# Related Rates Problem: Angle of Elevation

by aznshark4
Tags: angle, elevation, rates
 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 $$\theta$$ is changing when the angle is 30$$\circ$$" variables: x = ground distance of plane from the observer. $$\theta$$ = angle of elevation from observer to plane. given:altitude of plane is 5 miles. $$\frac{dx}{dt}$$ = -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 $$\theta$$, so the base equation for this problem is: tan $$\theta$$ = $$\frac{5}{x}$$ 3. The attempt at a solution I first found what x was, since I would need x to solve my problem: tan $$\theta$$ = $$\frac{5}{x}$$ tan 30$$\circ$$ = $$\frac{5}{x}$$ x = 5$$\sqrt{3}$$ Then, I found the derivative of both sides of the base equation, so my work looks like this: tan $$\theta$$ = $$\frac{5}{x}$$ sec2 $$\theta$$ $$\frac{d\theta}{dt}$$ = 5 (-x-2) $$\frac{dx}{dt}$$ $$\frac{d\theta}{dt}$$ = $$\frac{-5cos^2\theta}{x^2}$$ $$\frac{dx}{dt}$$ plugging in, the equation would turn into: $$\frac{d\theta}{dt}$$ = $$\frac{-5cos^230}{(5\sqrt{3})^2}$$ * -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: $$\frac{d\theta}{dt}$$ = $$\frac{miles}{miles^2}$$ * (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!
 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.
P: 14
 Quote by Billy Bob 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?

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.
 P: 14 I don't think I understand. When you take the derivative of tan x don't you always get sec2 x?
P: 394
 Quote by aznshark4 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.