(adsbygoogle = window.adsbygoogle || []).push({}); 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]

- sec
^{2}[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 was1/2 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!

**Physics Forums - The Fusion of Science and Community**

# Related Rates Problem: Angle of Elevation

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Related Rates Problem: Angle of Elevation

Loading...

**Physics Forums - The Fusion of Science and Community**