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Two Points Moving...

Date: 7/18/96 at 16:58:53
From: Bill Buchanan
Subject: Two Points Moving...

I would appreciate help in solving this problem.  If you can help me 
please explain the steps involved.

Point P is moving in the positive direction, along the y axis, at a 
constant velocity of  10 metres per second.  Point Q is moving with a 
constant speed of 11 metres per second (relative to the x-y plane) and 
always in the direction directly towards point P.  If at time t = 0, 
P is at the origin and Q is at the point (100,0) what is the equation 
of the locus of point Q ?

Thank you for your kind assistance. 

Michael Alger

Date: 7/19/96 at 8:25:51
From: Doctor Anthony
Subject: Re: Two Points Moving...

Draw the diagram I describe, and refer to it as we work through the 

We use the x and y axes in the usual way with P moving along OY and Q 
starting at (100,0) Let this point be Q0.  Because of ratio of speeds, 
after time t, the path distance Q0->Q will be 11*t and OP will be 
10*t, and if 's' is the distance along Q's path we have s/11 = OP/10 
or s = (11/10)*OP

Now the tangent at Q to the path will be pointing directly at P, so if 
dy/dx is the slope of the tangent and (x,y) the coordinates of Q we 
      (OP-y)/x = -dy/dx   

The minus sign because (OP-y)/x is positive, whereas the slope must be 
negative if y is increasing while x is decreasing.

       OP - y = -x(dy/dx)

           OP = y - x(dy/dx)   But s = (11/10)OP, so

            s = (11/10){y - x(dy/dx)}

We must try to get a differential equation in x and y, so will have to 
get rid of s.  Differentiate above equation with respect to x.

     ds/dx = (11/10){dy/dx - dy/dx - x(d^2y/dx^2)}

           = - (11/10)(x*d^2y/dx^2) .....(1)   

But dx/ds is the cos of the angle of slope, so ds/dx = sec of angle of 
slope so ds/dx = sqrt(1 + tan^2(angle of slope)) = sqrt(1 + (dy/dx)^2)

In this case because s increases while x decreases we have:

ds/dx = -sqrt(1 + (dy/dx)^2) ......(2)

Now to simplify the working, let p = dy/dx,  so  dp/dx = d^2y/dx^2 

Equating the right hand sides of equations (1) and (2)

    sqrt(1+p^2) = (11/10)(x*dp/dx)  Separating the variables we get

           dx/x = (11/10)(dp/sqrt(1+p^2))

         ln(kx) = (11/10)sinh^(-1)p     k is constant of integration.

Now x = 100 when p = 0  so ln(100k) = 0,  100k = 1,   k = 1/100

ln(x/100) = (11/10)sinh^(-1)p

p = sinh{ln(10/11)ln(x/100)} = sinh{ln(x/100)^(10/11)}

dy/dx = sinh{ln(x/100)^(10/11)}

      = (1/2){e^(ln(x/100)^(10/11) - e^(-ln(x/100)^(10/11)}

      = (1/2){(x/100)^(10/11) - (x/100)^(-10/11)}

    y = (1/2)*100*{(11/21)(x/100)^(21/11) - 11*(x/100)^(1/11)} + const

y = 0 when x = 100  so

   0 = (1/2)*100*{(11/21) - 11} + const

     = -11000/21 + const, so const = 11000/21

Equation of curve is :

  y = 50{(11/21)(x/100)^(21/11) - 11(x/100)^(1/11)} + 11000/21 

Q will meet P when x = 0, and this occurs at y = 11000/21 = 523.81 
meters from O

-Doctor Anthony,  The Math Forum
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Associated Topics:
High School Calculus

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