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# Derivatives and Rate of Change of Snow

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Early one morning it began to snow at a constant rate. At 7 AM a snowplow set off to clear a road. By 8 AM it had traveled 2 miles but it took two more hours for the snowplow to go another 2 miles. Assuming that the snowplow clears snow from the road at a constant rate (in cubic feet per hour), at what time did it start to snow?

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#### Solution Preview

Let h(t) the height of the snow at time t. Since we are given that it snows at constant rate, we have

h(t) = r(t-t_o), (1)

where r is the constant rate of snowing, and t_o is the time when it started to snow that we want to find.
Let x(t) be the position of the snow plow at time t, so that its speed is dx/dt.
The rate at which the plow removes the snow is (speed) * (snow height) and it is given to be constant, C,
so that we have
h(t) * dx/dt = C, (2)

From equations (1) and (2) we get

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#### Solution Summary

Derivatives and rate of change are applied to a word problem. The solution is detailed and well presented. The response received a rating of "5/5" from the student who originally posted the question.

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