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At noon on a certain day, a plane is $200$ miles south of another aircraft and flying north at $550$ mph, while the second plane is flying southwest at $600$ mph. How much later after this instant is their distance a minimum?

I can easily work this problem if the planes are at right angles to one another (Pythagorean Theorem) but these planes are flying at what is implied to be $45^o$ angle (due north towards origin and southwest from origin). My range is $[0,200]$ miles since they are at $200$ miles apart now at time$=0$.

The shape is a triangle with all three sides in movement, but it is not a right triangle except for at one instant in time. I thought about Law of Cosines but I don't know the angle measures for certain.

Can someone help get me started on this problem? Once I figure out their distance, I can work backwards to find the time it occurs.

Thank you!

fitzmerl duron
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user147485
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1 Answers1

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Lets say that that the airplane flying north is at $(0, -200) $ and the start. Its position is therefore $(0, 550t - 200)$ if $t$ is the number of hours.

Then the other airplane is starting at $(0,0)$ and its position is $(-\sqrt {\frac 12}*600t, -\sqrt{\frac 12}*600t)$.

And therefore $d(t) = \sqrt{(0-\sqrt {\frac 12}*600t)^2 + (550t-200--\sqrt {\frac 12}*600t)^2}$.

Now it's a max min problem.

fleablood
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