Show proof of point C in the given problem that involves Polar equation

  • #1
chwala
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Homework Statement
See attached. My interest is on part C (Highlighted in yellow)
Relevant Equations
polar equations
1712784862585.png


c
1712784886609.png



Parts (a) and (b) are okay ... though the challenge was on part (a)

My graph had a plot of r on the y-axis vs θ on the x-axis). The sketch of my graph looks like is shown below;
1712793020305.png

I suspect the ms had θ on the x-axis vs r on the y-axis.

I used the equation ##r=\sqrt{\dfrac {1}{θ^2+1}}## with various values of ##θ## in the given domain and ended up with a graph opening on the right side of the first and fourth quadrant which is different from the attached graph from the mark scheme.

The shapes were similar.

1712785051115.png



Now for part (c), the steps are quite straightforward, I just want to check why they used ##r\sin θ##. Most probably its the distance of the maximum point of the graph from the x-axis.

The other working steps are quite clear- they used the quotient rule and then sign change to wrap up the question.

I am aware that cartesian to polar form we have ##x = r \cos θ## and ##y = r \sin θ## and therefore to determine distance in the ##y## direction one has to use ##y = r \sin θ##. If this answers my own query then thanks in advance.



1712785469242.png
 
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  • #2
I don't follow your graph of ##r^2 = \frac 1 {\theta^2 + 1}##. If ##\theta = 0##, then ##r = \pm 1##, so you would get points on the horizontal axis 1 unit to the right and 1 unit to the left. Also, if ##\theta = \pi/2##, the values of r would be ##\pm\frac 1 {\sqrt{(\pi/2)^2 + 1}}##, and not ##\pm 1## as you show on your graph.
chwala said:
I suspect the ms had θ on the x-axis vs r on the y-axis.
That's not the way that polar graphs work. Angles are measured in the CCW direction from the positive x axis (the ray ##\theta = 0##), and r is measured out along the ray defined by the angle.
 
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  • #3
Mark44 said:
I don't follow your graph of ##r^2 = \frac 1 {\theta^2 + 1}##. If ##\theta = 0##, then ##r = \pm 1##, so you would get points on the horizontal axis 1 unit to the right and 1 unit to the left. Also, if ##\theta = \pi/2##, the values of r would be ##\pm\frac 1 {\sqrt{(\pi/2)^2 + 1}}##, and not ##\pm 1## as you show on your graph.
That's not the way that polar graphs work. Angles are measured in the CCW direction from the positive x axis (the ray ##\theta = 0##), and r is measured out along the ray defined by the angle.
Noted on the direction of r and the angle. Cheers.
 
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  • #4
Mark44 said:
I don't follow your graph of ##r^2 = \frac 1 {\theta^2 + 1}##. If ##\theta = 0##, then ##r = \pm 1##, so you would get points on the horizontal axis 1 unit to the right and 1 unit to the left. Also, if ##\theta = \pi/2##, the values of r would be ##\pm\frac 1 {\sqrt{(\pi/2)^2 + 1}}##, and not ##\pm 1## as you show on your graph.
That's not the way that polar graphs work. Angles are measured in the CCW direction from the positive x axis (the ray ##\theta = 0##), and r is measured out along the ray defined by the angle.
Mark, forgot your Latex after 16 years in PF???
 
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