MHB Best Way to Graph in Polar Coordinates

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To graph the polar equation r = 3 - 5 cos θ, it is recommended to plot several points to accurately represent the curve, especially if it is not a standard form like a circle or ellipse. Using software like Maxima can help visualize the curve effectively. The importance of marking points is emphasized for non-standard curves to ensure accuracy. After plotting, comparing the results with a reference graph can validate the correctness of the plotted points. This method enhances understanding of polar coordinate graphs.
Amer
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what is the best way to graph in polar coordinates say

r = 3 - 5 \cos \theta

is it to plot several points then make a curve between them or ?
 
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Re: Best way to Graph is polar

Amer said:
what is the best way to graph in polar coordinates say

r = 3 - 5 \cos \theta

is it to plot several points then make a curve between them or ?

Hi Amer,

Unless the curve given is a standard curve (circle, ellipse etc) or can be transformed into a standard form, we have to plot it by marking points. I drew your curve using Maxima.

http://imageshack.us/f/337/figure.png/
 
Re: Best way to Graph is polar

Sudharaka said:
Hi Amer,

Unless the curve given is a standard curve (circle, ellipse etc) or can be transformed into a standard form, we have to plot it by marking points. I drew your curve using Maxima.

http://imageshack.us/f/337/figure.png/

thanks so i need to plot some points. ok
 
Re: Best way to Graph is polar

Amer said:
thanks so i need to plot some points. ok

Yes. Try plotting it and see if it really fits the curve I have attached. :)
 

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There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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