anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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freckles
  • freckles
can you give an example of where you have trouble at
freckles
  • freckles
|dw:1437432457639:dw| arctan will output values for these quadrants where theta is between -pi/2 and pi/2
freckles
  • freckles
so to get the other quadrants we have to be a bit more clever but not too much more

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freckles
  • freckles
|dw: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
  • freckles
|dw: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
  • anonymous
anonymous
  • anonymous
i understand everything except the pi
freckles
  • freckles
Examples: 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
  • freckles
looking at your picture now
freckles
  • freckles
|dw:1437433116823:dw|
freckles
  • freckles
where R is |dw:1437433154036:dw|
freckles
  • freckles
so are you asking how they got the limits ?
anonymous
  • anonymous
the limits of the angle yes
freckles
  • freckles
do 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
  • freckles
|dw:1437433287671:dw|
freckles
  • freckles
|dw:1437433324784:dw|
freckles
  • freckles
|dw:1437433373914:dw|
freckles
  • freckles
that is what they have in your picture the top part of the circle
freckles
  • freckles
that is why they have theta is between 0 and pi
freckles
  • freckles
|dw:1437433484377:dw|
freckles
  • freckles
like does that answer your question?
anonymous
  • anonymous
but how do you see that it is in the second quadrant? and equals to pi?
freckles
  • freckles
well I drew it above
freckles
  • freckles
your R region is in the first quadrant and in the second quadrant
freckles
  • freckles
they 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
  • anonymous
okay 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
  • freckles
I 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
  • freckles
Well not reword the question but give a slightly different question by rewording it
freckles
  • freckles
anyways I hope that did help good luck
anonymous
  • anonymous
yes it's getting clearer and clearer now. thanks again!

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