Angle between function and axis

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Homework Help Overview

The discussion revolves around finding the angle between the function y=\sqrt{3}x and the x-axis, focusing on the relationship between the slope of the line and the angle it forms with the axis.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the concept of the angle formed by the function and the x-axis, questioning the relevance of specific points like (0,0) in determining the angle. They discuss the relationship between the slope of the line and the tangent of the angle, with some participants suggesting the use of trigonometric functions to find the angle.

Discussion Status

There is an ongoing exploration of the problem, with some participants providing hints and others questioning the assumptions made about the function's behavior. A few have identified the angle as \frac{\pi}{3} based on the slope, while others are clarifying misunderstandings regarding the function's representation.

Contextual Notes

Participants are navigating potential confusion between different interpretations of the function, particularly regarding the notation and the implications of evaluating the function at specific points.

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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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