



TIME
FOR SOME REVIEW 


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 





Example: 


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): 








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. 











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. 


So: 



D_{1} =
Distance going north 



S_{1} =
Speed going north 



D_{2} =
Distance going northeast 



S_{2} =
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
= 
D_{1} 
+ 
D_{2} 


S_{1} 
S_{2} 



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. 


Why? 


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... 


So: 


A = 
7 
B =
35 
C = 
475 


4 
4 









So 


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. 





NOW
A BREAK FOR A COMMERCIAL






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 
