The document summarizes a multi-part differential equations word problem about snow plows clearing roads during a snowstorm. 1) It reviews the standard snow plow problem and defines the variables and equations used. 2) It then solves the initial problem to determine when the snow started. 3) Additional parts of the problem address when two snow plows starting from different locations would crash into each other and if changing start times could prevent crashes. Equations are derived and solved to determine crash times and minimum separation distances between plows.

1. By
Patrick McCoy
With assistance
By
Dr. Danrun Huang

2.  Here in Minnesota snow plows are part of our every day life and seans such as
this are common every winter.
http://www.youtube.com/watch?v=z0cMI_gVVaI
 The snow plow problem has been covered many times in differential equations
and is a staple many differential equations text books.
 Before I can discuss the new parts of the problem I will need to give you an
over view of the old problem.
 The snow is coming down at a constant rate.
 As the snow gets deeper the plow goes slower.
 It is typically split in to two parts.
 Here is how the typical problem reads.
 (a)One morning it began to snow very hard and continued snowing steadily
throughout the day. A snowplow set out at 8:00 A.M. to clear a road, clearing 2 mi
by 11:00 A.M. and an additional mile by 1:00 P.M. At what time did it start
snowing?
 (b) One day it began to snow exactly at noon at a heavy and steady rate. A
snowplow left its garage at 1:00 P.M. and another one followed in its tracks at
2:00 P.M. At what time did the second snowplow crash into the first? Could the
crash have been avoided by dispatching the second snowplow at a later time?

3.  We want to know when it started snowing.
 Time is measured in hours
 x=x(t)= the distance the plow has traveled in t hours
 x(0)=0 x(3)=2 x(5)=3
 to<or = 0
 h=h(t)=the height of the snow at t hours and is
independent of the x
 H(to)=0 since this is when the snow started to fall.
 V(t)= where v(t) is the speed of the plow
 k = an unknown constant volume that the plow can remove
per unit time
 r=the rate of falling snow per unit volume per unit time

4.  Now since k is an constant and r is a constant
then k/r is also a constant I can rewrite
as where A= k/r
 So
all constants of integration have been
absorbed into C1
 This gives us 3 initial condition equations
 Using these initial conditions and substitution
and a lot of algebra I can find t0
 t0 =6.2915hours before the plow started out.

5.  We can then use this value to find the
unknown constant A
 A=5.1295
 This value will be useful for later.
 X(t)=5.1295ln(t-to)+C

6.  We want to know when the plows will crash and if the sending
the second plow out later will avoid a crash.
 t=0 at noon when it started snowing
 x=x(t)= the displacement of the first plow
 x(1)=0
 Y=y(t)= the displacement of the second plow
 Y(2)=0
 h1(t)=h1 = the height of the snow at time t at any point before the
first plow began plowing
 h2(t,y)=h2=the height of the snow when the second plow reaches
a particular location y
 where v(t) is the speed of the plow
 Solving we get x(t)=A ln(t)+c
 So x(1)=0=A ln(1)+c =>c=0
 k = an unknown constant volume that the plow can remove per
unit time
 r=the rate of falling snow per unit volume per unit time

7.  h2(t,y)=h2 isdependent how long it has been
since the first plow cleared the road.
 So at some future time lets say time T x(T)=A
ln(T)=y
 Solving
 So h2(t,y)=r(t-T)
 Then
 Using the inverse
 Thisis simply a first order differential
equation

8.  Solving this first order Differential equation
gives
 When t=2, y=0 so C=2
 This give the equation
 If we substitute our equation x(t)=A ln(t)=y
into this equation we get :
 So they crash at 2:43 P.M.

9.  We can use B.1 to solve B.2
 We already have an equation for t
 This time lets leave constant c in place
 Again substituting x(t)=A ln(t)=y into this
equation we get :
 Nomatter what a crash occurs because t is
never less than 0

10.  The problem
 One day it began to snow exactly at noon at a heavy and steady rate.
Snowplow #1 left its garage at 1:00 p.m. and moved east. After a
while Snowplow #2 left a different garage and moved south. Suppose
that the snowplows and the steady rate of snow are the same as that
in Problem 1, the garage of Snowplow #1 is 4 miles from the
intersection and the Garage of Snowplow #2 is 3 miles from the
intersection.
 (a) The two snowplows crash into each other at the intersection of two
perpendicular roads. At what time did Snowplow #2 leave its garage?
 (b) If snowplow #1 left its Garage at 2 p.m. and the two plows crash at the
intersection, at what time did Snowplow 32 leave its Garage? Why?
 (c) Snowplow #1 left its garage at 1 p.m. and Snowplow #2 left its garage at
1:15p.m. There is nothing to prevent the two plows from seeing each other
except the snow. Suppose the visibility of that day is only 1/10 miles. Could
the drivers on the two plows see each other?

11.  At what time did Snowplow #2 leave its
garage?
 t=0 at noon when it started snowing
 x=x(t)= the displacement of the first plow
 x(1)=0
 Y=y(t)= the displacement of the second plow
 From the first problem we already know
A=5.1295

12.  For the first plow similar to problem one
 The crash happens when
 For plow two we have
 So

13.  Now we know that in order for the two plows
to collide the two plows must have been the
same distance from the intersection when
the second plow started out.
 This
leads two possible ways to solve the
problem.

14.  Firstly if they were the same distance from the
intersection when plow two started out due to
the steady rate of snow fall then
 So
 Alternatively I could have solved it this way
 We also know that
 So the second plow left 1.22 hours after it
started snowing or 1:12 p.m.

15.  At what time did Snowplow #2 leave its
garage?
 t=0 at noon when it started snowing
 x=x(t)= the displacement of the first plow
 x(2)=0
 Y=y(t)= the displacement of the second plow
 From the first problem we already know
A=5.1295

16.  In this part of the problem let us do some
generalization
 If = the time the first plow left the garage
at distance a miles then since
 And for the second plow it is similar
 If = the time the first plow left the garage
at distance b miles then since

17.  And since they crash when
 Plugging in numbers from the problem
 or the second snow plow left at 2:25 p.m.

18.  Dr. Huang has come up with a crash theorem
for this problem.
 “ One day it began to snow exactly at noon at a
heavy and steady rate.”
 “Plow #1 left Garage #1 which is a miles from
the intersection at ta p.m.; Plow #2 left Garage
#2, which is b miles from the intersection at tb
p.m. Then the two plows crash at the
intersection of and only if ”
 Thus for this type of “crashing problem”
depends only of the difference a-b of the
distances.

19.  If you remember from part (b)
 However in this problem we are only
interested in absolute distance from the
intersection.
 Plugging in initial values we get.

20.  The reason we are only interested in the
absolute value of the distances is because
the distance between the two plows at any
time t is given by Pythagorean Theorem
which says the square of the distances
between the plows is and any
number squared is always positive.
 So if we let
 Then

21.  Now the minimum distance between the two
plows is when the derivative of the function
is =0
 So

22.  Now we know when they were the closest to
each other but remember they can only see
each other if they get within .1 miles of
each other.
 So
 Unfortunatelythe two plow drivers did not
get the pleasure of seeing each others ugly
mugs.

23.  Also
out of curiosity we can find how far
each one was from the intersection.