324 Higher Engineering Mathematics
When x = 2 and y = 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
5 2 1
2
Gradient
5
1
2
2
1
4
x
1
2
4
3
y
21
22
0
r 5 5
x 2 1y 2 2 2x
22y 5 3
Figure 30.2
Problem 10. Pressure p and volume v of a gas
are related by the law pv γ = k, where γ and k are
constants. Show that the rate of change of pressure
d p
dt
=−γ
p
v
dv
dt
Since pv γ = k, then p =
k
v γ = kv
−γ
d p
dt
=
d p
dv
×
dv
dt
by the function of a function rule
d p
dv
=
d
dv
(kv
−γ
)
= −γ kv
−γ −1
=
−γ k
v γ +1
d p
dt
=
−γ k
v γ +1 ×
dv
dt
Since k = pv
γ
,
d p
dt
=
−γ (pv γ )
v γ +1
dv
dt
=
−γ pv γ
v γ v 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
1. x 2 + y 2 + 4x − 3y + 1 = 0
2x + 4
3 −2y
2. 2y 3 − y + 3x − 2 = 0
3
1 − 6y 2
3. Given x 2 + y 2 = 9 evaluate
dy
dx
when
x =
√
5 and y = 2.
−
√
5
2
In Problems 4 to 7, determine
dy
dx
4. x 2 + 2x sin 4y = 0
−(x + sin 4y)
4x cos 4y
5. 3y 2 + 2x y − 4x 2 = 0
4x − y
3y + x
6. 2x 2 y + 3x 3 = sin y
x(4y + 9x)
cos y − 2x 2
7. 3y + 2x ln y = y 4 + x
1 − 2 ln y
3 +(2x/y) − 4y 3
8. If 3x 2 + 2x 2 y 3 −
5
4
y 2 = 0 evaluate
dy
dx
when
x =
1
2
and y = 1.
[5]
9. Determine the gradients of the tangents
drawn to the circle x 2 + y 2 = 16 at the point
where x = 2. Give the answer correct to 4
significant figures.
[±0.5774]
10. 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]
11. Determine the gradient of the curve
3x y + y 2 =−2 at the point (1,−2).
[−6]
When x = 2 and y = 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
5 2 1
2
Gradient
5
1
2
2
1
4
x
1
2
4
3
y
21
22
0
r 5 5
x 2 1y 2 2 2x
22y 5 3
Figure 30.2
Problem 10. Pressure p and volume v of a gas
are related by the law pv γ = k, where γ and k are
constants. Show that the rate of change of pressure
d p
dt
=−γ
p
v
dv
dt
Since pv γ = k, then p =
k
v γ = kv
−γ
d p
dt
=
d p
dv
×
dv
dt
by the function of a function rule
d p
dv
=
d
dv
(kv
−γ
)
= −γ kv
−γ −1
=
−γ k
v γ +1
d p
dt
=
−γ k
v γ +1 ×
dv
dt
Since k = pv
γ
,
d p
dt
=
−γ (pv γ )
v γ +1
dv
dt
=
−γ pv γ
v γ v 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
1. x 2 + y 2 + 4x − 3y + 1 = 0
2x + 4
3 −2y
2. 2y 3 − y + 3x − 2 = 0
3
1 − 6y 2
3. Given x 2 + y 2 = 9 evaluate
dy
dx
when
x =
√
5 and y = 2.
−
√
5
2
In Problems 4 to 7, determine
dy
dx
4. x 2 + 2x sin 4y = 0
−(x + sin 4y)
4x cos 4y
5. 3y 2 + 2x y − 4x 2 = 0
4x − y
3y + x
6. 2x 2 y + 3x 3 = sin y
x(4y + 9x)
cos y − 2x 2
7. 3y + 2x ln y = y 4 + x
1 − 2 ln y
3 +(2x/y) − 4y 3
8. If 3x 2 + 2x 2 y 3 −
5
4
y 2 = 0 evaluate
dy
dx
when
x =
1
2
and y = 1.
[5]
9. Determine the gradients of the tangents
drawn to the circle x 2 + y 2 = 16 at the point
where x = 2. Give the answer correct to 4
significant figures.
[±0.5774]
10. 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]
11. Determine the gradient of the curve
3x y + y 2 =−2 at the point (1,−2).
[−6]