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anonymous
 one year ago
hello
i wonder how you can decide the angle when changing to polar coordinates?
i know this:
x=rcos(delta)
y=rsin(delta)
delta= arctan (x/y)
but i find it very difficult to know the value of delta
anonymous
 one year ago
hello i wonder how you can decide the angle when changing to polar coordinates? i know this: x=rcos(delta) y=rsin(delta) delta= arctan (x/y) but i find it very difficult to know the value of delta

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freckles
 one year ago
Best ResponseYou've already chosen the best response.1can you give an example of where you have trouble at

freckles
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437432457639:dw arctan will output values for these quadrants where theta is between pi/2 and pi/2

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so to get the other quadrants we have to be a bit more clever but not too much more

freckles
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437432533105:dw that was a straight line so if you really meant to have the angle there you can adjust your theta by adding pi to it

freckles
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437432587064:dw same thing for that 3rd quadrant you will find the angle in the 1st and you add pi to get the one in the 3rd if that is the one you are meant to have

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i understand everything except the pi

freckles
 one year ago
Best ResponseYou've already chosen the best response.1Examples: EX 1: \[(x,y)=(1,1) \\ r=\sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2} \\ \theta=\arctan(\frac{y}{x})=\arctan(\frac{1}{1})=\arctan(1)=\frac{\pi}{4} \\ (r,\theta)=(\sqrt{2},\frac{\pi}{4}) \\ \] EX 2: \[(x,y)=(1,1) \\ r=\sqrt{(1)^2+1^2}=\sqrt{1^2+1^2}=\sqrt{1+1}=\sqrt{2} \\ \theta=\arctan(\frac{y}{x})=\arctan(\frac{1}{1})=\arctan(1) \\ \text{ this will give the angle in the 4th quadrant } \\ \text{ you want the one in the 2nd } \\ \text{ do } + \pi \\ \theta=\arctan(\frac{y}{x})=\arctan(1)+\pi=\frac{\pi}{4}+\pi=\frac{3\pi}{4} \\ (r, \theta)=(\sqrt{2}, \frac{3\pi}{4})\] EX 3: \[(x,y)=(1,1) \\ r=\sqrt{2} \\ \theta=\arctan(1) \text{ this will give you the one in the first quadrant } \\ \text{ we want the one in the 3rd } \\ \text{ do } +\pi \\ \theta=\arctan(1)+\pi =\frac{\pi}{4}+\pi=\frac{5\pi}{4} \\ (r,\theta)=(\sqrt{2},\frac{5\pi}{4}) \] EX 4: \[(x,y)=(1,1) \\ r=\sqrt{2} \\ \theta=\arctan(1)=\frac{\pi}{4} \\ (r, \theta)=(\sqrt{2},\frac{\pi}{4})\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1looking at your picture now

freckles
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437433116823:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.1where R is dw:1437433154036:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.1so are you asking how they got the limits ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the limits of the angle yes

freckles
 one year ago
Best ResponseYou've already chosen the best response.1do you know that looking at a quarter of a circle we would be looking from theta=0 to theta=pi/2 assuming we want the part in the 1st quadrant ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437433287671:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437433324784:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437433373914:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.1that is what they have in your picture the top part of the circle

freckles
 one year ago
Best ResponseYou've already chosen the best response.1that is why they have theta is between 0 and pi

freckles
 one year ago
Best ResponseYou've already chosen the best response.1dw:1437433484377:dw

freckles
 one year ago
Best ResponseYou've already chosen the best response.1like does that answer your question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but how do you see that it is in the second quadrant? and equals to pi?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1your R region is in the first quadrant and in the second quadrant

freckles
 one year ago
Best ResponseYou've already chosen the best response.1they gave us that we are suppose to look at these half top parts of x^2+y^2=1 and x^2+y^2=4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay i see, it's quite confusing in my opinion . I'm going to try some more examples. you are very polite for putting your time in my question, i really appreciate that!

freckles
 one year ago
Best ResponseYou've already chosen the best response.1I could reword the question Say it say to evaluate that same thing except R is the region in the first quadrant bounded by the circles x^2+y^2=1 and x^2+y^2=4 well here theta would be between 0 and pi/2

freckles
 one year ago
Best ResponseYou've already chosen the best response.1Well not reword the question but give a slightly different question by rewording it

freckles
 one year ago
Best ResponseYou've already chosen the best response.1anyways I hope that did help good luck

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes it's getting clearer and clearer now. thanks again!
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