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Discrepancy Between Related Rates Solution and Vector Solution

  1. Jun 10, 2012 #1
    Hi all!

    I was looking through my old calculus book and was looking at a problem on related rates. I looked at the problem that was in the related rates section and I saw what was an easier way of doing it. The problem was this:

    Car A is traveling west at 50 mi/hr and Car B is traveling north at 60 mi/hr. Both are headed for the intersection of the two roads (0,0). At what rate are the cars approaching each other when car A is at .3 mi (.3, 0) and car B is .4 mi(0, -.4) from the intersection.

    This problem is solved using related rates. The final answer is :

    dz/dt = 2(.3(50) + .4(60) = 78 mi/hr, exactly.

    The other way of doing this problem that I thought of using skips the calculus and uses vector addition and pythagorean theorem.

    I instead consider Car A is at rest and instead Car B is moving 50 mi/hr east in addition to its 60 mi/hr north. Then use pythagorean theorem to find net velocity.

    60^2 + 50^2 = 6100.
    √6100 = 78.1025 mi/hr.

    This leads me to the question of this thread: Why are these values not exactly the same?The related rates solution does not round, and the answer is exactly 78. The vector solution is extremely close, but it is not exactly 78. What is causing the slight error between these two solutions? Which one is the actual exact rate of change between the two cars?

    I have done changes to this problem and have reasoned through different scenarios. If Car B has its velocity always pointed at car A then the velocity is constant because it is a 1 dimensional kinematic problem.

    If car A is off from the velocity vector of Car B then the rate of change is not constant and will reach 0 around where Car B approaches Car A.

    Please help me restore my sanity! If this was performed in real life which result would be correct? Thank you!
    Last edited: Jun 10, 2012
  2. jcsd
  3. Jun 10, 2012 #2
    The discrepancy lies in the fact that you were asked for the velocity toward car A, which in the frame of car A being at rest lies at the origin. The velocity of car B toward car A should be the radial component of car B's velocity--this is the part that's coming toward A.
  4. Jun 10, 2012 #3


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    Hey Curtis15 and welcome to the forums.

    I agree with the analogy for using vectors and pythagoras' theorem, but I don't agree with how you made the calculation with the vectors.

    You are taking into account the velocities but not the positions which will affect the relative velocities of the two cars.

    Remember that you are calculating the change of the displacement of the hypotenuse instantaneously at t = 0. At t = 0, this displacement is calculated with pythagoras's theorem to give 0.5 miles (SQRT((-0.4)^2 + 0.3^2)).

    But the displacement to the origin is decreasing for both cars and the velocity not only corresponds to the instantaneous velocities of each car, but also the position since the relative velocity must factor this in (remember the relative velocity is the instantaneous rate of the change of the hypotenuse of the triangle defined by the two cars and the origin).

    With this information, can you now recalculate the relative instantaneous velocity at t=0?
  5. Jun 10, 2012 #4
    In the vector solution attempt, the velocity I found is directed radially toward Car A if it were at the orgin. Am i misinterpreting your response incorrectly?
  6. Jun 10, 2012 #5
    Car B's relative velocity vector and its relative position vector are not parallel to one another. 50/60 = 5/6, but .3/.4 = 3/4. Their slopes aren't even close. See what I mean?
  7. Jun 10, 2012 #6
    Yes! I completely see what you are talking about now. I had totally forgotten to check to see if the velocity component was directed parallel to the position.

    Thanks for the help!

    And thanks Chiro for taking the time to try to help me, but Muphrid gave me the insight I needed. Sorry!
  8. Jun 10, 2012 #7


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    The main thing is outcome: many ways to get there and not any one way is necessarily better than another.
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