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Angle between function and axis

  1. Feb 3, 2008 #1
    1. The problem statement, all variables and given/known data
    What angle is between function [tex]y=\sqrt{3}x[/tex] and Ox axis?


    2. Relevant equations
    For example is logicaly clear that angle between function y=x is 45 degrees or [tex]\frac{\pi}{4}[/tex]


    3. The attempt at a solution

    I just know that answer is [tex]\frac{\pi}{3}[/tex], but can't understand how to get it.
     
  2. jcsd
  3. Feb 3, 2008 #2

    CompuChip

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    Hint: the angle between the function and the x-axis is the same angle as between the tangent line at x = 0 and the x-axis (draw a picture to see why).
    Can you solve it now?
     
  4. Feb 3, 2008 #3
    If x=0 then y=0. I can't understand. More concretly, how to calculate it?
     
  5. Feb 3, 2008 #4

    Defennder

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    Why do you need to set x,y=0? The question is why the angle between the line graph and the axis is pi/3, not the angle between the point (0,0) and the x-axis, which doesn't make sense. You can see that the graph is a line right? Now, let theta be the angle between the line and the x-axis. Do you know of way to find theta using trigo? You'll have to draw a triangle to see it.
     
  6. Feb 3, 2008 #5

    HallsofIvy

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    The slope of a line, such as y= x, is than tangent of the angle between the line and the x-axis. As you said before, the angle between the line y= x and the x-axis is [itex]\pi/4[/itex]. tan([itex]\pi/4[/itex])= 1. What is the slope of y= [itex]\sqrt{3}[/itex] x? What angle has that tangent?
     
  7. Feb 3, 2008 #6
    [tex]\arctan\sqrt{3}=\frac{\pi}{3}[/tex]
     
  8. Feb 3, 2008 #7

    CompuChip

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    Sorry, I misread the question, I thought it said [tex]y = \sqrt{3x} = (3x)^{1/2}[/tex] instead of [tex]y = \sqrt{3}x = (3)^{1/2} \cdot x[/tex]. I had a picture in my mind of drawing the tangent line at the origin and then calculating the angle of that with the x-axis, which could of course be done at any point. But since the function is just a straight line, it doesn't matter in this case (y' does not depend on x)
     
  9. Feb 3, 2008 #8
    [tex]\tan\alpha=\frac{y}{x}=\frac{x\sqrt{3}}{x}=\sqrt{3}[/tex]
    [tex]\alpha=\arctan \sqrt{3}=\frac{\pi}{3}[/tex]
     
  10. Feb 3, 2008 #9

    CompuChip

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    Yep, that's the way to get it.
     
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