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# Calculus/Related Rates

Question
Hello.
I have presented the question and my thought process, I appreciate your input.

Two commercial airplanes are flying at 40,00ft along straight-line courses that intersect at right angles. Plane A is approaching the intersection at 442 knots. Plane B is approaching the intersection at 481 knots. At what rate is the distance between the planes changing when A is 5 nautical miles from the intersection and B is 12 nautical miles from the intersection.

I assumed I was using a^2 + b^2 = c^2 and taking the derivative of each with respect to time. I solved for dr/dt and got 614 knots. I believe I am miss guided in my train of thought it seems to simple:
A= 5nm (nautical miles per hour)  da/dt= 442knots
B= 12nm          db/dt= 481 knots
C 13nm          dr/dt= 614 knots

Thanks,
Alivia

The derivative is 2A(dA/dt) + 2B(dB/dt) = 2C(dC/dt).
If A is 5 nautical miles away and B is 12 nautical miles away.
From what is given, it has been seen that C = 13.

It is also given dA/dt = -442 nm/h and dB/Dt = -481 nm/h.

Using these, I get 2*5*(-442) + 2*12*(-481) = 2*13*(dC/dt).
From here, we can say that dC/dt = (-5*442 - 12*481)/13.

Looking back at your answer, I get the same for  C, except that it is negative since the distance is decreasing.  That is, dC/dt = -614 nautical miles per hour.
This means the are getting 614 nm/h closer together.

It asked for the rate at which the distance was changing,
and I would expect that to a negative speed since the distance is decreasing.

Questioner's Rating
 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment Thanks, I was preoccupied with the answer over looked that it would be negative.

Calculus

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