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How to solve the infamous “snowplow” ODE problem, very similar to Newton’s Law of Cooling ODE problems
Let’s try a famous historical problem, the “snowplow” problem.
It’s very similar to the Newton’s Law of Cooling problems we’ve previously done.
The problem is famous because you need to make a number of assumptions. It’s very strange to see such a problem with barely any information.
Example problem: One day it started to snow at a steady rate. A snowplow leaves at noon, going 2 miles the first hour and 1 mile the second hour. What time did it start snowing?
See how there’s barely any information in the problem statement. It explains that it started to snow at a certain time. A snowplow leaves at noon, but we’re working in one hour at a time, so let’s set 1:00 PM to be t = 0.
The solution is very easy, it’s meant to be a basic ODE question. Consider Newton’s Law of Cooling ODE as an example below:
\[\frac{{dT}}{{dt}} = – k\left( {T – {T_A}} \right)\]
It’s one of the most basic forms of an ODE. It basically tells us that the rate of change of temperature is proportional to the difference between the temperature and the surrounding temperature, multiplied by a constant. You don’t actually have to write the negative in the equation above, you could just merge the negative into k directly. k is just a “proportionality constant”.
Here is the same case, except distance is the dependent variable and time is the independent variable. The rate of change of distance is inversely proportional now to the difference between the current time and the time it started snowing, multiplied by a constant.
So, let y represent the distance, c represent a proportionality constant, and \({{t_o}}\) be the time it started snowing, and we can form an ODE:
\[\frac{{dy}}{{dt}} = \frac{c}{{t – {t_o}}}\]
Note that the time difference is in the denominator because of the inverse relation we have now (more time t since it started snowing means smaller y distance travelled), otherwise the ODE would’ve been exactly the same as Newton’s Law of Cooling!
Now just make two equations and solve two unknowns. When t = 0, y = 2, and when t = 1, y = 1, the way that we set up the question.
We don’t have “y” though, we have dy/dt, so we must integrate both sides of the ODE with respect to t.
\[y = c\ln \left( {t – {t_o}} \right)\]
We need to evaluate this from t = -1 to 0, and y = 2
\[2 = c\ln ( – {t_o}) – c\ln ( – {t_o} – 1)\]
Use logarithm laws to simplify. Subtracting logarithms means that you can put it into a fraction.
\[2 = c\ln \left( {\frac{{ – {t_o}}}{{ – {t_o} – 1}}} \right)\]
And we repeat this for y = 1, t goes from 0 to 1.
\[y = c\ln \left( {t – {t_o}} \right)\]
\[1 = c\ln \left( { – {t_o} + 1} \right) – c\ln \left( { – {t_o}} \right)\]
\[1 = c\ln \left( {\frac{{ – {t_o} + 1}}{{ – {t_o}}}} \right)\]
You could probably divide the equations by each other to cancel out c, I just substitute the above equation into the first one we got.
\[2 = \frac{1}{{\ln \left( {\frac{{ – {t_o} + 1}}{{ – {t_o}}}} \right)}}\ln \left( {\frac{{ – {t_o}}}{{ – {t_o} – 1}}} \right)\]
Solve for \({{t_o}}\), you can use an equation solver, graphing calculator, estimate the answer using iterative methods or Newton’s Method, but you’ll find that \({{t_o}}\) = -1.618.
We had set t = 0 to be 1:00 PM, so t = -1 is noon, so multiplying 0.618 by 30 to convert to minutes, we obtain that it started snowing 37 minutes before noon.