Calculus Distance Problems
Math-Prof HOME Calculus Table of Contents Ask A Question PREV NEXT

From Here to There
Distance Problems

 When we calculated the distance between two points before,
 we put the two points on a graph.
 It might have looked like:


 Then we found the distance using the Pythagorean theorem.



 Now we are going to talk about a problem with two related distances
 Some person (or thing) needs to go 10 miles north and 5 miles east.
 What's the quickest way?
 If the person can go the same speed in any direction, this one is too easy.
 The quickest way is a straight line to the point.
 With the Pythagorean theorem and a bit of trig we can find the distance traveled
  and the direction to go.


 We chose north to be along the Y axis.
 We didn't have to, but choosing the Y axis to point north rather than say, 
 southwest, made the problem easier to work.
 So far, this is review.
 But what if we said that the person didn't always move at the same speed.
 What if they could go 8 miles per hour straight north, 
 but only 6 miles per hour in any other direction?
 This problem is harder.
 It might still be the fastest to go straight for the point,
 but it MIGHT be faster to go north for a while and then cut over, like this.


 There are 3 special points on this diagram ...


 To make the problem as easy as possible to solve,
 place the whole thing on a graph.
 Set it up so that the starting point is at the origin (0,0) point.
 And the middle "special point" is along one of the axis.
 Like this:


 The coordinates of the three special points are (0,0), (0,Y), and (5,10).
 Because we put the middle point ON the Y axis, we only have one variable. 
 (Tricky, eh?)
 If that middle point had been somewhere out in the middle of the graph,
 we would have 2 variables. Keep that in mind as we go.
 The distance from the start to the middle point 
 and from the middle point to the end can be found using the Pythagorean theorem.
 From (0,0) to (0,Y):




 From (0,Y) to (5,10):



 There is an old equation you've probably seen before ...

 Distance = Rate (also known as speed) x Time

 We are looking for the shortest, that is minimum, time.
 So the first step is to solve the distance equation above for time.

Time = 


 We have two parts to our distance.
 One going straight north and the other going to the northeast.
 The distance going straight north may turn out to be zero,
 but that won't change the problem.
  D1 = Distance going north
  S1 = Speed going north
  D2 = Distance going northeast
  S2 = Speed going northeast

 (If you want to use the letter R for Rate instead of S for Speed that's fine)

 So the equation for the total time is:
Time = D1



S1 S2
 And we know that ...


 So the entire time is:


 We want to find the minimum time. 
 This can only happen at the point where the slope of the derivative equals zero
 or an endpoint.
 The endpoints are at Y = 0 and Y = 10.
 Because that is the least and most distance that you can travel in the Y direction
 that makes any sense.
 Now let's find the T' = 0 point.
 First convert the radicals to exponents of 1/2.
 Remember terms with radicals are inside/outside power chain rule type things.


 Now find the Y values where T' = 0 ...


 WOW! that was exciting, eh?. Well maybe not.
 And we STILL aren't done.
 But we are at a point where we can put these coefficients 
into the quadratic equation...

A =


B = -35

C =


4 4



 Y = 4.3305   or   Y = 15.6695

 Since Y = 15.669 is not in the values that make sense for our problem,
 it is not a candidate solution.
 Y = 4.3305, however, is.
 That makes our possible solutions:
  Y = 0    (an endpoint)
  Y = 4.3305  (the T' = 0 point)
  Y = 10  (the other endpoint)
 Plugging these three values into our time equation will give us our answer.
 Remember, we are looking for the minimum time.
 When Y = 0:



    T = 1.8634 hours
 When Y = 4.3305:


     T = 1.8012 hours
 When Y = 10:
    T = 2.0833 hours
 So the "slope equals zero" point gives us the shortest time.
 It's not a BIG difference, but less is less.
 If we really wanted to,
 we could find the second derivative of T 
 and check for the concavity at the point Y = 4.3305.
 When Y = 4.3305:
    T'' = 0.0096459
 The second derivative is positive (barely) so we do have a minimum.


 That problem was very nasty
 There were LOTS of places where mistakes could have been made.
 A few years ago, that would have just been the way it is.
 Today, technology has come to our rescue.
 For not much money you can buy computer programs
 that will work out all the nasty details.
 You still need to know how to set up the problem (the hardest part)
 But the software will save you from all that nasty manual computation.
 Unfortunately, it probably won't save you from having to do it
 on your Calculus tests.

   copyright 2005 Bruce Kirkpatrick

Math-Prof HOME Calculus Table of Contents Ask A Question PREV NEXT