312 Higher Engineering Mathematics
Problem 22. Find the equation of the normal to
the curve y = 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 line CD in Fig. 28.12 has the equation
y =−x − 1.
Problem 23. Determine the equations of the
tangent and normal to the curve y =
x 3
5
at the point
−1, −
1
5
Gradient m of curve y =
x 3
5
is given by
m =
d y
dx
=
3x 2
5
At the point
−1, −
1
5
, x = − 1 and m =
3(−1) 2
5
=
3
5
Equation of the tangent is:
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
5y + 1 = 3x + 3
or
5y − 3x = 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
Multiplying each term by 15 gives:
15y + 3 = −25x − 25
Hence equation of the normal is:
15y + 25x + 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 = 2x 2 at the point (1, 2)
(a) y = 4x − 2
(b) 4y + x = 9
2. y = 3x 2 − 2x at the point (2, 8)
(a) y = 10x − 12
(b) 10y + x = 82
3. y =
x 3
2
at the point
−1, −
1
2
(a) y =
3
2 x + 1
(b) 6y + 4x + 7 = 0
4. y = 1 + x − x 2 at the point (−2, −5)
(a) y = 5x + 5
(b) 5y + x + 27 = 0
5. θ =
1
t
at the point
3,
1
3
(a) 9θ + t = 6
(b) θ = 9t − 26
2
3 or 3θ = 27t − 80
28.6 Small changes
If y is a function of x, i.e. y = f (x), and the approximate change in y corresponding to a small change δx in
x is required, then:
δy
δx
≈
d y
dx
and δy ≈
dy
dx
· δx or δy ≈ f
(x) · δx
Problem 22. Find the equation of the normal to
the curve y = 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 line CD in Fig. 28.12 has the equation
y =−x − 1.
Problem 23. Determine the equations of the
tangent and normal to the curve y =
x 3
5
at the point
−1, −
1
5
Gradient m of curve y =
x 3
5
is given by
m =
d y
dx
=
3x 2
5
At the point
−1, −
1
5
, x = − 1 and m =
3(−1) 2
5
=
3
5
Equation of the tangent is:
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
5y + 1 = 3x + 3
or
5y − 3x = 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
Multiplying each term by 15 gives:
15y + 3 = −25x − 25
Hence equation of the normal is:
15y + 25x + 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 = 2x 2 at the point (1, 2)
(a) y = 4x − 2
(b) 4y + x = 9
2. y = 3x 2 − 2x at the point (2, 8)
(a) y = 10x − 12
(b) 10y + x = 82
3. y =
x 3
2
at the point
−1, −
1
2
(a) y =
3
2 x + 1
(b) 6y + 4x + 7 = 0
4. y = 1 + x − x 2 at the point (−2, −5)
(a) y = 5x + 5
(b) 5y + x + 27 = 0
5. θ =
1
t
at the point
3,
1
3
(a) 9θ + t = 6
(b) θ = 9t − 26
2
3 or 3θ = 27t − 80
28.6 Small changes
If y is a function of x, i.e. y = f (x), and the approximate change in y corresponding to a small change δx in
x is required, then:
δy
δx
≈
d y
dx
and δy ≈
dy
dx
· δx or δy ≈ f
(x) · δx