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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

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  1. freckles
    • one year ago
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    can you give an example of where you have trouble at

  2. freckles
    • one year ago
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    |dw:1437432457639:dw| arctan will output values for these quadrants where theta is between -pi/2 and pi/2

  3. freckles
    • one year ago
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    so to get the other quadrants we have to be a bit more clever but not too much more

  4. freckles
    • one year ago
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    |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

  5. freckles
    • one year ago
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    |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

  6. anonymous
    • one year ago
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  7. anonymous
    • one year ago
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    i understand everything except the pi

  8. freckles
    • one year ago
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    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: \[(x,y)=(1,-1) \\ r=\sqrt{2} \\ \theta=\arctan(-1)=\frac{-\pi}{4} \\ (r, \theta)=(\sqrt{2},\frac{-\pi}{4})\]

  9. freckles
    • one year ago
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    looking at your picture now

  10. freckles
    • one year ago
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    |dw:1437433116823:dw|

  11. freckles
    • one year ago
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    where R is |dw:1437433154036:dw|

  12. freckles
    • one year ago
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    so are you asking how they got the limits ?

  13. anonymous
    • one year ago
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    the limits of the angle yes

  14. freckles
    • one year ago
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    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 ?

  15. freckles
    • one year ago
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    |dw:1437433287671:dw|

  16. freckles
    • one year ago
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    |dw:1437433324784:dw|

  17. freckles
    • one year ago
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    |dw:1437433373914:dw|

  18. freckles
    • one year ago
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    that is what they have in your picture the top part of the circle

  19. freckles
    • one year ago
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    that is why they have theta is between 0 and pi

  20. freckles
    • one year ago
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    |dw:1437433484377:dw|

  21. freckles
    • one year ago
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    like does that answer your question?

  22. anonymous
    • one year ago
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    but how do you see that it is in the second quadrant? and equals to pi?

  23. freckles
    • one year ago
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    well I drew it above

  24. freckles
    • one year ago
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    your R region is in the first quadrant and in the second quadrant

  25. freckles
    • one year ago
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    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

  26. anonymous
    • one year ago
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    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!

  27. freckles
    • one year ago
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    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

  28. freckles
    • one year ago
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    Well not reword the question but give a slightly different question by rewording it

  29. freckles
    • one year ago
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    anyways I hope that did help good luck

  30. anonymous
    • one year ago
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    yes it's getting clearer and clearer now. thanks again!

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