What is the fastest cycling route?

Discussion in 'Math' started by Hymie, Jul 6, 2018.

  1. Raymond Genovese

    Active Member

    Mar 5, 2016
    You guys are going at it all wrong - just go straight lower left to upper right...after picking up one of these on alibangbay.

    Oh wait, the thrust for lift required on grass probably differs from that on the tarmac....dam you math nerds!:)
    wayneh and -live wire- like this.
  2. MrChips


    Oct 2, 2009

    \begin{align*}<br />
\frac{sin(\theta_1)}{sin(\theta_2)} &= \frac{v_1}{v_2} &= \frac{1}{2}<br />

    Now we have an equation with two unknowns. How do we solve for either angle?
    We need a second equation.
  3. WBahn


    Mar 31, 2012
    The geometry of the problem gives you the other equation. If x is the distance from the left edge to the point on the grass/tarmac boundary, then

    \tan(\theta_1) \; = \; \frac{x}{20 \; mi}

    \tan(\theta_2) \; = \; \frac{70 \; mi \; - \; x}{30 \; mi}
    MrChips likes this.
  4. MrChips


    Oct 2, 2009
    Easier said than done.
    Now we have three equations with three unknowns.
  5. WBahn


    Mar 31, 2012
    It's trivial to get it down to one equation in one unknown, namely x.

    Just use the problem geometry to write the sine of each angle in terms of x. Then take the ratio of the two sines and set them equal to 1/2. Then square everything and now you have the radicals gone. You are then left with a polynomial equation in x only.
    -live wire- likes this.
  6. MrAl

    Distinguished Member

    Jun 17, 2014
    Hello there,

    Here is another approach...

    Solve this polynomial for x:

    One of those solutions is the optimum value of x. It's not hard to spot because one of the solutions is negative and two others are imaginary, leaving just one that could be the solution (and it is)

    To arrive at that polynomial, we can just take the speed to be a constant 30 and the length of the bottom segment equal to 2 times it's actual length because then at a speed of 30 it will take twice as long for that segment, which is the actual time. What this does is it takes the speed out of the equation because then the speed multiplies out of the denominator and so disappears from the equation. So in effect it becomes a problem of length not time (x is a length).
    This is actually easy to set up because all we have to do is double the bottom length.
    Of course we still have to calculate the two lengths as:

    and that gives us two lengths L1 and L2, and we just multiply L1 by 2 for the equation to be minimized:

    I suspect we can also make the top segment half and assume the speed is a constant 15 but i did not try that.

    Another interesting approach is to just treat the two lengths as variables that vary with 'x' and then use an increment to 'x' and compare successive results. When a result is better than the previous result, we keep that but if worse we back up and then halve the increment. This is best done in a program. By hand i would probably want to use Newtons Method.
    Last edited: Jul 20, 2018 at 8:55 AM