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anonymous

  • 5 years ago

Hey can someone please explain to me how to figure out how to match slope fields to their graphs without the use of a calculator? for problem such as these: y'=-y b) y'=x-y c) y'=1/y

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  1. watchmath
    • 5 years ago
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    It really depends on how the give choice are look like. We can always try to extract some special feature of each equatio. For example for \(y]=-y\) we know that when \(y=0\) we have \(y'=0\). So along the x-axis the slope fields are horizontal line segments. But it is possible that the other graph has this feature. Then for example you look at \(y=1\). Here \(y'=-1\). So along the line \(y=1\) the slopes are all -1 and so on...

  2. anonymous
    • 5 years ago
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    no my question is how do you test for conditional convergence and absolute convergence

  3. anonymous
    • 5 years ago
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    please ignore the previous comment . so we plug zero in for y' zero

  4. anonymous
    • 5 years ago
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    can you please explain how to do it for y'=1/y

  5. watchmath
    • 5 years ago
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    for y'=1/y for example we can see that for y=0 the slope is undefined. So along the x-axis there is no slope field

  6. anonymous
    • 5 years ago
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    so we plug zero in for y ?

  7. watchmath
    • 5 years ago
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    yes, but 1/0 is undefined.

  8. anonymous
    • 5 years ago
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    but what i dont understand is my teacher wrote on the board for this problem y=0 y>0 y'= undefined y'>0 do you know how she came up with the inequality part

  9. watchmath
    • 5 years ago
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    I think what your techer meant was y' is undefined for y=0. And y' >0 for y>0

  10. anonymous
    • 5 years ago
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    do you know how she came up with that

  11. watchmath
    • 5 years ago
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    well, if y>0, then 1/y >0. Since y'=1/y. Then y' >0 too.

  12. anonymous
    • 5 years ago
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    so you have to look at each quadrant on the graph to figure it out

  13. watchmath
    • 5 years ago
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    Since this is about matching as I said earlier it depends on the given choices. Maybe by just analyzing for y>0 it is enough to see the the behavior of the right graph. If it is not enough my by you need an analysis for y<0. But basically you just plug in some certain values of y (or x) so that you can see a property that the right graph has but the others are not.

  14. anonymous
    • 5 years ago
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    okay . how would you do it for y'=x

  15. anonymous
    • 5 years ago
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    would you say x=0 and y'=0

  16. watchmath
    • 5 years ago
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    yes, for x=0 we have y'=0. But the other slope field might be has this feature too. So you need to look at for something else in that case. Say plug in x=1 to get y'=1. If the other slope field also have this feature then you need to look at for something else... and so on....

  17. anonymous
    • 5 years ago
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    so on the y axis you have horizontal lines got through it

  18. watchmath
    • 5 years ago
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    correct!

  19. anonymous
    • 5 years ago
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    i wanted to check with you one more so if you y'=4-y . you plugin zero for y an you get y=0 , y'=-4

  20. watchmath
    • 5 years ago
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    yes, but maybe the slope of -4 is not very easy to see. So instead plug in y=4 to get y'=0.

  21. anonymous
    • 5 years ago
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    okay i see so you can plug in whatever you want

  22. watchmath
    • 5 years ago
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    correct! :)

  23. anonymous
    • 5 years ago
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    also, if you want to check based on quadrants so the for example you explained to me y'=y you could say when y>0 ,y'> 0

  24. watchmath
    • 5 years ago
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    yes, you can do that as well :).

  25. anonymous
    • 5 years ago
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    hey do you think i can ask you another question because you have been very helpful

  26. anonymous
    • 5 years ago
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    hey do you think i can ask you another question because you have been very helpful

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