# Related rates problem - how a beam from a laser moves across the wall

The problem:

Let $$\theta$$ be the angle between the laser and the horizontal line which is at a 90 degree angle at the wall(see drawing). The laser point moves at a pace of 1cm/sec vertically on the wall between 45 degrees, and -45 degrees(-45 when the laser pointer is under the horizontal line). The laser is located 500 cm from the wall.

a) Find an expression for $$\theta(t)$$ for how $$\theta$$ measured in degrees must change with respect to time.

b) At the start, the laser moves from $$\theta$$ = -45 degrees, and moves to 45 degrees by the end. How long does it take for the laser to move from -45 degrees to 45 degrees?

c) what is the angle-velocity of $$\theta'(t)$$ ? Measured in degrees.

d) What is the angle-velocity when the laser pointer passes the midpoint of the wall? Ok, so far I've drawn the problem, and dx/dt is 1cm/s, correct?

I'm having trouble solving a), but I think I have b) solved:

b) = tan 45 * 500 = 1 * 500 = 500. So x is 500cm long between 0 degrees and 45 degrees. Taking it from -45 to 45 means 500*2 = 1000cm, correct? This means the laser should use 1000 seconds, or 16 minutes and 40 seconds to go from -45 degrees to 45 degrees.

c) Now I guess I have to find $$d\theta/dt$$?

Am I on the right track, or have I misunderstood?

So finding $$d\theta/dt$$, I tried:

tan$$\theta$$ = x/500

sec^2$$\theta$$*$$d\theta/dt$$ = 1/500 * dx/dt

$$d\theta/dt$$ = 1/500 * (1/sec^2 $$\theta$$) * dx/dt

$$d\theta/dt$$ = 1/500 * cos^2 $$\theta$$ * dx/dt

= 1/500 * cos^2(500/1000) * 1 # since dx/dt = 1

0.769 rad * 180/pi = approx 0.088 degree/sec. Is this correct?

You are not too far off.

assuming the laser moves from bottom to the top once (it's not clear what happens after it reaches the top), we can express $$x=-500+t, 0 \le t \le 1000$$; we can also express $$\tan \theta=\frac{x(t)}{500}, \theta(t)=\frac{180^\circ}{\pi}\arctan \frac{-500+t}{500}$$

To find $$\theta'(t)$$, you can simply take a derivative of expression obtained in a).
$$\theta'(t)=\frac{180^\circ}{500\pi}\frac{500^2}{500^2+(-500+t)^2}$$, $$\theta'(500)=\frac{180^\circ}{500\pi}$$.
Your way would also work but for some reason you used $$\cos^2 \frac{500}{1000}$$ whereas $$\theta$$ is clearly zero at the midpoint.