Related Rates Problem: Angle of Elevation

[aznshark4]

Homework Statement

"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 traveling towards observer, distance between them decreases).

Homework 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 $$\theta$$ = $$\frac{5}{x}$$

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!

 

[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.

 

[aznshark4]

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]

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]

I don't think I understand. When you take the derivative of tan x don't you always get sec² x?

 

[Billy Bob]

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]

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