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course Mth 279
April 12 around 10:40am.
q_a_07________________________________________
110214
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Question: `q001. Solve the equation
dP/dt = .02 P ( 1 - P / 100 )
for P(0) = 10.
What is the limiting value of P, as t increases?
The solution is given in the notes for 110214, but you should work this out yourself without reference to the notes.
Determine how long it would take this function to reach 50% of its limiting value, and how long it would then take to get halfway from that value to its limiting value.
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Your solution:
P(0) = 10 = 100(1 + (1/A)e^(-0.02*10))
(1/[1+(1/A)]) = 10/100
(A/A+1) = 10/100
A = 1/9
P(t) = 50(1/[1+9e^(-0.02t)]) solve for t
Ln(P/100-P) = 0.02t+c
(P/100-P) = Ae^(0.02t)
P = Ae^(0.02t)*(100-P)
P = 100Ae^(0.02t) - P(Ae^0.02t)
P + P(Ae^0.02t) = 100Ae^(0.02t)
P(1 + Ae^0.02t) = 100Ae^(0.02t)
P = 100(A/[A+e^-0.02t]) or 100(1/[1+1/Ae^-0.02t])
Int (dP/P(100-P)) = Int (0.0002 dt)
(1/P(100-P)) = (A/P) + (B/100-P)
(1/P(100-P)) = (100A-AP+BP)/(P(100-P))
100A + P(B-A) = 1
After integrating:
0.01 ln (P/(100-P)) = 0.0002t + c
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The last line of the solution given in the notes is
-.01 ln( P / (100 - P) ) = .0002 t + c.
We rearrange this as follows:
ln(P / (100 - P) ) = -.02 t + c
P / (100 - P) = e^(-.02 t + c) = A e(-.02 t), where A > 0.
P = 100 A e^(-.02 t) - P A e^(-.02 t)
P + P A e^(-.02 t) = 100 A e^(-.02 t)
P ( 1 + A e^(-.02 t) ) = 100 A e^(-.02 t)
P = 100 A e^(-.02 t) / ( 1 + A e^(-.02 t) )
P = 100 / (1 + (1/A) e^(.02 t) ).
Since 1 / A, for A > 0, can be any positive number we can rewrite this as
P = 100 / (1 + A e^(-.02 t) ).
The limiting value of this function, as t -> infinity, is 100, since e^(-.02 t) approaches zero at t gets large.
P(0) = 10 implies A = 1/9, so our solution is
P = 100 / (1 + 1/9 e^(-.02 t) ).
To determine when P reaches 50% of its limiting value of 100, we solve the equation
50 = 100 / (1 + 1/9 e^(-.02 t) )
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Question: `q002. Solve the equation
dP/dt = .02 P ( 1 - P / 100 ) + 3
for P(0) = 10
and explore how the solution differs from the solution to the preceding problem.
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Your solution:
You would solve it the exact same way except you have to add the 3 into it.
Would the 3 be the t in the equation or would you just simply add it on to the answer?
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`@& Separating variables we get
dP / (.02 P ( 1 - P / 100) - 3) = dt.
The denominator expands to
- P^2 / 5000 + P / 50 - 3, so the left-hand equation can be written as
dP / (- P^2 / 5000 + P / 50 + 3)
Multiplying through our equation
by 1/5000 we get
dP / (-P^2 + 100 P + 15000) = .0002 dt.
-P^2 + 100 P +15000 = 0 when
P = (-100 +- sqrt( 100^2 + 60000 ) ) / 2 = -50 +- 50 sqrt(7) = -50 ( 1 +- sqrt(7)).
Partial fractions yield
A / (P + 50 (1 - sqrt(7) ) + B / (P + 50 ( 1 + sqrt(7) ). When placed over the common denominator the numerator becomes
A ( P + 50 ( 1 + sqrt(7) ) + B ( P + 50 ( 1 - sqrt(7) )
= (A + B) P + 50 ( A + B + (A - B) sqrt(7) )
so that
A = -B
and
50 ( A + B + (A - B) sqrt(7) ) = -1
Since A = -B we get
100 A sqrt(7) = -1
so that
A = -sqrt(7) / 700 and
B = sqrt(7) / 700.
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@& Integrating A / (P - 50 (1 - sqrt(7)) and B / (P - 50 (1 + sqrt(7)) we get
sqrt(7) / 700 * ( ln((P - 50 (1 + sqrt(7)) - ln (P - 50 (1 - sqrt(7)) )
This is a good exercise in algebra and a real test of our ability to integrate using partial fractions.
However at this point we can approximate. sqrt(7) is about 2.6 so sqrt(7) / 700 is about .004, 50 * (1 + sqrt(7) ) is about 180, 50 * (1 - sqrt(7)) is about -80, so our integral is about
.004 ln (( P - 180) / (P + 80) ).
Setting this equal to .0002 t + c we have
.004 ln (( P - 180) / (P + 80) ) = .0002 t + c
so that
ln (( P - 180) / (P + 80) ) = .05 t + c
( P - 180) / (P + 80) = A e^(.05 t), A > 0.
Solving for P:
P - 180 = A e^(.05 t) ( P + 80 )
P ( 1 - A e^(.05 t) ) = 80 A e^(.05 t) + 180
P = ( 80 A e^(.05 t) + 180 ) / ( 1 - A e^(.05 t) )
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Question: `q003. If an object of mass m experiences a drag force equal to k * v when it is moving at velocity v, and if the only other force acting on the object is a constant force F_applied, then what differential equation governs its velocity as a function of time? What is the solution of this equation?
What is the limiting value of this solution as t approaches infinity? Can you explain why this makes sense?
Can you make sense of the following statement?
'The rate at which velocity changes is proportional to the difference between the velocity and the limiting velocity, so the solution is expected to be an exponential function of this difference'.
Also, the statement could use specific wording. Can you provide it?
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Your solution:
Int (m*dv/F1-kv) = Int (1 dt)
= (-m/k) * (ln |F1 - kv|) = t+c
= ln (|F1 - kv|) = (-k/m)*(t) + c
= F1 - kv = Ae^((-k/m)*t)
V = (F1/k) - (A/k)*(e^((-k/m)*t))
Lim (as t->infitite) v(t) = F1/k
F1 = kv = -F_air
F_net = 0
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Question: `q004. If an object of mass m experiences a drag force equal to k * v^2 when it is moving at velocity v, and if the only other force acting on the object is a constant force F_applied, then what differential equation governs its velocity as a function of time?
What is the solution of this equation?
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Your solution:
Mv = F1 - kv^2
(dv/F1-kv^2) = dt
(dv/F1*(1-(k/F1)*v^2)) = dt
(dv/(1-sqrt(k/F1)*(1+sqrt(k/F1)*v) = F1*dt
(1/(1-bv)*(1+bv)) = (A/1-bv) * (B/1+bv)
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Question:
`q005. If velocity v is a function v(t) of clock time t, then acceleration is dv/dt.
If velocity is a function v(x) of its position x, then acceleration, being dv/dt, would be expressed (using the chain rule) as
dv/dt = dv/dx * dx/dt.
In terms of motion, what is the meaning of dx/dt?
What therefore is another way to write dx/dt?
How can we therefore express dv/dt in terms of just the two variables v and t?
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Your solution:
A = dv/dt
dv/dt = dv/dx * dx/dt
If v is a function of f, then (dv(x)/dx)
(dv(v)/dt) = (dv/dx)*(dx/dt) = (dy/dx)*(v)
(dy/dt) = (dy/dx)*(v) OR v(dv/dx)
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&#Good responses. See my notes and let me know if you have questions. &#