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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?
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: 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: We now find time t 1 to walk distance AP.(time = distance / speed). Time t 2 to walk distance PC is given by 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 |