Direction fields: Don't understand them.

[MitsuShai]

Question: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/q1.jpg
Graph1: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/q1-1.jpg
Graph2: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/q1-2.jpg
Graph3: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/q1-3.jpg
Graph4: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/q1-4.jpg


I know that the arrows tell you what function matches what graph, but I don't understand how. I'm trying to figure it out, but I completey don't understand it. How can you tell what graphs match the function just by looking at the function?

 

[tiny-tim]

Hi MitsuShai!

Easy … the derivative (dy/dx) is the slope of the arrow.

So the simplest way is to look at each equation to see where the slope is 0, ∞, and ±1.

 

[MitsuShai]

Hi MitsuShai!

Easy … the derivative (dy/dx) is the slope of the arrow.

So the simplest way is to look at each equation to see where the slope is 0, ∞, and ±1.

ok so like for the first question, you plug the number 0, infinity and 1, -1 into the function and so at zero the function should be at zero and when the function is going to infinity the arrows should be going to inifinity, and when the function is at 1, the arrows should be set at one, so the first function represents the first graph. Am I doing this correctly?

And when you plug those same numbers into the second function you will always get zero, and as the function goes to infinity, so should the arrows and this should be graph #2? Though the same pattern occurs for the third function...

 

[tiny-tim]

ok so like for the first question, you plug the number 0, infinity and 1, -1 into the function and so at zero the function should be at zero and when the function is going to infinity the arrows should be going to inifinity, and when the function is at 1, the arrows should be set at one, so the first function represents the first graph. Am I doing this correctly?

No.

I don't understand what you mean by, for example, "when the function is at 1".

You should give the values of x and y at which f is 0 ∞ or ±1, and check the direction of the arrows at those (x.y) points.

(and no, the first equation is not for the first graph)

 

[MitsuShai]

No.

I don't understand what you mean by, for example, "when the function is at 1".

You should give the values of x and y at which f is 0 ∞ or ±1, and check the direction of the arrows at those (x.y) points.

(and no, the first equation is not for the first graph)

so you're saying to integrate the function first to get it in terms of y then plug in 0, infinity or +/-1 for y to get x.

 

[tiny-tim]

No, no integration is involved.

Find the points (x,y) at which the given dy/dx is 0 ∞ or ±1.

 

[MitsuShai]

No, no integration is involved.

Find the points (x,y) at which the given dy/dx is 0 ∞ or ±1.

ok so for example the first one:
0=x^2/y^2
x=0
y=0

infinity= x^2/y^2
x=infinity
y=infinty

1=x^2/y^2
x=y

your looking for a graph that has a horizontal line on 0 and goes to infinity. Graph#4?

 

[tiny-tim]

ok so for example the first one:
0=x^2/y^2
x=0
y=0

infinity= x^2/y^2
x=infinity

No!

(For a start, 0/0 and ∞/∞ don't exist, so that's no information at all)

What is x²/y² along x = 0 (the y-axis)?

What is x²/y² along y = 0 (the x-axis)?

What is x²/y² along x = y?

What is x²/y² along x = -y?

 

[MitsuShai]

No!

(For a start, 0/0 and ∞/∞ don't exist, so that's no information at all)

What is x²/y² along x = 0 (the y-axis)?

What is x²/y² along y = 0 (the x-axis)?

What is x²/y² along x = y?

What is x²/y² along x = -y?

ok I think I got it now:
0=x^2/y^2
c=x^2/y^2
y^2c=x^2
y= sqrt(x^2/c)
c=0, y=0
c=1
y=x^2
c=-1
y=-x^2
c=infinity
y=0
like this? so it should be graph 1.

 

[tiny-tim]

Sorry, that makes no sense.

Get some sleep! :zzz:

(and in the morning, answer the questions I asked before:

What is x²/y² along x = 0 (the y-axis)?

What is x²/y² along y = 0 (the x-axis)?

What is x²/y² along x = y?

What is x²/y² along x = -y?)

 

[MitsuShai]

Sorry, that makes no sense.

Get some sleep! :zzz:

(and in the morning, answer the questions I asked before:

What is x²/y² along x = 0 (the y-axis)?

What is x²/y² along y = 0 (the x-axis)?

What is x²/y² along x = y?

What is x²/y² along x = -y?)

How can I answer those questions?
0^2/y^2?

 

[MitsuShai]

you have two unknown variables how is it possible to solve for them, with only one known variable?
dy/dx=0/y^2
y^2dy/dx=0

 

[MitsuShai]

1. (0,1)
dy/dx=x^2/y^2=0/1=0
when x=0, dy/dx=0
when x=1= dy/dx= 1/y^2 -asymptote on x-axis
when y=1: dy/dx= x^2
x=-1, dy/dx= -1/y^2
x=+/-infinity dy/dx=infinity
so this resembles graph 1?

3. 0=x-y
y=x, so there should be a horizonal line at 0,0
when x=0
dy/dx=0-y=-y, so as slope increases y decreases
y=0
dy/dx=x-0=x, so as slope increases x increases
slope: (0,1): dy/dx=0-1=-1

so question #3 is graph #4, right?


2. 0=x^2-y^2
y^2=x^2
slope: (0,1)- dy/dx=0-1=-1, horizonal line at (0,1)
x=0--> dy/dx=-y^2
y=0 dy/dx= x^2
x=1-->dy/dx= 1-y^2
y=1--->dy/dx=x^2-1
x=infinty dy/dx=infinty
y=infinty dy/dx= -infinity
so function 2 is graph 3


then function 4 is graph 2

 

[tiny-tim]

(just got up :zzz: …)

1. (0,1)
dy/dx=x^2/y^2=0/1=0
when x=0, dy/dx=0

Yes, so along the y-axis, the arrows must all be horizontal.

when x=1= dy/dx= 1/y^2 -asymptote on x-axis
when y=1: dy/dx= x^2
x=-1, dy/dx= -1/y^2
x=+/-infinity dy/dx=infinity
so this resembles graph 1?

The lines y = 1, x = -1, and the regions x = ±∞, aren't going to be helpful.

Just carry on with the lines I suggested …

y = 0, x = y, and x = -y.

(and no, it's not graph 1 anyway)

 

[MitsuShai]

(just got up :zzz: …)


Yes, so along the y-axis, the arrows must all be horizontal.


The lines y = 1, x = -1, and the regions x = ±∞, aren't going to be helpful.

Just carry on with the lines I suggested …

y = 0, x = y, and x = -y.

(and no, it's not graph 1 anyway)

x=2 dy/dx= 4/y^2
y=2= dy/dx= x^2/4
(1,0) dy/dx= 1/0=0 horizontal line
graph 2

 

[MitsuShai]

I was able to see a tutor today and got help on this. Anyways, thanks for helping me Tiny Tim :)