110214
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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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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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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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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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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Question: `q006. If an object of constant mass m is subject to a net force proportional to the inverse square of its position r, then we have
F_net = k / r^2.
By Newton's Second Law this is equal to m a, where a is the second derivative with respect to time of the position function. Thus
m r '' = k / r^2.
Is this equation linear? Is it first-order? Can you see how to solve it?
Now r ' is just the velocity v of the object, so r '' is just dv/dt. So our equation could be written
m dv/dt = k / r^2.
This looks good, because now appears to be a first-order equation, though still nonlinear. However it's not as simple as it appears. There are three variables, v, r and t, involved in this equation.
We can get around this. v ' means dv/dt. If we regard v as a function of r, then we have
dv/dt = dv/dr dr/dt
and since v is the derivative dr/dt of the position function, we have
dv/dt = v dv/dr.
Replacing dv/dt with v dv/dr we get the equation
m v dv/dr = k / r^2.
Now we have only the variables r and v, and we recognize the equation as a separable first-order equation.
Solve this equation for m = 1, k = -5 and initial condition v(r_0) = v_0, where r_0 = 100 and v_0 = 0.
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