Question

Let It Snow The “snowplow problem” is a classic and appears in many differential equation’s texts, but it was probably made famous by Ralph Palmer Agnew: One day it started snowing at a heavy and steady rate. A snowplow started out at noon, going 2 miles the first hour and 1 mile the second hour. What time did it start snowing? Find the textbook Differential Equations, Ralph Palmer Agnew, McGraw-Hill Book Co., and then discuss the construction and solution of the mathematical model.

Short answer

Answer:

It started snowing approximately 37 minutes before noon.

Step-by-step solution

Define the derivative of a function

The derivative of a function of a real variable in mathematics describes the sensitivity of a function value (output value) to changes in its argument (input value).

Calculus uses derivatives as a fundamental tool. When a derivative of a single-variable function exists at a given input value, it is the slope of the tangent line to the function's graph at that point.

Determine the mathematical model.

Let the assumptions be as follows:,

k- The "heavy and steady rate" the snow is falling.

t- The time parameter in hours such that noon is at .

s-The time it started snowing.

h(t)-The function of the height of snow.

x(t)-The function of the distance travelled by the snowplow.

If $h\text{'}\left(t\right)=k$, then $h\left(t\right)=kt+c$.

As $h\left(s\right)=0$, then $c=-ks$so that $h\left(t\right)=k\left(t-s\right)$.

Determine the required time

The rate at which the plough clears the snow is given by the cross section of the snow in the road times its velocity, which means $c=wh\left(t\right)x\text{'}\left(t\right)$. Hence, $x\text{'}\left(t\right)=\frac{c}{wk\left(t-s\right)}$, which tells that $x\left(t\right)=\frac{c}{wk}In\left|t-s\right|+d$for some constant d. As $x=0$, $d=-\frac{c}{wkk}In\left|s\right|$and $x\left(t\right)=\frac{c}{wk}In\frac{\left|t-s\right|}{\left|s\right|}$.

The fact is $x\left(1\right)=2$and $x\left(2\right)=3$to get the equations $2=\frac{c}{wk}In\frac{\left|1-s\right|}{\left|s\right|}$and $3=\frac{c}{wk}In\frac{\left|2-s\right|}{\left|x\right|}$, so that

$$\begin{gathered} 3In\frac{\left|1-s\right|}{\left|s\right|}=2In\frac{\left|2-s\right|}{\left|s\right|} \\ 3In\left|1-s\right|-2In\left|2-s\right|-In\left|s\right|=0 \\ \frac{{\left|1-s\right|}^3}{{\left|2-s\right|}^2\left|S\right|}=1 \end{gathered}$$

The solution is $s\approx -0.618$, which means that it started snowing approximately 37 minutes before noon.