# Optimization / Related Rate Searchlight

I can't seem to figure out how to set up this problem. I'm a bit confused on the relationship between the variables here; I've seen a similar problem involving a searchlight but the angle of $\theta$ at which the beam leaves the light was given. Without that I'm not sure how to manipulate the information given.

The problem reads: A searchlight is 100m from the nearest point on a straight highway. As it rotates the searchlight casts a horizontal beam that intersects the highway in a point. If the light revolves at a rate of pi/6 rad/s, find the rate at which the beam sweeps along the highway as a function of $\theta$. For what value of $\theta$ is this rate minimized?

I don't need the problem solved for me, I can do that, I just don't know how to set this up. The relationship is confusing to me. Thanks.

Draw a picture. The road might as well be the $x$-axis, and the searchlight could be at $(0,100)$.

Label the position of the serchlight $S$. The nearest point on the road to $S$ might as well be called $O$. At a certain time $t$, the searchlight illuminates the point $X=X(t)$ on the road. Put $X$ say to the right of $O$. Call the $x$-coordinate of $X$ by the name $x=x(t)$.

Let $\theta=\theta(t)$ be the angle the beam has travelled through in time $t$. We measure angle with respect to the line $ON$.

For the picture, let $\theta=\angle XON$. Then $\tan\theta=\dfrac{x}{100}$.

Over to you! We know $\frac{d\theta}{dt}$ and want $\frac{dx}{dt}$.

Added: We could now solve for $x$ in terms of $\theta$. I prefer to differentiate immediately. Using the Chain Rule, we get $$\sec^2\theta \frac{d\theta}{dt}=\frac{1}{100}\frac{dx}{dt}.$$ On the assumption the rotation is counterclockwise, we get that $\frac{d\theta}{dt}=\frac{\pi}{6}$. That gives $$\frac{dx}{dt}=\frac{100\pi}{6}\sec^2\theta.$$ The minimum value of $\sec^2\theta$ is reached at $\theta=0$ (and $2n\pi$ for any positive integer $n$). As the intuition readily verifies, the minimum rate of change of $x$ is when $x=0$, that is, when the beam illuminates the nearest point $N$ on the road.

• Apologize for posting this way. I ended up with the $tan \theta$ expression you have above, I'm just not sure I understand how to relate the information. It doesn't seem like enough to me to get a derivative. Do I isolate x, differentiate both sides, and then solve for dx/dt? – user109868 Nov 19 '13 at 4:10
• I will add to my post. Left it at the point I did because you specified you did not want the problem solved. – André Nicolas Nov 19 '13 at 6:13
• Thank you, that is the result I had gotten but I don't have the answer given to me so I was not able to check if I did it right. It may be a type of question I'm tested on so I wanted to make sure I can take the right steps. – Jesse Anne Nov 19 '13 at 19:15
• You are welcome. Typical related rates: Find a simple relationship between the two objects, differentiate immediately. – André Nicolas Nov 19 '13 at 19:19
• This one threw me a bit because the other optimization questions we were given were much, much more involved. I figured what I ended up with was too simple to possibly be the answer. :) Thanks again. – Jesse Anne Nov 19 '13 at 19:21