Maximum Angle of Elevation for Lighthouse Light?

[undrcvrbro]

Homework Statement

The axis of a light in a lighthouse is tilted. When the light points east, it is inclined upward at 7 degree(s). When it points north, it is inclined upward at 2 degree(s). What is its maximum angle of elevation?

Homework Equations

The Attempt at a Solution

Duf(0, 0) = gra.f(0, 0)* u


and the maximum slope will be

grad. f(0, 0) = sqrt(fx(0, 0)^2 + (fy(0, 0)^2))

fx(0, 0) = tan 9; fy(0,0)= tan2
And therefore

abs(grad. f(0, 0)) = ((tan9)^2 + (tan2)^2)
The maximum elevation is therefore given by

arctan ((tan 9)^2 + (tan2)^2) = 66 degrees

For some reason my final answer is wrong. Any idea where I made a mistake?

 

[HallsofIvy]

I really have no idea what you are doing there. Why would the maximum slope be given by "grad. f(0, 0) = sqrt(fx(0, 0)^2 + (fy(0, 0)^2))"? What is f?

I certainly don't see how, from knowing the elevation at two directions, ninety degrees apart, are "2" and "9" you get "66" for the maximum! It's not too difficult to see that, if we take \theta as the maximum angle of elevation and measure \phi around the circle from the direction of elevation, the angle of elevation in direction \phi is periodice with period 2\pi, starts at \theta, is 0 at \phi= \pi/2, -\theta at \phi= \pi, 0 againat \phi= 3\pi/2 and \theta again at \phi= 2\pi.

In other words the elevation at angle \phi is \theta cos(\phi).

Let \phi_0 be the angle corresponding to "east". Then \theta cos(\phi_0)= 7. "North" will now be \phi_0+ \pi/2 so \theta cos(\phi_0+ \pi/2)= 2.

Solve those equations for \theta.

 

[cartonn30gel]

\theta[/itex] as the maximum angle of elevation and measure \phi around the circle from the direction of elevation, the angle of elevation in direction \phi is periodice with period 2\pi, starts at \theta, is 0 at \phi= \pi/2, -\theta at \phi= \pi, 0 againat \phi= 3\pi/2 and \theta again at \phi= 2\pi.

In other words the elevation at angle \phi is \theta cos(\phi).

I think saying that "elevation at angle \phi is \theta cos(\phi)" is a very good assumption but not exactly correct. I encountered a very similar problem some time ago and I think the correct method is a little different. It is close to a cosine curve but not exactly<br /> <br /> Doing it your way Hallsoflvy, I get 7.280109 degrees. Doing it my way, I get 7.282863 degrees. Do you have the answer to that many significant digits undrcvrbro? If what I got is correct, I'll explain what you need to do.<br /> <br /> PS: I can't get my quote and some latex to work properly. Sorry..