y+ 3 =− 25 x− 25

Henceequation of the normalis:

15 y+ 25 x+ 28 = 0

Now try the following exercise

Exercise 124 Further problems on

tangents and normals

For the curves in problems 1 to 5, at the points

given, find (a) the equation of the tangent, and (b)

the equation of the normal.

1. y= 2 x^2 at the point (1, 2)

[

(a)y= 4 x− 2

(b) 4 y+x= 9

]

2. y= 3 x^2 − 2 xat the point (2, 8)
   [
   (a)y= 10 x− 12
   (b) 10 y+x= 82

]

3. y=

x^3

2

at the point

(

− 1 ,−

1

2

)

[

(a)y=^32 x+ 1

(b) 6 y+ 4 x+ 7 = 0

]

4. y= 1 +x−x^2 at the point (−2,−5)
   [
   (a)y= 5 x+ 5
   (b) 5 y+x+ 27 = 0

]

5. θ=

1

t

at the point

(

3 ,

1

3

)

[

(a) 9 θ+t= 6

(b)θ= 9 t− 2623 or 3θ= 27 t− 80

]

28.6 Small changes

Ifyis a function ofx,i.e.y=f(x), and the approxi-

mate change inycorresponding to a small changeδxin

xis required, then:

δy

δx

≈

dy

dx

andδy≈

dy

dx

·δx or δy≈f′(x)·δx