Wikipedia:Reference desk/Archives/Mathematics/2016 March 20

= March 20 =

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A ship sailing east at 15 km/hr passed a point A at 11 AM. A second ship sailing south at 10 km/hr passed A at 9 AM. How fast were the ships separating at 1 P.M.?49.126.255.48 (talk) 12:34, 20 March 2016 (UTC)


 * Hello Lalitpur, Nepal! Here one way to approach your exercise.  Ask yourself the following:  At 1 P.M., how many hours has it been since the east-sailing ship passed point A?  Using that, how far east of point A will it have traveled?  Similarly, how many hours has the south-sailing ship been sailing since passing point A and how far south will it have traveled in that time?  Now draw a diagram showing the position of the two ships at 1 P.M. relative to point A and then calculate the distance between the ships.  (See Pythagorean theorem.)  Once done, you may wish to look back at the problem to see if you could have done it in your head.  This particular exercise was constructed to make the calculation easy via the use of a special right triangle.
 * For problems like this you are expected to assume a flat earth, and if the distances involved are small, as they are in this problem, and point A is not at an extreme latitude, then your answer will be close to what you would get if you took the Earth's curvature into account. For larger distances or extreme latitudes, you would need to know the latitude of point A and whether the first ship was sailing a rhumb line (constantly due east) or a great circle route which carried it due east as it passed point A. -- ToE 13:37, 20 March 2016 (UTC)  [My apologies to the questioner for misreading the question as being the much simpler one of "How far were the ships separated at 1 P.M.?" -- ToE 21:52, 20 March 2016 (UTC)]
 * Oops! So if the questioner did work out the simpler exercise as I initially suggested, they would have wound up with a diagram showing the east-sailing ship 30 km east of point A and the south-sailing ship 40 km south of point A at 1 P.M.  From that point, I offer two methods to calculate the separation speed at that time, the first using simple calculus.  Let x be the distance from point A to the east-sailing ship, and let y be the distance from point A to the south-sailing ship.  At time 1 P.M., x = 30 km and y = 40 km.  Moreover, dx/dt = 15 km/hr and dy/dt = 10 km/hr.  If you let s be the distance between the ships, then the Pythagorean theorem tells you s2= x2 + y2.  Differentiate both sides of the equation with respect to time to get 2s·ds/dt = 2x·dx/dt + 2y·dy/dt.  Solve for ds/dt = (x·dx/dt + y·dy/dt)/s.  Plug in the values and there is your answer.  And the numbers do turn out to be easy enough to do in your head!
 * This question could also be asked in the context of a pre-calculus physics class where the student is expected to use vectors but not differentiation. In that case, let's consider the the motion of the south-sailing ship with respect to the east-sailing ship.  Choose the convenient coordinate system relative to the east-sailing ship (so that ship is at rest in this coordinate system) with the positive x-axis pointing west and the positive y-axis pointing south.  (Other choices will work as well, but this way we get to use only positive values.)  The displacement vector s of the south-sailing ship with respect to the east-sailing ship is (30 km, 40 km).  The  velocity vector v of the south-sailing ship with respect to the east-sailing ship is (15 km/hr, 10 km/hr).  (That is, in addition the the 10 km/hr southward component, the south-sailing ships is moving 15 km/hr west with respect to the east-sailing ship, even though it is moving neither east nor west with respect to the globe.)  To determine the separation speed of the two ships, find the component of the velocity vector v parallel to the displacement vector s by taking the dot product of v with the unit vector parallel to s, namely v·(s/|s|), which gives you the same calculation as when we used the derivative.  If you haven't learned about dot products yet, you can determine this scalar projection by using a bit of trigonometry.
 * So between these two solutions and the one suggested below by PiusImpavidus, which do you prefer? (And if you tell us what you got as an answer, we can confirm your result, though you could also use StuRat's method to approximate an answer as a check.) -- ToE 03:42, 21 March 2016 (UTC)


 * I may well be overthinking this, but it seems to me that the problem as stated by the OP is more complicated than I thought at first glance, or than the first two replies recognize. Assuming the flat earth case: It seems to me that it cannot be done in one's head, nor is the speed of separation time-independent and distance-independent. The eastbound ship (E) is 30 km east at 1PM, 60 km east at 3 PM, and 90 km east at 5 PM. The southbound ship (S) is 40 km south at 1PM, 60 km south at 3PM, and 80 km south at 5PM. Thus their distance apart is 50 km at 1 PM, 60sqrt(2) km at 3 PM, and 10sqrt(145) km at 5 PM. So the rate of separation is increasing over time: from 34.8528 km per 2 hours (from 1:00 to 3:00) to 35.5631 km per two hours (from 3:00 to 5:00). Since the rate of separation is changing, this looks like a calculus problem to me. Loraof (talk) 19:09, 20 March 2016 (UTC)


 * Yes, but you can get a reasonable approximation by looking at the distances a half second before and half second after 1 PM, and get the rate by dividing the change in distance by one second (or 1/3600th of an hour, if the answer is to be in km/hr). StuRat (talk) 19:28, 20 March 2016 (UTC)


 * Didn't react at once because this struck me as a homework exercise, but given the discussion above I shall give some input anyway. Not a complete answer though.
 * You can easily get a formula for the position of the first ship: p(t)=15×(t-11). Similar for the second ship, moving at right angle. Use Pythagoras to calculate the separation as a function of time, which is the square root of some quadratic polynomial of time. Differentiate that one and evaluate at 13 o'clock. PiusImpavidus (talk) 19:56, 20 March 2016 (UTC)