448 Higher Engineering Mathematics
Problem 8. (a) The variation of resistance,
R ohms, of an aluminium conductor with
temperature θ ◦ C is given by
dR
dθ
= α R, where α
is the temperature coefficient of resistance of
aluminium. If R = R 0 when θ = 0 ◦ C, solve the
equation for R. (b) If α = 38 ×10 −4 / ◦ C, determine
the resistance of an aluminium conductor at 50 ◦ C,
correct to 3 significant figures, when its resistance
at 0 ◦ C is 24.0 .
(a)
dR
dθ
= α R is of the form
d y
dx
= f ( y)
Rearranging gives: dθ =
d R
α R
Integrating both sides gives:
dθ =
d R
α R
i.e.
θ =
1
α
ln R + c,
which is the general solution.
Substituting the boundary conditions R = R 0 when
θ = 0 gives:
0 =
1
α
ln R 0 + c
from which c = −
1
α
ln R 0
Hence the particular solution is
θ =
1
α
ln R −
1
α
ln R 0 =
1
α
( ln R − ln R 0 )
i.e. θ =
1
α
ln
R
R 0
or αθ = ln
R
R 0
Hence e
αθ
=
R
R 0
from which, R = R 0 e
αθ
(b) Substituting α = 38 ×10 −4 , R 0 = 24.0 and θ = 50
into R = R 0 e αθ gives the resistance at 50 ◦ C, i.e.
R 50 = 24.0 e (38×10 −4 ×50) = 29.0 ohms
Now try the following exercise
Exercise 178 Further problems on
equations of the form
dy
dx
= f ( y)
In Problems 1 to 3, solve the differential
equations.
1.
d y
dx
= 2 +3y
x =
1
3
ln(2 + 3y) + c
2.
d y
dx
= 2 cos 2 y
[tan y = 2x + c]
3. ( y 2 + 2)
d y
dx
= 5y, given y = 1 when x =
1
2
y 2
2
+ 2 ln y = 5x − 2
4. The current in an electric circuit is given by
the equation
Ri + L
di
dt
= 0,
where L and R are constants. Show that
i = I e
− Rt
L , given that i = I when t = 0.
5. The velocity of a chemical reaction is given by
dx
dt
= k(a − x), where x is the amount transferred in time t , k is a constant and a is
the concentration at time t = 0 when x = 0.
Solve the equation and determine x in terms
of t .
[ x = a(1 − e −kt )]
6. (a) Charge Q coulombs at time t seconds
is given by the differential equation
R
dQ
dt
+
Q
C
= 0, where C is the capacitance in farads and R the resistance in
ohms. Solve the equation for Q given
that Q = Q 0 when t = 0.
(b) A circuit possesses a resistance of
250 ×10 3 and a capacitance of
8.5 × 10 −6 F, and after 0.32 seconds
the charge falls to 8.0 C. Determine
the initial charge and the charge after
1 second, each correct to 3 significant
figures.
[(a) Q = Q 0 e
−t
CR (b) 9.30 C, 5.81 C]
7. A differential equation relating the difference
in tension T , pulley contact angle θ and coefficient of friction μ is
dT
dθ
= μT . When θ = 0,
T = 150 N, and μ = 0.30 as slipping starts.
Determine the tension at the point of slipping
when θ = 2 radians. Determine also the value
of θ when T is 300 N.
[273.3 N, 2.31 rads]
Problem 8. (a) The variation of resistance,
R ohms, of an aluminium conductor with
temperature θ ◦ C is given by
dR
dθ
= α R, where α
is the temperature coefficient of resistance of
aluminium. If R = R 0 when θ = 0 ◦ C, solve the
equation for R. (b) If α = 38 ×10 −4 / ◦ C, determine
the resistance of an aluminium conductor at 50 ◦ C,
correct to 3 significant figures, when its resistance
at 0 ◦ C is 24.0 .
(a)
dR
dθ
= α R is of the form
d y
dx
= f ( y)
Rearranging gives: dθ =
d R
α R
Integrating both sides gives:
dθ =
d R
α R
i.e.
θ =
1
α
ln R + c,
which is the general solution.
Substituting the boundary conditions R = R 0 when
θ = 0 gives:
0 =
1
α
ln R 0 + c
from which c = −
1
α
ln R 0
Hence the particular solution is
θ =
1
α
ln R −
1
α
ln R 0 =
1
α
( ln R − ln R 0 )
i.e. θ =
1
α
ln
R
R 0
or αθ = ln
R
R 0
Hence e
αθ
=
R
R 0
from which, R = R 0 e
αθ
(b) Substituting α = 38 ×10 −4 , R 0 = 24.0 and θ = 50
into R = R 0 e αθ gives the resistance at 50 ◦ C, i.e.
R 50 = 24.0 e (38×10 −4 ×50) = 29.0 ohms
Now try the following exercise
Exercise 178 Further problems on
equations of the form
dy
dx
= f ( y)
In Problems 1 to 3, solve the differential
equations.
1.
d y
dx
= 2 +3y
x =
1
3
ln(2 + 3y) + c
2.
d y
dx
= 2 cos 2 y
[tan y = 2x + c]
3. ( y 2 + 2)
d y
dx
= 5y, given y = 1 when x =
1
2
y 2
2
+ 2 ln y = 5x − 2
4. The current in an electric circuit is given by
the equation
Ri + L
di
dt
= 0,
where L and R are constants. Show that
i = I e
− Rt
L , given that i = I when t = 0.
5. The velocity of a chemical reaction is given by
dx
dt
= k(a − x), where x is the amount transferred in time t , k is a constant and a is
the concentration at time t = 0 when x = 0.
Solve the equation and determine x in terms
of t .
[ x = a(1 − e −kt )]
6. (a) Charge Q coulombs at time t seconds
is given by the differential equation
R
dQ
dt
+
Q
C
= 0, where C is the capacitance in farads and R the resistance in
ohms. Solve the equation for Q given
that Q = Q 0 when t = 0.
(b) A circuit possesses a resistance of
250 ×10 3 and a capacitance of
8.5 × 10 −6 F, and after 0.32 seconds
the charge falls to 8.0 C. Determine
the initial charge and the charge after
1 second, each correct to 3 significant
figures.
[(a) Q = Q 0 e
−t
CR (b) 9.30 C, 5.81 C]
7. A differential equation relating the difference
in tension T , pulley contact angle θ and coefficient of friction μ is
dT
dθ
= μT . When θ = 0,
T = 150 N, and μ = 0.30 as slipping starts.
Determine the tension at the point of slipping
when θ = 2 radians. Determine also the value
of θ when T is 300 N.
[273.3 N, 2.31 rads]