- #1

- 95

- 0

## Homework Statement

A snowplow can remove snow at a constant rate (in cb. ft/min). One day, there was no snow on the ground at sunrise, but sometime in the morning it began snowing at a steady rate. At noon, the plow began to remove snow. It had cleared 2 miles of snow between noon and 1 PM, and 1 more mile of snow between 1 PM and 2 PM. What time did it start snowing?

## Homework Equations

## The Attempt at a Solution

At first I tried to do the problem without Calculus and I found an answer, however my answer does not match the correct solution answer and I am having trouble finding the flaw in my non-Calculus solution. Can someone help me find the flaw in my solution?

__My Flawed Solution:__Let [itex]f[/itex] be the rate of snowfall and [itex]r[/itex] be the rate of snow removal. Like the AoPS solution, let noon be the time [itex]h[/itex] hours after the snow started falling. Let the width of the snowplow be [itex]w[/itex].

So in first hour after noon, the snowplow plowed [itex]r[/itex] amount of snow. In the second hour, it also plowed [itex]r[/itex] amount of snow. What is important is that it plowed the same amount of snow in the first and second hours.

Additionally, the instantaneous height of snow that is plowed at time [itex]t[/itex] after the snow started falling is [itex]tf[/itex] and thus the shape of all the snow that has been removed to be a trapezoid in the first and second hours.

The trapezoid that is plowed in the first hour has height [itex]2\text{ miles}[/itex] and bases [itex]hf[/itex] and [itex](h+1)f[/itex]. The trapezoid that is plowed in the second hour has height [itex]1\text{ mile}[/itex] and bases [itex](h+1)f[/itex] and [itex](h+2)f[/itex] and since the amount of snow that is plowed in the first hour is equal to the amount of snow plowed in the second hour, we must have the volumes of both trapezoids equal. So we have

[itex]w(2\text{ miles})\frac{hs+(h+1)s}{2}=w(1\text{ mile})\frac{(h+1)s+(h+2)s}{2} [/itex]

[itex]s2(2h+1)=s(2h+3) [/itex]

[itex]4h+2=2h+3[/itex]

[itex]2h=1[/itex]

[itex]h=\frac12.[/itex]

Thus the snow started falling exactly half an hour before noon which is [itex]11:30[/itex] AM, which does not match the correct solution.

__Correct Solution:__Observe that the instantaneous velocity of the snowplow is inversely proportional to the time after the snow started falling, so [itex]\frac{\text{d}{s}}{\text{d}t}=\frac{k}{t}[/itex] for some constant [itex]k[/itex] where [itex]s[/itex] is the position of the snowplow and [itex]t[/itex] is the time after the snow started falling. Thus we have

[itex]

\frac{\text{d}{s}}{\text{d}t}=\frac{k}{t}\implies s(t)=k\ln t+C

[/itex]

for some constants [itex]C[/itex] and [itex]k[/itex].

Let noon be the time [itex]h[/itex] hours after the snow started falling, so we have

[itex]s(h+1)-s(h)=k(\ln h+1-\ln h)=2\text{ miles}[/itex]

and

[itex]

s(h+2)-s(h+1)=\ln h+2 -\ln h+1 =1\text{ mile}

[/itex]

so

[itex]2k(\ln h+2-\ln h+1)=k(\ln h+1-\ln h) [/itex]

[itex]2\ln\frac{h+2}{h+1}=\ln\frac{h+1}{h}[/itex]

[itex]\left(\frac{h+2}{h+1}\right)^2=\frac{h+1}{h}[/itex]

[itex]h^2+h-1=0[/itex]

[itex]h=\frac{-1+\sqrt{5}}{2}\approx 0.618[/itex]

so it started snowing [itex]0.618[/itex] hours before noon which is about 11:23 AM.