{"id":2801,"date":"2020-04-06T22:53:22","date_gmt":"2020-04-07T03:53:22","guid":{"rendered":"https:\/\/laurentlessard.com\/bookproofs\/?p=2801"},"modified":"2020-04-06T23:49:28","modified_gmt":"2020-04-07T04:49:28","slug":"when-did-the-snow-start","status":"publish","type":"post","link":"https:\/\/laurentlessard.com\/bookproofs\/when-did-the-snow-start\/","title":{"rendered":"When did the snow start?"},"content":{"rendered":"

This week’s Riddler Classic<\/a> is a neat calculus problem:<\/p>\n

\nOne morning, it starts snowing. The snow falls at a constant rate, and it continues the rest of the day.<\/p>\n

At noon, a snowplow begins to clear the road. The more snow there is on the ground, the slower the plow moves. In fact, the plow’s speed is inversely proportional to the depth of the snow \u2014 if you were to double the amount of snow on the ground, the plow would move half as fast.<\/p>\n

In its first hour on the road, the plow travels 2 miles. In the second hour, the plow travels only 1 mile.<\/p>\n

When did it start snowing?\n<\/p><\/blockquote>\n

Here is my solution:
\n[Show Solution]<\/a><\/p>\n

\n

We’ll assume the snow starts at $t=0$ hours. So after $t$ hours, the depth of snow on the ground is proportional to $t$. Since the plow’s speed $v$ is inversely proportional to the amount of snow on the ground, we have $v = \\tfrac{c}{t}$, where $c>0$ is a constant of proportionality. We’ll assume the constant is in units of miles so that $v$ is in miles per hour.<\/p>\n

Suppose the plow starts at $t=x$ hours. During the first hour, the plow travels 2 miles. Since distance is the integral of velocity with respect to time, we have:
\n\\[
\n2\\text{ miles} = \\int_{x}^{x+1} v\\,\\mathrm{d}t = \\int_x^{x+1} \\frac{c}{t}\\,\\mathrm{d}t = c\\,\\log\\left( \\frac{x+1}{x} \\right)
\n\\]During the second hour, the plow travels 1 mile. So we also have:
\n\\[
\n1\\text{ mile} = \\int_{x+1}^{x+2} v\\,\\mathrm{d}t = \\int_{x+1}^{x+1} \\frac{c}{t}\\,\\mathrm{d}t = c\\,\\log\\left(\\frac{x+2}{x+1} \\right)
\n\\]Dividing the first equation by the second, the units and constants of proportionality cancel. Simplifying, we obtain:
\n\\[
\n2\\,\\log\\left(\\frac{x+2}{x+1}\\right) = \\log\\left( \\frac{x+1}{x} \\right)
\n\\]Exponentiating both sides to cancel out the logs,
\n\\[
\n\\left(\\frac{x+2}{x+1}\\right)^2 = \\frac{x+1}{x}
\n\\]Finally, clearing fractions and simplifying, we obtain:
\n\\[
\nx^2+x-1 = 0
\n\\]Since we must have $x\\gt 0$, we can discard the negative solution to this equation, and keep only the positive one. This leads us to:
\n\\[
\nx = \\frac{\\sqrt{5}-1}{2} \\approx 0.618034
\n\\]Incidentally, this is the inverse of the Golden Ratio<\/a>, though that’s more of a coincidence than anything meaningful… If the plow started at noon, then the snow started $x$ hours before noon, which is 11:22.55 AM (11 hours, 22 min, 55 sec).\n<\/div>\n","protected":false},"excerpt":{"rendered":"

This week’s Riddler Classic is a neat calculus problem: One morning, it starts snowing. The snow falls at a constant rate, and it continues the rest of the day. At noon, a snowplow begins to clear the road. The more snow there is on the ground, the slower the plow moves. In fact, the plow’s … Continue reading “When did the snow start?”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_monsterinsights_skip_tracking":false,"footnotes":""},"categories":[7],"tags":[28,4,2],"class_list":["post-2801","post","type-post","status-publish","format-standard","hentry","category-riddler","tag-calculus","tag-integration","tag-riddler"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t