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.
- y= 2 x^2 at the point (1, 2)
[
(a)y= 4 x− 2
(b) 4 y+x= 9
]
- y= 3 x^2 − 2 xat the point (2, 8)
[
(a)y= 10 x− 12
(b) 10 y+x= 82
]
- y=
x^3
2
at the point
(
− 1 ,−
1
2
)
[
(a)y=^32 x+ 1
(b) 6 y+ 4 x+ 7 = 0
]
- y= 1 +x−x^2 at the point (−2,−5)
[
(a)y= 5 x+ 5
(b) 5 y+x+ 27 = 0
]
- 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