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Uniform Relative Motion in 3D

  1. Mar 2, 2010 #1
    This is not homework. I am reviewing all of the calculus that I have forgotten.

    Howard Anton's Calculus w/Analytical Geometry, 5th ed., section 4.1, prob #32:

    A police helicopter is flying due north at 100km/hr and at a constant altitude of 1/2mi. Below, a car is traveling west on a highway at 75km/hr. At the moment that the helicopter passes over the highway the car is 2mi east of the helicopter.
    a)How fast is the distance between the car and the helicopter changing at the moment the helicopter crosses the highway?

    The Complete Solutions Manual to Anton's 5th ed. Calculus, Prepared by Albert Herr, provides the following solution for prob 32, sec4.1. I quote it verbatim, except I describe the diagram:

    Let x, y, and z be the distances shown in the figure. [the diagram labels the point on the highway where the helicopter crosses as O, the point 1/2mi above O as P, the position of the helicopter as H, the position of the car as C, x=CO, y=HP, z=HC]. Find dz/dt evaluated at x=2, y=0 given that dx/dt=-75 and dy/dt=-100. Because triangle OPC is a right triangle, it follows that PC has length (x2+(1/2)2)1/2; but triangle HPC is also a right triangle so
    z^2=((x2+(1/2)2)1/2)2+y2=x2+y2+1/4
    and 2zdz/dt=2xdx/dt+2ydy/dt+0, dz/dt=(1/z)(xdx/dt+ydy/dt).
    Now, when x=2 and y=0, z2=22+02+1/4=17/4, z=(171/2)/2
    so dz/dt [at x=2, y=0] = (2/(171/2))(2(-75)+0(-100))=-300/(171/2).

    This published solution doesn't seem right to me. The helicopter's 100km/hr northerly rate should contribute to the rate of change of the distance between the helicopter and the car:

    Let the point O in the above solution be the origin and let the positive x-axis extend from O to C and beyond. Let the z-axis be vertical. Therefore, C=(2,0,0) and the car is moving in the negative x direction at 75km/hr and H=(0,0,1/2) and the helicopter is moving in the positive y direction at 100km/hr. The distance between the two points H and C seems to me to be irrelevant. The rate at which the distance between H and C is changing with time, d/dt(C-H)=dC/dt-dH/dt.
    let C=(xc,yc,zc) and H=(xh,yh,zh).
    Then dC/dt=(dxc/dt,dyc/dt,dzc/dt)=(-75,0,0) and dH/dt=(dxh/dt,dyh/dt,dzh/dt)=(0,100,0).
    dC/dt-dH/dt=(-75,-100,0)= the rate of change of the position of the car relative to the helicopter, or in the frame of reference in which the helicopter is stationary. If we had started with H-C rather than C-H then we would have (75,100,0) as the rate of change of the position of the helicopter with respect to the car.
    Either way, this velocity, (-75,-100,0) in the helicopter's frame, or (75,100,0) the the car's reference frame, =the rate of change with time of the distance between the two objects, and has a magnitude=(752+1002)1/2=125km/hr and is decreasing in both reference frames.

    Yet, since published solutions manuals are usually checked by several people, I'm hesitant as to whether or not my solution is correct.

    Please let me know whether you think my solution, the published solution, or some other solution is correct. No opinions, please. Supply precise reasoning. Thanks in advance to all responses.
     
  2. jcsd
  3. Mar 2, 2010 #2

    arildno

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    1. The position vectors of the two objects as functions of time are:
    C(t)=(2-75t,0,0)
    H(t)=(0,100t,1/2)

    (You mix together km/h and distances in miles, I haven't bothered to explore that issue)

    2. The DISTANCE function D(t) is therefore:
    [tex]D(t)=||C(t)-H(t)||=\sqrt{(2-75t-0)^{2}+(0-100t)^{2}+(0-\frac{1}{2})^{2}[/tex]

    3. The rate of change is now found by computing dD/dt
     
  4. Mar 3, 2010 #3
    Thx for ur very clear response. The published solution is evidently correct.
     
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