can anyone solve my assignment? it is very hard to do... please read the question below?

Question

You happen to sight someone very adorable at a distance 100m from the junction at the junction. At a given instant that someone is walking straight across speed of v m/s. You hope to catch up her by walking speed at 2v m/s, always towards the target. What is ur path? (actually it is a curve line) You have to consider setting up a differential equation. You can simplify the solutions of the differential equation by differentiating and then solve the first order differential equations. (adapted from American Math. monthly 3942, 1941

Answer 1

♠ the origin: her t=0, x=y=0, her speed =v*j;
me t=0, x=-100m, y=0, my speed =2v*i;
♣ for t>0; her position h=v*t*j; my position r=x*i+y*j; my speed = 2v*(h-r)/|h-r| = x’*i+y’*j; vector h-r = -x*i +(v*t –y)*j and |h-r| =√(x²+(v*t-y)²); thus
♦ x’ = -2v*x/|h-r|; y’= 2v*(v*t-y)/|h-r|; y’/x’ =(dy/dt)/(dx/dt) =dy/dx =y`;
as well as -y`=p=(y-v*t)/x, where y’ is derivative by t, while y` is derivative by x;
☻so x’=-2v/√(1+p²); y’=-2v*p/√(1+p²);
☺ as p=(y-vt)/x, thence y=xp+vt and y’=x’p+xp’+v or y’-x’p =xp’+v or 0=xp’+v or p’=-v/x; p’/x’ =p`= {see☻} =0.5*√(1+p²) /x;
♥ thus dp/√(1+p²) =0.5*dx/x; integration comes: ln|√(1+p²)-p| = 0.5*ln|xC|;
or √(1+p²) = p+√(xC); square: 1+p² =p² +2p*√(xC) +xC, hence
p=0.5(1-xC)/√(xC)=-y` and C=-0.01 (see ♠);
thus dy= 0.5dx*(-1+xC)/√(xC);
♥ once more integration comes:
y(x)= -0.5* ∫dx/√(xC) +0.5* ∫dx*√(xC) = (xC/3-1)* √(x/C)+D;
☼ finally: y(x) = (xC/3-1)*√(x/C)+D, where x<=0, C=-0.01, D=200/3, (see ♠)
thus [-100,0] is my start point and [0, 200/3] is my final point.

Answer 2

I think this can be done without differential equations. Here is my approach:

Draw a line from your starting point to the target, with starting point at the origin. Express the relations in polar coordinates. The line to the target is r, and the angle to the target is theta. His position along r is vr*t. His angle theta is arctan(y/x0) = arctan(vy*t/x0). (Here vr is 2v and vy = v and x0 = 100m). Then use coordinate converstion x=rcos(theta) and y=rsin(theta) to get parametric equations for y and x in terms of t. Convert those to y(x). Here is what I came up with: http://img266.imageshack.us/img266/5679/rtangletracking1fn4.png

y = x^2/[200^2 - x^2]

The curve looks like this http://img402.imageshack.us/img402/5820/rtangletrackgraphcopyxc6.png

I'm not sure this is it, but it seems right.

Answer 3

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