Angle between function and axis

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The angle between the function y = √3x and the x-axis is π/3. This angle can be determined by calculating the slope of the line, which is √3, and using the arctangent function. The tangent of the angle formed with the x-axis is equal to the slope, leading to the conclusion that α = arctan(√3) = π/3. The discussion clarifies that the angle is derived from the slope of the line rather than any specific point on the graph. Understanding the relationship between the slope and the angle is crucial for solving similar problems.
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Homework Statement


What angle is between function y=\sqrt{3}x and Ox axis?


Homework Equations


For example is logicaly clear that angle between function y=x is 45 degrees or \frac{\pi}{4}


The Attempt at a Solution



I just know that answer is \frac{\pi}{3}, but can't understand how to get it.
 
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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?
 
If x=0 then y=0. I can't understand. More concretly, how to calculate it?
 
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.
 
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 \pi/4. tan(\pi/4)= 1. What is the slope of y= \sqrt{3} x? What angle has that tangent?
 
\arctan\sqrt{3}=\frac{\pi}{3}
 
Defennnder said:
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.
Sorry, I misread the question, I thought it said y = \sqrt{3x} = (3x)^{1/2} instead of y = \sqrt{3}x = (3)^{1/2} \cdot x. 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)
 
\tan\alpha=\frac{y}{x}=\frac{x\sqrt{3}}{x}=\sqrt{3}
\alpha=\arctan \sqrt{3}=\frac{\pi}{3}
 
Yep, that's the way to get it.
 

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