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Math 263b

Sept. 24

12 Applications of Linear Equations Equations Objectives: • Identify the correct linear equations to be solved • Apply A l th the linear li equations ti in i th the given i problems • Solve the coefficients and find the particular solution

Applications of Linear Equations

Electric Circuit

Linear Equation

Mixing Problem

1

Math 263b

Sept. 24

Electric Circuits

RL Circuit

Electric Circuit

RC Circuit

Variables used Electric Circuits • i - current • q – charge (coulomb) dq Relationship : i = • t – time (second)

dt

Voltage drop : di 1. Across an inductor, L : VL = L

dt

2. Across a resistor, R : VR = Ri = R 3. Across a capacitor, C : VC =

dq dt

1 q C

2

Math 263b

Sept. 24

Type 1 : RL Circuit

Kirchoff’s Second Law

V L + VR = E ( t ) L

di + Ri = E ( t ) Where L and R are constants dt

L E

R

Type 2 : RC Circuit

Kirchoff’s Second Law

VR + VC = E ( t ) R

dq 1 + q = E ( t ) Where C and R are constants dt C

E

R

C

3

Math 263b

Sept. 24

Kirchoff’s Second Law

Example 1

A generator having emf 100 volts is connected in series with 10 ohms resistor and inductor 2 henries. If the switch K is i closed l d att time ti t = 0, 0 sett up a diff differential ti l for f th the current and determine the current at time t.

Solution L E

R

RL circuit with E(t)=100V, R=10 Ω Ω,, L=2H di L + Ri = E ( t ) dt

2

Example 1 (cont.) Solution

Integrating factor ∫ p( t )dt ∫ 5 dt

∫

=e = e5t

I ( t ) i = I ( t ) q ( t ) dt

di + 5i = 50 dt

Kirchoff’s Second Law

di + 5i = 50 dt di + p (t ) i = q (t ) dt

I (t ) = e

di + 10i = 100 dt

∫

ie 5t = e 5t .50 dt

ie 5t = i=

50e 5t +c 5 10e 5t + c e

5t

t=0 0, i = 0

i = 10 + ce −5t −5 0 0 = 10 + ce ( ) c = −10

i ( t ) = 10 − 10e −5t

4

Math 263b

Sept. 24

Kirchoff’s Second Law

Example 2

A decaying emf E = 200e −5t is connected in series with a 20 ohms resistor and 0.01 farad capacitor. Assuming q = 0 att t = 0. 0 Find Fi d the th charge h and d currentt att any time. ti If th the charge reaches a maximum, calculate its value and determine when it is reached.

Solution

RC circuit with E(t)=200e-5tV, R=20 Ω Ω,, C=0.01F

E

R

dq 1 dq 1 200e −5 t + q = E (t ) + q= dt C dt 20 ( 0.01 ) 20 dq 1 + 20 q = 200e −5 t dt 0.01 R

C

Example 2 (cont.) Solution

Kirchoff’s Second Law

dq q + 5q 5 = 10e 10 −5 t dt

dq + p (t ) q = q (t ) dt Integrating factor ∫ p( t )dt ∫ 5 dt

I (t ) = e

=e = e5t

∫

I ( t ) q = I ( t ) q ( t ) dt

∫ = 10 dt ∫

qe 5t = e 5t .10e −5t dt qe 5t

qe 5t = 10t + c 10t + c q= e 5t

t=0 0, q = 0

q = 10te −5t + ce −5t −5 0 0 = 0 + ce ( )

c =0 q ( t ) = 10te −5t

Charge at time t

5

Math 263b

Sept. 24

Example 2 (cont.) Solution

Kirchoff’s Second Law

DONT’ FORGET! Product Rule q ( t ) = 10te −5t DONT dq = 10t −5e −5 t + e −5 t ( 10 ) i (t ) = dt = 10e −5 t ( 1 − 5t ) Current at time t

(

)

