1. ## Related Rates Problem

An airplane is flying at a constant speed and altitude, on a line that will take it directly over a radar station located on the ground. At the instant the airplane is 60,000 feet from the station, an observer in the station notes that its angle of elevation is 30 degrees and is increasing at a rate of 0.5 degrees per second. Find the speed of the airplane.

So I know this involves taking the derivative, but I can't formulate a general equation that relates the known variables: the distance from the station (x), the angle of elevation ( $\Theta$ ), the rate that the angle is increasing( $d \Theta /dt$), and the speed ( $dx/dt$ ).

2. let x = horizontal distance from the station

z = direct distance from the station

first, you know that $x^2 + 60000^2 = z^2$

$2x \frac{dx}{dt} = 2z \frac{dz}{dt}$

so ...

$\frac{dz}{dt} = \frac{x}{z} \cdot \frac{dx}{dt} = \cos{\theta} \cdot \frac{dx}{dt}$

$x = z\cos{\theta}$

$\frac{dx}{dt} = -z\sin{\theta} \cdot \frac{d\theta}{dt} + \cos{\theta} \cdot \frac{dz}{dt}$

$\frac{dx}{dt} = -z\sin{\theta} \cdot \frac{d\theta}{dt} + \cos^2{\theta} \cdot \frac{dx}{dt}
$

$\frac{dx}{dt} - \cos^2{\theta} \cdot \frac{dx}{dt} = -z\sin{\theta} \cdot \frac{d\theta}{dt}
$

$\frac{dx}{dt}(1 - \cos^2{\theta}) = -z\sin{\theta} \cdot \frac{d\theta}{dt}$

$\frac{dx}{dt} \cdot \sin^2{\theta} = -z\sin{\theta} \cdot \frac{d\theta}{dt}$

$\frac{dx}{dt} = -z\csc{\theta} \cdot \frac{d\theta}{dt}$

from here, remember that the airplane's speed is the absolute value of the last expression (ignore the negative). also, remember that $\frac{d\theta}{dt}$ has to be in radians per second, so you'll have to convert. finally, remember that $\csc{\theta}$ = hypotenuse/opposite side.

3. So I know this involves taking the derivative, but I can't formulate a general equation that relates the known variables: the distance from the station (x), the angle of elevation ( ), the rate that the angle is increasing( ), and the speed ( ).
You have used x for 2 different values: the distance from the station and the distance travelled. Since the plane is not travelling directly away from the station, these values will differ. I will use r for the distance to the station and x for the distance travelled by the plane. Since the plane is travelling horizontally, $x = r\cos(\theta)$
$\frac{dx}{dt} = \frac{dr}{dt}\cos(\theta)-r\sin(\theta)\frac{d\theta}{dt}$
We have a value for everything here except $\frac{dr}{dt}$, so we go looking for that.
We know that $r^2 = x^2+y^2$ where y is the constant altitude. Using implicit differentiation,
$2r\frac{dr}{dt} = 2x\frac{dx}{dt}+2y\frac{dy}{dt}$
but $\frac{dy}{dt} = 0$ since y is constant, so it just remains to substitute for $\frac{dr}{dt}$ and solve for $\frac{dx}{dt}$ and substitute the actual values.

let x = horizontal distance from the station

z = direct distance from the station

first, you know that
The distance from the station is 60 000, not the distance from the ground.