anonymous
  • anonymous
Match the slope field with the differential equation.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
https://i.gyazo.com/27a0d30b69bc4a86b3419438b826eac5.png
anonymous
  • anonymous
@jim_thompson5910 Hook a brotha up with an explaination?
baru
  • baru
mind if i try :p ?

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jim_thompson5910
  • jim_thompson5910
go ahead @baru
anonymous
  • anonymous
@baru Please do ;)
baru
  • baru
look at graph C, it is clearly the slope field of a circle \[x^2+y^2=r^2\] differentiate that
anonymous
  • anonymous
2x+2y=2r
baru
  • baru
'r' is a constant so RHS=0
anonymous
  • anonymous
RHS?
baru
  • baru
right hand side
baru
  • baru
2x+2y=0
baru
  • baru
sorry 2x+2y\(\frac{dy}{dx}=0\)
anonymous
  • anonymous
Ohh because we are differentiating by x
anonymous
  • anonymous
So what's next? Separation of variables?
baru
  • baru
just re-arrange to get \(\frac{dy}{dx}=?\) and compare with the options
baru
  • baru
@Ephemera
anonymous
  • anonymous
It would be -x/y
anonymous
  • anonymous
So that means it would be C for the first one, correct? Due to the circlish look
baru
  • baru
yep, now we have eliminated graph C and first option
baru
  • baru
now look at graph B, the slope field cutting the y axis is perfectly horizontal, or in other words, dy/dx=0 at x=0 which option fits this info?
anonymous
  • anonymous
dy/dx=x/y?
baru
  • baru
:) yes it cant be any of the other options, because they have x in the denominator, and division by zero is not defined (the first option would have fit, but we already eliminated that)
baru
  • baru
followed?
anonymous
  • anonymous
Yeah
anonymous
  • anonymous
I went ahead and attempted the other two
anonymous
  • anonymous
Got C D B A in that order
baru
  • baru
what logic did you use to solve the remaining?
anonymous
  • anonymous
For A I got dy/dx=y/x
anonymous
  • anonymous
Where at (0,0) the slope is positive
baru
  • baru
:) nice that looks right
anonymous
  • anonymous
Thanks for the help.
baru
  • baru
woops, you have got the right answer, but \(\frac{0}{0}\) is not defined so we cannot eliminate based on that
jim_thompson5910
  • jim_thompson5910
this is what I got
1 Attachment
baru
  • baru
@jim_thompson5910 i agree @Ephemera graph D looks like the upward hyperbola \[\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\] where and b are constants
baru
  • baru
*a and b
baru
  • baru
use the same procedure as for the circle and find dy/dx
anonymous
  • anonymous
Oh ok....I should've chosen a different point to test as well should've known it was undetermined which would mislead to wrong answer. Thanks for the help.
baru
  • baru
@jim_thompson5910 you have a better method? rather than guessing the graph?
jim_thompson5910
  • jim_thompson5910
why not solve for y and see what forms you get for example... \[\Large \frac{dy}{dx} = -\frac{x}{y}\] \[\Large y dy = -x dx\] \[\Large \int y dy = \int -x dx\] \[\Large \frac{y^2}{2} = -\frac{x^2}{2}+C\] \[\Large y^2 = -x^2+2C\] \[\Large y^2 = -x^2+C\] \[\Large x^2 + y^2 = C\] Equation for a circle. So \(\Large \frac{dy}{dx} = -\frac{x}{y}\) matches with C. The slope field is basically a group of concentric circles
jim_thompson5910
  • jim_thompson5910
slopefield D is actually a bunch of lines (not hyperbolas)
jim_thompson5910
  • jim_thompson5910
B is the hyperbolic one
baru
  • baru
hmm...yep, y'=y/x is diff eq for the special case hyperbola of a=b,not any hyperbola, my mistake :P
baru
  • baru
\(y^2-x^2=a^2\\2yy'-2x=0\\y'=x/y\) hmmm, i'm confused, straight lines and the spcl case hyperbola fit, any clue in the graph to suggest one over the other?
baru
  • baru
nevermind xD i see my error

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