324 Higher Engineering Mathematics
Whenx=2andy=3,
dy
dx
=
1 − 2
3 − 1
=
− 1
2
Hence the gradients of the tangents are±
1
2
The circle having the given equation has its centre at
(1, 1) and radius
√
5 (see Chapter 13) and is shown in
Fig. 30.2 with the two gradients of the tangents.
Gradient
521
2
Gradient
51
2
142 x
1
2
4
3
y
21
22
0
r^55
x 21 y 222 x
22 y 53
Figure 30.2
Problem 10. Pressurepand volumevof a gas
are related by the lawpvγ=k,whereγandkare
constants. Show that the rate of change of pressure
dp
dt
=−γ
p
v
dv
dt
Sincepvγ=k,thenp=
k
vγ
=kv−γ
dp
dt
=
dp
dv
×
dv
dt
by the function of a function rule
dp
dv
=
d
dv
(kv−γ)
=−γkv−γ−^1 =
−γk
vγ+^1
dp
dt
=
−γk
vγ+^1
×
dv
dt
Sincek=pvγ,
dp
dt
=
−γ(pvγ)
vγ+^1
dv
dt
=
−γpvγ
vγv^1
dv
dt
i.e.
dp
dt
=−γ
p
v
dv
dt
Now try the following exercise
Exercise 130 Further problems on implicit
differentiation
In Problems 1 and 2 determine
dy
dx
- x^2 +y^2 + 4 x− 3 y+ 1 = 0
[
2 x+ 4
3 − 2 y
]
- 2y^3 −y+ 3 x− 2 = 0
[
3
1 − 6 y^2
]
- Givenx^2 +y^2 =9evaluate
dy
dx
when
x=
√
5andy=2.
[
−
√
5
2
]
In Problems 4 to 7, determine
dy
dx
- x^2 + 2 xsin4y= 0
[
−(x+sin4y)
4 xcos4y
]
- 3y^2 + 2 xy− 4 x^2 = 0
[
4 x−y
3 y+x
]
- 2x^2 y+ 3 x^3 =siny
[
x( 4 y+ 9 x)
cosy− 2 x^2
]
- 3y+ 2 xlny=y^4 +x
[
1 −2lny
3 +( 2 x/y)− 4 y^3
]
- If 3x^2 + 2 x^2 y^3 −
5
4
y^2 =0evaluate
dy
dx
when
x=
1
2
andy=1. [5]
- Determine the gradients of the tangents
drawn to the circlex^2 +y^2 =16 at the point
wherex=2. Give the answer correct to 4
significant figures. [± 0 .5774] - Find the gradients of the tangents drawn to
the ellipse
x^2
4
+
y^2
9
=2 at the point where
x=2. [± 1 .5]
- Determine the gradient of the curve
3 xy+y^2 =−2 at the point (1,−2). [−6]