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### When Did It Start Snowing?

```
Date: 02/17/99 at 13:44:48
From: Anonymous
Subject: Differential Equations

The question is: Convert the following system into a differential
equation.

A snowblower throws 30 cu. ft. of snow per minute. At some point in the
morning it starts snowing, and it snows evenly and steadily all day.
A woman starts clearing the sidewalk with the snowblower at noon. She
travels 200 ft in the first hour and 100 ft. in the second hour. When
did it start snowing?

Thanks!
```

```
Date: 02/17/99 at 17:36:17
From: Doctor Anthony
Subject: Re: Differential Equations

We let DV be the volume of snow moved per minute by the snowblower.
Let DH be the rate at which height of snow is increasing per minute.
Both DV and DH are constants.  w = width of the blower.

At time t after snow began falling, the depth will be DH*t.

Let x = the distance travelled by the snowblower. Then we have:

Volume of snow moved in time dt at time t is

DV*dt = DH*t*w*dx

DV*(dt/t) = DH*w*dx    and integrating we have:

DV*ln(t) = DH*w*x + const   At t=t0   x = 0

DV*ln(t0) = 0 + const   and so const = DV*ln(t0)

DV*ln(t/t0) = DH*w*x
ln(t/t0) =(DH/DV)*w*x

and we can combine (DH/DV)*w into a single constant k.

ln(t/t0) = kx

When x = 200 m, t=t0+60 (minutes)  and t/t0 = 1+60/t0

So,

ln(1+60/t0) = 200k  and  k = (1/200)ln(1+60/t0).....(1)

When x = 300,  t = t0+120   and t/t0 = 1+120/t0

So       ln(1+120/t0) = 300k    and from (1)

= (3/2)*ln(1+60/t0)

ln(1+120/t0)  = ln(1+60/t0)^(3/2)

1+120/t0 = (1+60/t0)^(3/2)

This has solution  t0 = 30(sqrt(5)-1)

= 37.08 minutes

So we conclude that it started snowing 37.08 minutes before noon.
That would be at about 11:23 A.M.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Calculus

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