When the charges reaches maximum, i ( t ) = 10e −5t ( 1 − 5t ) = 0

≠0

Example 2 (cont.) Solution

1 − 5t = 0 1 t= 5

dq =0 dt

Kirchoff’s Second Law

Then use Second Derivative Test q ′′ ( t ) =

⎛1⎞ q ′′ ⎜ ⎟ = −50e −1 < 0 ⎝5⎠

d ⎡ 10e −5 t ( 1 − 5t ) ⎤ ⎦ dt ⎣

q ′′ ( t ) = − 50e −5 t ( 1 − 5t ) − 50e −5 t

This is maximum value

q ′′ ( t ) = − 50e −5 t ( 1 − 5 t + 1 )

Therefore, the maximum charge is

q ′′ ( t ) = −50e −5 t ( 2 − 5t )

t=

1 5

⎛1⎞ q ′′ ⎜ ⎟ = ⎝5⎠

⎛1⎞

⎛1⎞ −5 ⎜ ⎟ ⎛ −50e ⎝ 5 ⎠ 2

⎛ 1 ⎞⎞ ⎜ − 5⎜ ⎟⎟ ⎝ 5 ⎠⎠ ⎝

⎛1⎞ ⎛ 1 ⎞ −5 ⎜ ⎟ q ⎜ ⎟ = 10 ⎜ ⎟ e ⎝ 5 ⎠ ⎝5⎠ ⎝5⎠ −1 = 2e

6

Math 263b

Sept. 24

Mixing Problems rin > rout

Mixing Problem rin = rout

rin < rout

Mixing Problems Variables used • Q(t) : the quantity of substance in the tank at time t. • Rate in : the rate at which substance enter the tank • Rate out : the rate at which substance leave the tank dQ : rate at which the quantity of substance in • dt the tank change • rate of flow in : the rate at which substance flows into the tank • Rate R t of f fl flow outt : th the rate t att which hi h substance b t fl flows out of the tank • Volume of solution : volume of solution in the tank.

7

Math 263b

Sept. 24

Mixing Problem

Formula

concentration =

quantity of subs tan ce ,Q ( t ) volume of solution

volume of solution = volume of solution in tan k +

( rin − rout ) t rate in = ( c in )( rin )

dQ = rate in − rate out dt

Example 3

⎛ Q (t ) ⎞ rate out = ⎜ ⎟ r ⎜ V ⎟ ( out ) ⎝ ⎠

Mixing Problem

A tank is filled with 10 liter of brine in which is dissolved 5 grams of salt. salt. Brine containing 3 grams per liter of salt enters t the th tank t k att 2 liter lit per minute i t and d th the mixture i t leaves at the same rate. rate. i) find the amount of salt in the tank at any time t. ii) how much salt is present when t approach infinity?

Solution

t = 0 ,Q = 5

V = 10 + ( 2 − 2 ) t

V = 10 rate in = ( 3 )( 2 ) = 6

⎛ Q (t ) ⎞ Q (t ) rate out = ⎜ ⎟ 2 = ⎜ 10 ⎟ ( ) 5 ⎝ ⎠

8

Math 263b

Sept. 24

Mixing Problem

Example 3 (cont.)

i) find the amount of salt in the tank at any time t. ii) how much salt is present when t approach infinity?

Solution dQ = rate in − rate out dt dQ Q =6− dt 5 dQ Q + =6 dt 5 dQ + p (t )Q = q (t ) dt

Example 3 (cont.)

I (t ) = e

1

∫ 5 dt

=e

1 t 5 Qe

1 t 5

∫ I ( t ) q ( t ) dt = e .66 dt ∫

I (t )Q = 1 t 5 Qe

1 t 5 Qe

Q=

1 t 5

1

=

6e 5 1

=

1 t 5 30 e

1 t 5 30 e

+C

1 t e5

t

Q = 30 + Ce

+C

+C

1 − t 5

5

Mixing Problem

i) find the amount of salt in the tank at any time t. ii) how much salt is present when t approach infinity?

