WORKSHOP 2·11
Solutions
Warm-up
Here are two possible modes of transportation in the classroom.
Mode A: wall-walking. In this mode, students move along a wall with their feet parallel to the wall and
touching (or as close as possible to touching) the wall. You walk by putting one foot in front of the other,
heel-to-toe.
Mode B: room-walking. In this mode, students move in a straight line through the interior of the room.
You walk sideways, by putting one foot beside the other like a tripped out crab. It should be slower than
Mode A.
Nominate one group member for each mode to be the walker. Measure the speed of the two modes in
metres/second, and the length and width of the room.
Main Problem
You want to go from one corner of the room to the corner diagonally opposite, as quickly as possible.
From your starting corner, you must room-walk in a straight line to some point on the other side, and
then wall-walk along the wall to the corner which is your destination. Find which point you should aim
for when room-walking.
Solution:
First we need a diagram of the trajectory
x
y
W
L
L−x
We label the distance traveled parallel to the wall (mode A) by x and the distance traveled acroos the interior
of the classroom (mode B) by y. Let va and vB be the walking speeds for modes A and B respectively, then
the time T it takes for the student to walk from one corner of the classroom to the other is
T =
x
y
+
.
vA
vB
We would like this to be a function of a single variable x, so we need to describe y using only x and measurable
constants. We may do this using the Pythagorean theorem
p
y = W 2 + (L − x)2 ).
Our time equation then becomes
x
T (x) =
+
vA
p
W 2 + (L − x)2
.
vB
The last thing we need is to find an interval of possible values for x. The lower bound is x = 0, when we only
walk using mode B and the upper bound is x = L when we use mode A to walk the length of the classroom
and mode B along its width.
The next step is to find any critical point of T (x) lying in our interval [0, L]. We then compute the first
derivative
L−x
1
p
.
−
T 0 (x) =
vA
vB W 2 + (L − x)2
Critical points are where the first derivative is equal to zero, so we set
1
L−x
p
=
vA
vB W 2 + (L − x)2
p
vA (L − x) = vB W 2 + (L − x)2
2
2
vA
(L − x)2 = vB
(W 2 + (L − x)2 )
2
W2
vB
2 − v2
vA
B
vB W
L−x= p 2
2
vA − vB
vB W
x=L− p 2
.
2
v A − vB
(L − x)2 =
Note that this is strictly less than L and we will need to verify if it is greater than zero when including our
measurements for L, W , vA and vB . Also note that we only keep the positive part when simplifying the
squares because we want that L − x ≥ 0.
Now suppose our measurements give L = 12m, W = 10m, vA = 0.7m/s, vB = 0.3m/s, then our critical point
is
0.3 · 10
x = 12 −
= 7.5m,
0.72 − 0.32
which lies in our interval. To minimize time within our interval we need to compare the values of T (x) at
the endpoints and critical points
√
20 61
≈ 52.1texts,
T (0) =
3
T (7.5) ≈ 47.3s,
1060
T (L) =
≈ 50.5s.
21
That makes our critical point the global minimum in our interval. You would need to walk using mode A
for 7.5 meters then walk straight to the opposite corner using mode B.
There is more than one way to verify that our critical point is a global minimum. If we had shown it was
also a local minimum, since there are no other critical point inside the interval it would have to be global as
the function would be decreasing before and increasing after. To verify this we may use the first derivative
test at x = 0 and x = L
1
L
√
≈ −1.1s/m
−
vA
vB W 2 + L2
1
T 0 (L) =
> 0.
vA
T 0 (0) =
We indeed have our function decreasing before and increasing after our critical point and it is a local
minimum.
Another way to check this is using the second derivative rule at our critical point. The second derivative is
given by
p
vB W 2 + (L − x)2 − (L − x) √ vB2(L−x) 2
W +(L−x)
.
T 00 (x) =
2 (W 2 + (L − x)2 )
vB
Using a common denominator, the numerator becomes
vB (W 2 + (L − x)2 ) − vB (L − x)2 = vB W 2 > 0,
so our function is concave up and we do have a local minimum.