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?

enter image description here

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.

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


1 Answer 1


You are not too far off.
Let's start with a):

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 $\tan \theta=\frac{x(t)}{500}, \theta(t)=\frac{180^\circ}{\pi}\arctan \frac{-500+t}{500}$

You are correct about b).

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.

  • $\begingroup$ Ah, makes sense, thanks! $\endgroup$
    – Lars
    Oct 26, 2022 at 10:14

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