Related Rates - Not getting answer in book

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Homework Help Overview

The problem involves related rates in the context of a plane flying horizontally over a radar station. The original poster is trying to determine the rate at which the distance from the plane to the station is increasing when the plane is 2 miles away from the station.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply the Pythagorean theorem and differentiate to find the rate of change of distance. Some participants suggest differentiating implicitly and question the interpretation of the distance measurement.

Discussion Status

Participants are exploring different interpretations of the problem, particularly regarding the meaning of "2 miles away." Some guidance has been provided on how to approach the differentiation, and there is acknowledgment of a potential misunderstanding in the setup.

Contextual Notes

There is a noted confusion about whether the 2 miles refers to horizontal distance or the actual distance from the plane to the radar station, which impacts the calculations. The participants are working within the constraints of the problem as presented in the original post.

GeoMike
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The problem given in my book is:

A plane flying horizontally at an altitude of 1mi and a speed of 500 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 2mi away from the station.

I set up and solved the problem this way:
dx/dt = 500 mi/h
x = 2mi
y = 1mi (constant)

Distance:
s2 = y2 + x2

s2 = 1 + x2

s = (1 + x2)1/2

d/dt = d/dt[sqrt(1 + x2)1/2]

ds/dt = (1/2)(1 + x2)-1/2 * (2x(dx/dt))

ds/dt = (1 + x2)-1/2 * (x(dx/dt))

ds/dt = [tex]\frac{x(dx/dt)}{(1 + x^2)^{1/2}}[/tex]

Substiting:
ds/dt = [tex]\frac{(2)(500)}{(1 + (2)^2)^{1/2}}[/tex]

ds/dt = [tex]\frac{1000}{\sqrt{5}}[/tex]

The final answer:
ds/dt = [tex]200\sqrt{5}[/tex] mi/h

However the back of my book has: [tex]250\sqrt{3}[/tex] mi/h

What am I doing wrong?
Thank you,
-GM-
 
Last edited:
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don't solve for s, just leave it at [tex]s^2=1+x^2[/tex] and differentiate implicitly to get

[tex]2s\frac{ds}{dt}=2x\frac{dx}{dt}[/tex]

solve for ds/dt

[tex]\frac{ds}{dt}=\frac{x}{s}\frac{dx}{dt}[/tex]
 
You work was correct that far, however, since [tex]s=\sqrt{x^2+1}[/tex] anyhow. The error is the sqrt(5) should be sqrt(3): 2 mi away is straight line distance.
 
You have misinterpreted the meaning of "2 mi away from the station".

It is s that equals 2, not x!

Thus, you have:
[tex]2^{2}=y^{2}+x^{2}\to{x}=\sqrt{3}[/tex]
 
AH! Thank you!
I understand now, I was taking 2mi as horizontal distance.

Thanks!
-GM-
 

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