Problem : You decide to walk from point A (see figure below) to point C. To the south of the road through BC, the terrain is difficult and you can only walk at 3 km/hr. However, along the road BC you can walk at 5 km/hr. The distance from point A to the road is 5 km. The distance from B to C is 10 km. What path you have to follow in order to arrive at point C in the shortest ( minimum ) time possible?
Interactive Tutorial
We first try to understand the problem using the applet below. There are several possible paths one can follow to go from A to C. On the left panel of the applet, is shown possible paths: You may walk from point A to a certain point P, somewhere on the road between B and C, and continue along the road to get to point C. The question is: What is the position of point P that will minimize the time taken to go from A to C?
APPLET
Use the mousse to press and drag point P. What you are doing here is changing distance BP = x. On the right panel you have the time plotted against x. As you can see there seem to be one value of x for which the time is smallest (minimum). You may also plot the whole graph using the "on" and "off" buttons above it.
The total time t taken from A to C is calculated as follows:
t = distance AP / 3 km/hr + distance PC / 5 km/hr
Analytical Tutorial
We now look at a solution using derivatives and other calculus concepts. Let distance BP be equal to x. Let us find a formula for the distances AP and PC. Using Pythagorean theorm, we can write:
distance AP = sqrt(5^{ 2} + x^{ 2})
distance PC = 10  x
We now find time t_{ 1} to walk distance AP.(time = distance / speed).
t_{ 1} = distance AP / 3 = sqrt(5^{ 2} + x^{ 2}) / 3
Time t_{ 2} to walk distance PC is given by
t_{ 2} = distance PC / 5 = (10  x) / 5
The total time t is found by adding t_{ 1} and t_{ 2}.
t = sqrt(5^{ 2} + x^{ 2}) / 3 + (10  x) / 5
we might consider the domain of function t as being all values of x in the closed interval [0 , 10]. For values of x such that point P is to the left of B or to the right of c, time t will increase.
To find the value of x that gives t minimum, we need to find the first derivative dt/dx (t is a functions of x).
dt/dx = (x/3) / sqrt(5^{ 2} + x^{ 2})  1/5
If t has a minimum value, it happens at x such that dt/dx = 0.
(x/3) / sqrt(5^{ 2} + x^{ 2})  1/5 = 0
Solve the above for x. Rewrite the equation as follows.
5x = 3sqrt(5^{ 2} + x^{ 2})
Square both sides.
25x^{ 2} = 9(5^{ 2} + x^{ 2})
Group like terms and simplify
16x^{ 2} = 225
Solve for x (x >0 )
x = sqrt(225/16) = 3.75 km.
dt/dx has one zero. The table of sign of the first derivative dt/dx is shown below.
The first derivative dt/dx is negative for x < 3.75, equal to zero at x = 3.75 and positive for x >3.75. Also the values of t at x = 0 and x = 10 (the endpoints of the domain of t) are respectively 3.6 hrs and 3.7 hrs. The value of t at x = 3.75 is equal to 3.3 hrs and its is the smallest. The answer to our problem is that one has to walk to point P such BP = 3.75 km then procced along the road to C in order to get there in the shortest possible time.
Exercises
1  Solve the same problem as above but with the following values.
solution to the above exercise
x = 6.26 km (rounded to 2 decimal places).
More references on
calculus problems
