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An airplane passes over an airport at noon traveling $540$ mi/hr due north. At 1:00pm, another airplane passes over the same airport at the same elevation traveling due east at $580$ mi/hr. Assuming both airplanes maintained their (equal) elevations, how fast is the distance between them changing at 2:15pm?

I know $dx/dt= 540$ mi/hr and $dy/dt=580$ mi/hr. And that I need to use the Pythagorean theorem but I don't know what to use for $x,y,z$.

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closed as off-topic by colormegone, Antonios-Alexandros Robotis, Daniel W. Farlow, choco_addicted, Shahab Mar 16 '16 at 2:53

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Start thinking about the scenario at 1pm with the airport at the origin.

The airplane traveling north (along the y-axis) has been traveling for an hour already so it is 540 miles north of the airport. For every additional hour it will have traveled another 540 miles. So at 1pm its distance from the airport is given by $$y=540+540t$$ where $t$ is measured in hours.

Now the second plane is traveling east (along the x-axis). It's distance from the airport is given by $$x=580t.$$

Now for any positive time $t$ we have a right triangle so we can use the pythagorean theorem to find the length of the hypotenuse as a function of $t$, which is the distance D between the planes. So since the hypotenuse, i.e. D, is given by the Pythagorean Theorem as $D=\sqrt{x^2+y^2}$, we see that the distance between the planes is $$D=\sqrt{(580t)^2+(540+540t)^2}.$$

Now to find the rate of change you simply need to take the derivative of $D$ with respect to $t$: $$ \frac{dD}{dt}=\frac{d}{dt}\sqrt{(580t)^2+(540+540t)^2} .$$

That is the set up and I think you can take it from there.

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