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I took a test yesterday, and would like to know how to answer this specific question on the exam:

One airplane flew over an airport at the rate of $300$ mi/hr. Ten minutes later another airplane flew over the airport at $240$ mi/hr. If the first airplane was flying west and the second flying south (both at the same altitude), determine the rate at which they were separating $20$ minutes after the second plane flew over the airport.

I know that the pythagorean theorem should be used: $x^2 + y^2 = z^2$; but I don't know what to use for the $x, y$, and $z$ values.

$dy/dt = 240$ mi/hr

$dx/dt = 300$ mi/hr

Audrey
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    Let the position of the first plane be x and the position of the second be y. You're given dx/dt and dy/dt and you can solve for the initial x and y values. Then use the Pythagorean theorem to find the distance between the two planes and differentiate with respect to time. – Pranav Marathe Jan 29 '15 at 23:31
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    To attract answers to your question, please add some context and background information. For example, where did you encounter this problem (e.g. a book, class, real-life)? Please also show your attempt; seeing your work helps us help you. If this is homework, please read this post. – apnorton Jan 29 '15 at 23:46
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    @anorton Thank you for your help. I added some background information. – Audrey Jan 30 '15 at 00:03
  • @Audrey note that if one plane flies into west you can view that as a vector pointing to the left, that is in negative x-direction on a plane (-something,0). The plane that flies to the south, you can view as vector pointing down, or negative y-direction on a plane (0, -somethingelse). so you can apply pythagorean theorem something^2+somethingelse^2=result^2... – Loreno Heer Jan 30 '15 at 00:22

2 Answers2

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Paths of the planes

$p_1(t) = (t \cdot 300, 0)$

$p_2(t) = (0,(t-1/6) \cdot (240))$

$s(t) = p_1(t) - p_2(t)$

$\left.\left(\frac{d}{dt} \|s(t)\| \right) \right\vert_{t=.5} = ...$

Loreno Heer
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You have $x^2 + y^2 = z^2$ your goal is to find $\frac {dz}{dt}$

$2z\frac {dz}{dt} = 2x\frac {dx}{dt} + 2y\frac {dy}{dt}$

Now you will need to find the positions of the planes at the appointed time, to find $x,y,z$

$x$ is the distance the west-bound airplane travels in 30 minutes.

$y$ is the distance the south-bound airplane travels in 20 minutes.

Use the Pythagorean theorem above to find $z.$

distance = speed $\times$ time

You will need to convert units because the speed is in mph, and time is in minutes.

And $\frac {dx}{dt}, \frac {dy}{dt}$ are the speeds of the planes.

Doug M
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