Inverse Trig Function Help

J

jacksondesu

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Feb 13, 2010
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A 15- foot wide highway sign is placed 10 feet from a road, perpendicular to the road. A spotlight at the edge of the road is aimed at the sign


a. Express theta as a function of the distance x from Point A to the spotlight.
b. How far from Point A should the spotlight be placed so that the angle theta is as large as possible?


for a. I got theta= tan[sup:3gywwovn]-1[/sup:3gywwovn](25/x) - tan [sup:3gywwovn]-1[/sup:3gywwovn](10/x)

but i am not sure about part b.

I found an answer from another site using derivatives, but since this is pre-calc we have not learned this as of yet.
My first guess was to solve for x, but I am not sure as to how.
Please help.
 
I saw a similar one, but since we have not learned about derivatives, I was thinking there would be another way to solve.
 
They probably want you to graph it. That is usually the method in a pre-calc course when maximizing something like this.

As an aside, let the distance from the point on the road nearest the sign be c and let d be the length of the sign.

Therefore, the distance that maximizes the angle theta is \(\displaystyle \sqrt{c^{2}+cd}\). In this case, c=10 and d=15.

I bet this general formula matches your case if you use c=10 and d=15.
 
Thanks for taking the time to provide the light but the bulb is still off.
I cant make the connection as to how you got the formula to maximize theta.
Perhaps I am overthinking the process. Can you step by step me or at least give me a hint.
Have I missed a crucial chapter?
 
Just use your calculator or other graphing utility to graph \(\displaystyle {\theta}=tan^{-1}(\frac{25}{x})-tan^{-1}(\frac{10}{x})\).

Then, look at the graph or the one I provided. That is what I graphed. See the high point on the graph?. That is the

value of x that maximizes the angle theta. Enter that value of x in to the above formula and you have the max angle.

The x value is \(\displaystyle 5\sqrt{10}\approx 15.81\). I think you are over thinking it :wink:

Remember back when you found the max by finding the vertex of a parabola?. It's like that.
 
Here's a shot from my TI
 

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Eureka!!!
The 20 watts are glowing.
I am using a TI-84 and using the table I see the answer.
Thanks a lot.
For my edification....like my coach says....how did you know it was 5*square root of 10?
Thanks again for the insight.
 
Actually, I saw they were equivalent just by osmosis.

Another thing is to use the derivative of the arctan equation we derived, setting to 0 and solving for x.

This leads to the quadratic \(\displaystyle -15x^{2}+3750=0\)

\(\displaystyle x=5\sqrt{10}\)
 
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