4U Graphing question (2 Viewers)

[bell531]

OK, so this was one of the questions from my 4U half yearly:




Sketch x^3 + y^3 - 3xy = 0





I had no idea what it was, and took a random guess at it. What does this look like? and, how is this meant to be done?

 

[MC Squidge]

ue implicit differentiation

 

[fuckit1991]

yes implictily diffentiate...graph is attached

 

[Trebla]

yes implictily diffentiate...graph is attached

What graphing program did that picture come from?

 

[azureus88]

yes implictily diffentiate...graph is attached

you sure you can accurately sketch it using cartesian form? How do find asymptotes?

 

[jet]

What graphing program did that picture come from?

Looks like the mac utility 'Grapher'. Its the one I use.

 

[bell531]

yes implictily diffentiate...

I did so, and got:





dy/dx = (y - x^2) / (y^2 - x)



But then what do I do with this result?

 

[alakazimmy]

I did so, and got:





dy/dx = (y - x^2) / (y^2 - x)



But then what do I do with this result?

You find you horizontal tangents and your vertical tangents. Then you mark those points in.
Then, find your x and y intercepts, which should just be (0,0)

Then, you find out what happens when x -> negative infinite.
And the same for when x -> positive infinite.

once you have these points, essentially, you'll be connecting the dots up.

 

[Iruka]

You find out where the tangent is vertical or horizontal.

Edit: got beaten to it. You could also note that the function is symmetrical in x and y, which means that the line y=x is an axis of symmetry

 

[azureus88]

how would you find out what happens when x goes to infinity?

 

[fuckit1991]

What graphing program did that picture come from?

yes it was grapher on the mac...excellent program

 

[iSplicer]

yes it was grapher on the mac...excellent program

Could you please recommend a good one for Vista

 

[fuckit1991]

Could you please recommend a good one for Vista

to be honest i haven't seen anything equivalent to grapher on windows. on my xp laptop i use graphmatica or graphfunc but i am not sure if they are compatible with vista...although graphfunc is online and only needs java.

 

[alakazimmy]

how would you find out what happens when x goes to infinity?

Rearrange the equation so y's are all on one side.

x³ = -y³ + 3xy

Then divide everything by x³
So:
1 = (-y³ + 3xy)/x³
= -(y/x)³ + 3y/x²


Then take the limits of x going to infinity.
The 2nd term will become insignificant, so -(y/x)³ will tend to 1 as x tends to infinity.

Hence, y = -x

 

[Iruka]

^ Its not quite that simple. The asymptote is x+y+1=0.

Foliumd

If you use the parameterisation given on that webpage,

x = 3t/(1 + t<SUP>3</SUP>), y = 3t<SUP>2</SUP>/(1 + t<SUP>3</SUP>) ,

you have to analyze what happens as t goes to -1. You can see that dy/dx approaches -1, also, which shows that the asymptote has the form x+y=k. You then need to use the parameterised form of the coordinates and l'Hopital's rule to figure out what is going on. (That is, figure out the limit of x+y as t approaches -1).

To the OP, when you get your half yearly back, I would be interested to know how your teacher proposes that you analyze this problem using only 4 unit techniques.

 

[azureus88]

^ Its not quite that simple. The asymptote is x+y+1=0.

Foliumd

If you use the parameterisation given on that webpage,

x = 3t/(1 + t<SUP>3</SUP>), y = 3t<SUP>2</SUP>/(1 + t<SUP>3</SUP>) ,

you have to analyze what happens as t goes to -1. You can see that dy/dx approaches -1, also, which shows that the asymptote has the form x+y=k. You then need to use the parameterised form of the coordinates and l'Hopital's rule to figure out what is going on. (That is, figure out the limit of x+y as t approaches -1).

To the OP, when you get your half yearly back, I would be interested to know how your teacher proposes that you analyze this problem using only 4 unit techniques.

can't you just go y=tx from the parametric equation and as t approaches -1, asymptote is y=-x ?

 

[Iruka]

can't you just go y=tx from the parametric equation and as t approaches -1, asymptote is y=-x ?

Have a close look at the graph from the graphing utility. The asymptote is not y=-x. It is y=-x-1. As t approaches -1 both x and y go to infinity, but the question is, in what fashion.

 

[bell531]

^ Its not quite that simple. The asymptote is x+y+1=0.

Foliumd

If you use the parameterisation given on that webpage,

x = 3t/(1 + t<SUP>3</SUP>), y = 3t<SUP>2</SUP>/(1 + t<SUP>3</SUP>) ,

you have to analyze what happens as t goes to -1. You can see that dy/dx approaches -1, also, which shows that the asymptote has the form x+y=k. You then need to use the parameterised form of the coordinates and l'Hopital's rule to figure out what is going on. (That is, figure out the limit of x+y as t approaches -1).

To the OP, when you get your half yearly back, I would be interested to know how your teacher proposes that you analyze this problem using only 4 unit techniques.

Yeah, I'll post the solutions, but that should take a bit more than two weeks. It seems as though no one here is certain.

 

[hermand]

Yeah, I'll post the solutions, but that should take a bit more than two weeks. It seems as though no one here is certain.

hahahahaha bellman, you should have seen my graph, it was pretty cool. kind of just a squiggle. i think i tried to supress the memories of that exam. *shudders*.

 

[untouchablecuz]

dy/dx = (y - x^2) / (y^2 - x)
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for horizontal asymptotes we require points where dy/dx = 0

i.e. y - x^2 = 0 => y = x^2

sub this into the original equation of the proposed curve and find the points satisfying the equation
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for vertical asymptotes we require points where dy/dx is undefined; vetical tangent

i.e y^2 -x = 0 => x = y^2

sub this into the original equation of the proposed curve and find the points satisfying the equation
---------------------------------------------------------------------------------------
as a side note: we can see from the equation that it is symmetrical about the line y = x (subbing in y = x yields the same equation)
---------------------------------------------------------------------------------------
all you do now is "connect the dots" (mark points with horiz. and vert. tangents with horiz. or vert. lines) so that what you have found is displayed in the graph

edit: crap, reading the thread now, people have beat me to it