Match the slope field with the differential equation.

- anonymous

Match the slope field with the differential equation.

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- anonymous

https://i.gyazo.com/27a0d30b69bc4a86b3419438b826eac5.png

- anonymous

@jim_thompson5910 Hook a brotha up with an explaination?

- baru

mind if i try :p ?

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## More answers

- jim_thompson5910

go ahead @baru

- anonymous

@baru Please do ;)

- baru

look at graph C, it is clearly the slope field of a circle
\[x^2+y^2=r^2\]
differentiate that

- anonymous

2x+2y=2r

- baru

'r' is a constant so RHS=0

- anonymous

RHS?

- baru

right hand side

- baru

2x+2y=0

- baru

sorry
2x+2y\(\frac{dy}{dx}=0\)

- anonymous

Ohh because we are differentiating by x

- anonymous

So what's next? Separation of variables?

- baru

just re-arrange to get
\(\frac{dy}{dx}=?\)
and compare with the options

- baru

@Ephemera

- anonymous

It would be -x/y

- anonymous

So that means it would be C for the first one, correct? Due to the circlish look

- baru

yep, now we have eliminated graph C and first option

- 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

dy/dx=x/y?

- 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

followed?

- anonymous

Yeah

- anonymous

I went ahead and attempted the other two

- anonymous

Got
C
D
B
A
in that order

- baru

what logic did you use to solve the remaining?

- anonymous

For A I got dy/dx=y/x

- anonymous

Where at (0,0) the slope is positive

- baru

:) nice
that looks right

- anonymous

Thanks for the help.

- baru

woops, you have got the right answer, but \(\frac{0}{0}\) is not defined
so we cannot eliminate based on that

- jim_thompson5910

this is what I got

##### 1 Attachment

- 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

*a and b

- baru

use the same procedure as for the circle and find dy/dx

- 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

@jim_thompson5910 you have a better method? rather than guessing the graph?

- 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

slopefield D is actually a bunch of lines (not hyperbolas)

- jim_thompson5910

B is the hyperbolic one

- baru

hmm...yep, y'=y/x is diff eq for the special case hyperbola of a=b,not any hyperbola, my mistake :P

- 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

nevermind xD i see my error

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