Drawing Graphs: Concentric Circles & Straight Lines

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SUMMARY

This discussion focuses on the mathematical representation of concentric circles and straight lines through the origin (0,0). The equation for concentric circles is derived from ln(x² + y²) = constant, leading to the conclusion that the graph represents a circle centered at the origin with a radius of √(e^constant). For straight lines, the equation arctan(y/x) = constant results in the line equation y = (tan(constant))x, indicating a straight line through the origin with a slope of tan(constant). The discussion clarifies that ln(0) is undefined and x=0 does not yield valid solutions for these graphs.

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  • Understanding of logarithmic functions, specifically ln(x)
  • Familiarity with Cartesian coordinates and graphing
  • Knowledge of trigonometric functions, particularly arctan and tan
  • Basic concepts of injective functions in mathematics
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Raees
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Hi, how would I go about drawing these two graphs?

31e4a231b2f01490ad29b8db02cbfd4c.png


and

3bbb0c98927abe3e79236b39a7587ef4.png
The first one would be concentric circles with the centre at (0,0).
The second one would be straight lines through (0,0).

Is this correct?
Also, what happens at ln(0) = constant for the first graph and x = 0 for the second graph?

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Correct.
Raees said:
Also, what happens at ln(0) = constant for the first graph
That is not possible. x=y=0 is never a solution. Same for x=0 in the second case.

For "interesting" functions f(x)=constant usually means x has one or a somewhat special set of values for a given constant. If f is injective then it is equivalent to x=constant.
 
For the first one, ln(x^2+ y^2)= constant, taking the exponential of both sides, x^2+ y^2= e^{constant}. Yes, the graph of that is a circle with center at the origin and radius \sqrt{e^{constant}}.

For the second one, arctan(y/x)= constant, taking the tangent of both sides, y/x= tan(constant) so y= (tan(constant))x. Since the tangent of a constant is also a constant, this is a straight line through the origin with slope tan(constant).
 
HallsofIvy said:
For the first one, ln(x^2+ y^2)= constant, taking the exponential of both sides, x^2+ y^2= e^{constant}. Yes, the graph of that is a circle with center at the origin and radius \sqrt{e^{constant}}.

For the second one, arctan(y/x)= constant, taking the tangent of both sides, y/x= tan(constant) so y= (tan(constant))x. Since the tangent of a constant is also a constant, this is a straight line through the origin with slope tan(constant).

Thanks, that helps a lot!
 

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