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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!

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# Discrepancy Between Related Rates Solution and Vector Solution

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