
#1
Sep1909, 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:
given:
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:
Then, I found the derivative of both sides of the base equation, so my work looks like this:
plugging in, the equation would turn into:
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! 



#2
Sep2009, 09:44 AM

P: 394

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. 



#3
Sep2009, 01:08 PM

P: 14

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? 



#4
Sep2009, 02:40 PM

P: 394

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.




#5
Sep2009, 08:37 PM

P: 14

I don't think I understand. When you take the derivative of tan x don't you always get sec^{2} x?




#6
Sep2009, 11:16 PM

P: 394

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 xaxis 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. 



#7
Sep2109, 09:13 AM

P: 394

Aha. 30 radians per hour is 1/2 radian per minute. I bet that's what they did.



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