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Homework Help: Maximum angle of elevation

  1. Oct 26, 2009 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations

    3. 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?
  2. jcsd
  3. Oct 26, 2009 #2


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    Science Advisor

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

    In other words the elevation at angle [itex]\phi[/itex] is [itex]\theta cos(\phi)[/itex].

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

    Solve those equations for [itex]\theta[/itex].
  4. Oct 26, 2009 #3
    I think saying that "elevation at angle [tex]\phi[/tex] is [itex]\theta cos(\phi)[itex]" 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

    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.

    PS: I can't get my quote and some latex to work properly. Sorry..
    Last edited: Oct 26, 2009
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