Related rates - finding hypotenuse of triangle

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SUMMARY

The discussion focuses on solving a related rates problem involving a plane flying at an altitude of 3 miles and a speed of 480 miles per hour, passing over a radar station. The key equation used is the Pythagorean theorem, represented as a² = b² + c², where 'a' is the hypotenuse. The correct interpretation of the distance from the plane to the station is crucial; initially misinterpreted as the horizontal distance, it is clarified that the distance is the hypotenuse. The final calculated rate at which the distance from the plane to the radar station is increasing is approximately 317 miles per hour.

PREREQUISITES
  • Understanding of related rates in calculus
  • Familiarity with the Pythagorean theorem
  • Ability to differentiate equations with respect to time
  • Knowledge of basic trigonometric concepts
NEXT STEPS
  • Review related rates problems in calculus textbooks
  • Practice solving problems using the Pythagorean theorem in real-world scenarios
  • Learn how to apply implicit differentiation in related rates
  • Explore additional examples of related rates involving motion and distance
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Students studying calculus, particularly those focusing on related rates, as well as educators looking for examples to illustrate these concepts in a practical context.

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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!
 
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How did "4 mi away from the station" become only the horizontal component of how far the plane is away from the station?
 
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!
 

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