Question

g A rotating light is placed 3 meters from a wall. Let W be the point on the wall that is closest to the light. Suppose the light completes a rotation every 15 seconds. Use an inverse trigonometric function to determine how fast the beam of light is moving along the wall when the tip of the light is 1 meter from W. Express your answer in meters per minute.

Answer 1

Answer:

$v=\frac{80}{3}\pi \frac{m}{min}$

Step-by-step explanation:

In order to start solving this problem, we can begin by drawing a diagram of what the problem looks like (see attached picture).

From that diagram, we can see that we have a triangle we can analyze. We'll call the angle between the vertical line and the slant line $\theta$. And we'll call the distance between W and the point we are interested in l.

Now, there are different things we need to calculate before working on the triangle. For example, we can start by calculating the angle $\theta$.

From the diagram, we can see that:

$tan \theta = \frac{l}{3}$

when solving for $\theta$ we will get:

$\theta = tan^{-1} (\frac{l}{3})$

we could use our calculator to figure this out, but for us to get an exact answer in the end, we will leave it like that.

Next, we can calculate the angular velocity of the beam. (This is how fast the beam is rotating).

We can use the following formula:

$\omega = \frac{2\pi}{T}$

where T is the period of the rotation. This is how long it takes the beam to rotate once. So the angular velocity will be:

$\omega = \frac{2\pi}{3} \frac{rad}{s}$

Next, we can take the relation we previously got and solve for l, so we get:

$tan \theta = \frac{l}{3}$

$l = 3 tan \theta$

Now we can take its derivative, so we get:

$dl = 3 sec^{2} \theta d\theta$

and we can divide both sides of the equation into dt so we get:

$\frac{dl}{dt} = 3 sec^{2} \theta \frac{d\theta}{dt}$

in this case $\frac{dl}{dt}$ represents the velocity of the beam on the wall and $\frac{d\theta}{dt}$ represents the angular velocity of the beam, so we get:

$\frac{dl}{dt} = 3 sec^{2} (tan^{-1} (\frac{l}{3})) (\frac{2\pi}{15})$

we can simplify this so we get:

$\frac{dl}{dt} = (\frac{2\pi}{5})sec^{2} (tan^{-1} (\frac{l}{3}))$

we can use the Pythagorean identities to rewrite the problem like this:

$\frac{dl}{dt} = (\frac{2\pi}{5})(1+tan^{2} (tan^{-1} (\frac{l}{3})))$

and simplify the tan with the $tan^{-1}$ so we get:

$\frac{dl}{dt} = (\frac{2\pi}{5})(1+(\frac{l}{3})^{2})$

which simplifies to:

$\frac{dl}{dt} = (\frac{2\pi}{5})(1+(\frac{l^{2}}{9}))$

In this case, since l=1, we can substitute it so we get:

$\frac{dl}{dt} = (\frac{2\pi}{5})(1+(\frac{1}{9}))$

and solve the expression:

$\frac{dl}{dt} = (\frac{2\pi}{5})(\frac{10}{9})$

$\frac{dl}{dt} = \frac{20}{45}\pi$

$\frac{dl}{dt} = \frac{4}{9}\pi \frac{m}{s}$

now, the problem wants us to write our answer in meters per minute, so we need to do the conversion:

$\frac{dl}{dt} = \frac{4}{9}\pi \frac{m}{s} * \frac{60s}{1min}$

$velocity = \frac{80}{3} \pi \frac{m}{min}$