Solution Q = 30 +

1 − t 5 Ce

t = 0 ,Q = 5 5 = 30 +

C = − 25

1 − (0 ) 5 Ce

Q = 30 − 25 e

1 − t 5

ii) T approach infinity lim Q = 30 − 25 e

−

1 t 5

=0

t→∞

= 30

9

Math 263b

Sept. 24

Example 4

Mixing Problem

A tank filled with 10 liters of brine in which dissolved 20 gm of salt. salt. Brine containing 4 gm/liter of salt enters the t k att 2 liter/min tank lit / i and d th the mixture i t lleaves att 1 liter/min. lit / i . liter/min Find the amount of salt in the tank after 40 minutes.

Solution t = 0 ,Q = 20

rate in = ( 4 )( 2 ) = 8

Example 4 (cont.) Solution dQ = rate in − rate out dt dQ Q =8− dt 10 + t dQ Q + =8 dt 10 + t dQ + p (t )Q = q (t ) dt

V = 10 + ( 2 − 1 ) t V = 10 + t

⎛ Q (t ) rate out = ⎜ ⎜ ⎝ 10 + t

⎞ Q (t ) ⎟ (1) = ⎟ 10 + t ⎠

Mixing Problem I (t ) = e

1

∫ 10 + t dt

=e

ln( 10 + t )

= 10 + t

∫ I ( t ) q ( t ) dt Q ( 10 + t ) = 8 ( 10 + t )dt ∫ I (t )Q =

8t 2 +C 2 80 t + 4 t 2 + C Q= 10 + t

Q ( 10 + t ) = 80 t +

10

Math 263b

Sept. 24

Example 4 (cont.) Solution

Mixing Problem

t = 0 ,Q Q = 20 20 =

0+0+C 10

C = 200 Q=

80 t + 4 t 2 + 200 10 + t

Example 5

Q=

80 t + 4 t 2 + 200 10 + t

After t = 40 minutes 80 ( 40 ) + 4 ( 40 ) + 200 2

Q=

10 + 40

Q = 196 gm

Mixing Problem

A tank initially contains 40 liters of pure water water.. A solution with 1 gm/liter of salt is added into the tank at 2 lit / i . The liter/min. liter/min Th mixture i t l leaves th tank the t k att the th rate t of f 3 liter/min.. Find the quantity of salt remained in the solution liter/min at any time t.

Solution

rate in = ( 1 )( 2 ) = 2

t = 0 ,Q = 0

V = 40 + ( 2 − 3 ) t V = 40 − t ⎛ Q (t ) rate out = ⎜ ⎜ ⎝ 40 − t

⎞ 3Q ( t ) ⎟(3) = ⎟ 40 − t ⎠

11

Math 263b

Sept. 24

Mixing Problem

Example 5 (cont.) Solution

I (t ) = e

dQ = rate in − rate out dt

3

∫ 40 − t dt

=e =e

dQ 3Q =2− dt 40 − t

−3 ln( 40 − t )

ln( 40 − t )

−3

=

1

( 40 − t )

3

∫ I ( t ) q ( t ) dt 2 = ∫ ( 40 − t ) dt

I (t )Q = Q

dQ 3Q + =2 dt 40 − t

( 40 − t )3

dQ + p (t )Q = q (t ) dt

( 40 − t )

Q 3

3

=

1

( 40 − t )2

3 40 − t ) ( Q (t ) = ( 40 − t )2

+ c ( 40 − t )

3

Mixing Problem

Example 5 (cont.) Solution Q ( t ) = ( 40 − t ) + c ( 40 − t )

+c

3

t = 0 ,Q = 0

0 = 40 + c ( 40 )

c=

− 40

( 40 )

3

=−

3

1 1600

Q ( t ) = ( 40 − t )

3 40 − t ) ( −

1600

Amount of salt at any time t

12