Register to reply

Related Rates Problem: Angle of Elevation

by aznshark4
Tags: angle, elevation, rates
Share this thread:
aznshark4
#1
Sep19-09, 09:07 PM
P: 14
1. The problem statement, all variables and given/known data
"An airplane flies at an altitude of 5 miles toward a point directly over an observer. The speed of the plane is 600 miles per hour. Find the rate at which the angle of elevation [tex]\theta[/tex] is changing when the angle is 30[tex]\circ[/tex]"

variables:
  • x = ground distance of plane from the observer.
  • [tex]\theta[/tex] = angle of elevation from observer to plane.

given:
  • altitude of plane is 5 miles.
  • [tex]\frac{dx}{dt}[/tex] = -600mi/h (because the plane is travelling towards observer, distance between them decreases).


2. Relevant equations
logic:

the situation creates a right triangle with base x and height 5 mi; hypotenuse is not necessary in this problem. Angle of elevation is [tex]\theta[/tex], so the base equation for this problem is:

tan [tex]\theta[/tex] = [tex]\frac{5}{x}[/tex]


3. The attempt at a solution
I first found what x was, since I would need x to solve my problem:
  • tan [tex]\theta[/tex] = [tex]\frac{5}{x}[/tex]

  • tan 30[tex]\circ[/tex] = [tex]\frac{5}{x}[/tex]

  • x = 5[tex]\sqrt{3}[/tex]

Then, I found the derivative of both sides of the base equation, so my work looks like this:
  • tan [tex]\theta[/tex] = [tex]\frac{5}{x}[/tex]

  • sec2 [tex]\theta[/tex] [tex]\frac{d\theta}{dt}[/tex] = 5 (-x-2) [tex]\frac{dx}{dt}[/tex]

  • [tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{-5cos^2\theta}{x^2}[/tex] [tex]\frac{dx}{dt}[/tex]

plugging in, the equation would turn into:
  • [tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{-5cos^230}{(5\sqrt{3})^2}[/tex] * -600 = 30


however, the answer the book gave was 1/2 radians, and also, if you look at the way the units interact in the equation:

[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{miles}{miles^2}[/tex] * (miles/hour) = hour-1

the answer I had was in "nothing per hour" when the answer should be in "radians per hour".

What did I do wrong in this problem? Thanks in advance!
Phys.Org News Partner Science news on Phys.org
New model helps explain how provisions promote or reduce wildlife disease
Stress can make hard-working mongooses less likely to help in the future
Grammatical habits in written English reveal linguistic features of non-native speakers' languages
Billy Bob
#2
Sep20-09, 09:44 AM
P: 393
You and I are getting the same answer, so either we both are wrong, or your book is wrong.

Radians is a dimensionless unit, so "nothing per hour" and "radians per hour" are the same.
aznshark4
#3
Sep20-09, 01:08 PM
P: 14
Quote Quote by Billy Bob View Post
You and I are getting the same answer, so either we both are wrong, or your book is wrong.

Radians is a dimensionless unit, so "nothing per hour" and "radians per hour" are the same.
yes, thanks for the confirmation!

Yet, I also forgot to note this: when I convert the 30 (supposedly degrees, because I was working in degrees) to radians, I got .5236 which rounds to approximately 1/2 radians. However, since I got the answer in "radians per hour", I wasn't supposed to convert, which brings me to another question:

If it doesn't matter which format (radians or degrees) you use for the angle (because tan of any format always becomes a ratio without any units), why does the change in angle always stay 30?? isn't it supposed to change according to whatever unit you use in the first place?

Billy Bob
#4
Sep20-09, 02:40 PM
P: 393
Related Rates Problem: Angle of Elevation

Because the derivative of tan x is only sec^2 x if you are using radians. If you use degrees, you must use the chain rule which gives a factor of pi/180 or 180/pi.
aznshark4
#5
Sep20-09, 08:37 PM
P: 14
I don't think I understand. When you take the derivative of tan x don't you always get sec2 x?
Billy Bob
#6
Sep20-09, 11:16 PM
P: 393
Quote Quote by aznshark4 View Post
Yet, I also forgot to note this: when I convert the 30 (supposedly degrees, because I was working in degrees) to radians, I got .5236 which rounds to approximately 1/2 radians.
No, you were really working in radians, you just wrote "30" instead of pi/6, but your computations were really in radians.

Your answer was 30 radians per hour.

--

Explanation of derivatives in degrees: Sketch a graph of a sinusoid function with amplitude 1 and period 360, i.e. put tickmarks on the x-axis at 180 and 360, with the x intercepts at 0, 180, 360, etc. There is no way the slope at the origin is 1.

This function is y=sin(pi x/180). Its derivative wrt x is cos(pi x/180) * pi/180. The derivative at x=0 is pi/180.

You don't need to know this to do your problem. Just work in radians.
Billy Bob
#7
Sep21-09, 09:13 AM
P: 393
Aha. 30 radians per hour is 1/2 radian per minute. I bet that's what they did.


Register to reply

Related Discussions
Related Rates Problem Calculus & Beyond Homework 4
Related Rates Problem Calculus & Beyond Homework 5
Related rates problem Calculus & Beyond Homework 19
Related rates problem another one Calculus & Beyond Homework 2
Calculas Related Rates and Angle Problem Calculus & Beyond Homework 3