Related rates of a rotating light atop a moving car The question I am struggling with first:  Police are chasing a criminal down a narrow street at a speed of 90kmh.  If the blue light atop the car is rotating counterclockwise at a rate of one rotation per second and the buildings are only 3 meters from the car on the right, how fast is the beam moving on the wall at the instant when it is already six meters ahead of its source?
Through looking at several examples, I understand that I should use tan(theta)=y/x with x being constant at 3.  I have also converted everything to meters/minute.  That gives dy/dx at 5,400,000 (90kmh to meters/minute). d(theta)/dt would be 120pi (2pi per second times 60 seconds in a minute).  I am supposed to be getting an answer of 429.29 kmh.  Am I misunderstanding and the hypotenuse should reflect the "six meters ahead of its source"?  I have been fighting with this for DAYS.  Thank you in advance!!
 A: I do hope I don't get in trouble for doing your homework!
Take a look at the sketch and maybe this will make sense.police chase scene triangle
$$d\theta/dt=1\cdot\cfrac{rev}{sec}\cdot\cfrac{2\pi}{rev}=2\pi/s$$ The governing equation is $tan\theta=\cfrac{x}{3}$.
When differentiated:$$\frac{d(tan\theta)}{d\theta}\cdot\frac{d\theta}{dt}=\frac{d\left(\cfrac{x}{3}\right)}{dx}\cdot\left(\frac{dx}{dt}\right)$$
$$sec^{2}\theta\frac{d\theta}{dt}=\frac{1}{3}\cdot\frac{dx}{dt}$$
Before we can solve for $dx/dt$, we will need to find either $\theta$ or $sec\theta$.
Using the triangle relationship of our sketch, we can say $tan\theta=\cfrac{3}{6}$.
Therefore, $\theta=atan2(6,3)$ I haven't shown this numerically so as not to lose any decimal places.
Then we can plug in values.$$sec^{2}(atan2(6,3))\cdot(2\pi)=\frac{1}{3}\frac{dx}{dt}$$
$$\frac{dx}{dt}=94.2478\,m/s$$
Now add the forward speed of the car, which is $25\frac{m}{s}$ and convert the whole thing to km/hr.  When geogebra did it, I got exactly the answer you provided, which unconverted is 119.247779 m/s.
