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/
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