Related rates - finding hypotenuse of triangle

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Homework Statement


A plane flying horizontally at an altitude of 3 mi and a speed of 480 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 4 mi away from the station. (Round your answer to the nearest whole number.)

Homework Equations


##a^2=b^2+c^2## where ##a## is the hypotenuse of a triangle.

The Attempt at a Solution


I started by relating the given variables as follows
  • Altitude is a constant of 3 mi
  • Horizontal distance of the plane will be ##x##. We measure when ##x=4## miles.
  • Distance from the plane to the radar station will be ##y##. We measure when ##y^2=4^2+3^2 \longrightarrow y=5## miles.
  • Change in horizontal distance will be ##\frac{dx}{dt}## We are given that this is 480 miles/hour.
We have ##y^2=x^2+3^2##. Taking the derivative with respect to time gives ##2y\times\frac{dy}{dt}=2x\times\frac{dx}{dt}+0##. Substituting in known values gives: ##2(5)\times\frac{dy}{dt}=2(4)(480) \longrightarrow \frac{dy}{dt}=\frac{4\times480}{5}=384##. Yet this is not the answer.

Where am I going wrong? I've actually drawn out the triangle and variables, and I'm fairly stuck as to which part I'm messing up. Any insight will be appreciated!
 
on Phys.org
phinds said:
How did "4 mi away from the station" become only the horizontal component of how far the plane is away from the station?
Looks like I totally misread the question. I thought that the problem statement stated 4 to be the distance traveled in the x direction, not the distance between the plane and the station. That's cleared everything up.

Using the above information, the value for ##x## is ##\sqrt{4^2-3^2}=\sqrt{7}##. Plugging the correct values into my above derivative gives the correct answer to this problem: ##2(4)\times\frac{dy}{dt}=2(\sqrt{7})(480)\longrightarrow\frac{dy}{dt}=\frac{\sqrt{7}\times480}{4}\approx317##.

Thanks for pointing that out!