Find the DE modelling the value of $x$

Question

Early one morning it starts to snow. At $7AM$ a snowplow sets off to clear the road. By $8AM$, it has gone $2$ miles. It takes an additional $2$ hours for the plow to go another $2$ miles. Let $t = 0$ when it begins to snow, let $x$ denote the distance traveled by the plow at time $t$. Assuming the snowplow clears snow at a constant rate in cubic meters/hour:

a) Find the DE modelling the value of $x$.

b) When did it start snowing?

At the moment, I'm stuck on (a) with no idea how to solve this. Since the snowplow clears snow at a constant rate, I was thinking $\dfrac{dx}{dt} = kx$; in other words, the change in the distance travelled by the plow with respect to time is proportional to the distance travelled multiplied by some constant of proportionality? But apparently this is incorrect.

The problem set and solution can be found here (problem 7), but I do not understand reasoning behind the solution.

I would greatly appreciate it if people could please take the time to help me understand how to solve this.

Answer 1

Tell your teacher he must take a good course in DE and physics, for he has to include the following totally unrealistic assumptions for making the problem solvable:

• The snow falls at constant rate (ok),
• The snowplow clears at constant rate (ok, indicated),
• No snow falls over the clear road (not ok at all).

Hence:

• Snowplow starts clearing at $t=0$,
• Snow starts to fall at $t=t_0<0$,
• $v_i(t)=k_1(t-t_0)$ is the snow fallen (snow in) over an uncleaned road (see below),
• $v_o(t)=k_2t$ is the snow cleared (snow out),
• The 4m route is half cleared at $t=1$, hence at that moment the snow fallen is $v_i(1)=k_1(1-t_0)$ and the snow cleared is $v_o(1)=k_2=\frac 12 v_i(1)=\frac 12 k_1(1-t_0)$,

After $t=1$, your teacher unknowingly assumed the snow stopped falling at the cleared part of the road, so:

• The 4m route is full cleared at $t=3$, hence at that moment, the snow fallen is $v_i(3)=\frac12 k_1(3-t_0)$ (note the $\frac12$ factor) and the snow cleared is $v_o(3)=3k_2$
• Equalling both volumes: $$v_i(3)=\frac12 k_1(1-t_0)+\frac12 k_1(3-t_0)=v_o(3)=3k_2=3\frac 12 k_1(1-t_0)$$
• Leading to: $$t_0=-1$$ which is 7-1AM=6AM.

Of course, the proper solution with the proper assumptions are:

• The snow falls at constant rate,

• The snowplow clears at constant rate.

• Hence the correct equation is $$v_i(3)=k_1(3-t_0)=v_o(3)=3k_2=\frac 32 k_1(1-t_0)$$

• Leading to $$t_0=-3$$ which means the snow started to fall at 7-3AM=4AM, which is a more correct result.

Under the correct case, the DE of the snowplow walking can be obtained as:

• The volume of snow fallen is $v_i(t)=k_1(t-t_0)=4w h(t)$ for a road of width $w$, and length 4 miles,
• The differential volume of snow cleared is $dv_o(t)=k_2dt=wh(t)dx$, hence $\frac{dx}{dt}=\frac{k_2}{wh(t)}$
• Hence the differential equation of the snow cleared is $\frac{dx}{dt}=\frac{k_2}{wh(t)}=\frac{4k_2}{k_1(t-t_0)}$ and $x(t)=\frac{4k_2}{k_1}\ln(t-t_0)-c_0$
• For $t=0, x=0$ $x(t)=4\frac{k_2}{k_1}\ln(\frac{t-t_0}{-t_0})$
• For $t=1,x=2$, $x(1)=2=\frac{4k_2}{k_1}\ln(\frac{1-t_0}{-t_0}) \to k_1=2k_2\ln(\frac{1-t_0}{-t_0})$
• And for $t=3,x=4$, then $x(3)=4=4\frac{k_2}{k_1}\ln(\frac{3-t_0}{-t_0}) \to \frac{k_2}{k_1}=\frac{1}{\ln(\frac{3-t_0}{-t_0})}$
• Finally: $$x(t)=4\frac{\ln(\frac{t-t_0}{-t_0})}{\ln(\frac{3-t_0}{-t_0})} \\ \blacksquare$$
• The solutions works only for $t_0=-1$.

Answer 2

After many hours, I managed to develop a good understanding of the solution given by the instructor.

Let $t$ be time in hours and $x$ be distance travelled at $t$.

Part a)

As $t$ increases and the distance of the plow, $x(t)$, increases ($x(t)$ is a monotonically increasing function), the amount of snow on the path ahead of the plow accumulates and becomes greater than the amount of snow that was on the previously-travelled paths. This means that it will take longer for the plow to travel forward as time increases. Therefore, there is an inverse relationship between $\dfrac{dx}{dt}$ and $x(t)$:

$\dfrac{dx}{dt} = k \cdot \dfrac{1}{t} \forall t \not= 0$

Part b)

$\dfrac{dx}{dt} = k \cdot \dfrac{1}{t} \forall t \not= 0$

$\implies dx = k \ dt$

$\implies \int 1 \ dx = k \int \dfrac{1}{t} \ dt$

$\implies x = k \ln(|t|) + C = k \ln(t) + C$, since $t \ge 0$.

Let $t = t_1$ at $7$ A.M.

$\therefore t = t_1 + 1$ at $8$ A.M., and $t = t_1 + 3$ at $10$ A.M.

$2 = x(t_1 + 1) - x(t_1) = k \ln(t_1 + 1) - k\ln(t_1)$

$= k[\ln(t_1 + 1) - \ln(t_1)]$

$= k \ln \left( \dfrac{t_1+1}{t_1} \right) \forall t_1 \not= 0$

$4 = x(t_1 + 3) - x(t_1) = k \ln(t_1 + 3) - k\ln(t_1)$

$= k[\ln(t_1 + 3) - \ln(t_1)]$

$= k \ln \left( \dfrac{t_1 + 3}{t_1} \right) \forall t_1 \not= 0$

$\therefore 2 k \ln \left( \dfrac{t_1 + 1}{t_1} \right) = k \ln \left( \dfrac{t_1 + 3}{t_1} \right) \forall t_1 \not= 0$

$\implies 2 \ln \left( \dfrac{t_1 + 1}{t_1} \right) = \ln \left( \dfrac{t_1 + 3}{t_1} \right)$

$\implies \left( \dfrac{t_1 + 1}{t_1} \right)^2 = \dfrac{t_1 + 3}{t_1}$

$\implies \dfrac{(t_1 + 1)^2}{t_1^2} = \dfrac{t_1 + 3}{t_1}$

$\implies \dfrac{t_1^2 + 2t_1 + 1}{t_1^2} = \dfrac{t_1 + 3}{t_1}$

$\implies t_1^2 + 2t_1 + 1 = t_1^2 + 3t_1$

$\implies t_1 = 1$

$\therefore t = 0 = t_0$ at $6$ A.M.