Is the graph of x3+3xy+y3=1 a line or a curve?
Using Autograph, plot the graph x3  3 xy  y 3  1
4
y
2
x
–4
–2
2
4
–2
This gives the graph shown above. It appears to be the line x  y  1 .
Is this a mistake?
One obvious problem
It is clear that the point  1, 1 satisfies x3  3 xy  y 3  1 but the line shown
goes nowhere near that point! Maybe this is an isolated point, a bit like if you
2
2
think about  x  1   y  1  0 ; that has only one point, namely  1, 1 .
Question 1
Substitute y  1  x in x3  3 xy  y 3  1 . What does this show?
Question 2
Can you find any other points besides  1, 1 that satisfy x3  3xy  y 3  1 but
not x  y  1 ?
© MEI 2007
An algebraic approach
Look for the factor x  y  1 in the left hand side of x3  3xy  y 3  1  0 :
0  x 3  3 xy  y 3  1
0   x  y  1  x 2  yx  y 2  x  y  1
So far, this seems to give two curves, one corresponding to x  y  1  0 and
the other corresponding to x 2  yx  y 2  x  y  1  0 . The trouble is that second
bit doesn’t appear to show up on the graph. Perhaps no values of x and y
satisfy x 2  yx  y 2  x  y  1  0 ?
We’ll try to prove this by completing the square.
2

3
 y 1  3 2 3
0   x  y  1  x 2   y  1 x  
 y  y 


2
4 
 2  4

2

y 1  3
2
0   x  y  1   x 
  y  1 



2  4


y 1  3
2

But  x 
   y  1  0 ,since we’re adding two squares, for all values of
2  4

x and y , taking the value 0 only when x  y  1 .
2
It follows that the graphs of x3  3xy  y 3  1 and x  y  1 are the same except
for the inclusion of the point  1, 1 in the former.
y 1  3
2

If you replace  x 
   y  1 with any other expression involving
2  4

squares added together then your curve wil just be the straight line x  y  1 .
2
However, if you went to the trouble of multiplying out the brackets you
probably wouldn’t end up with anything as concise as x3  3 xy  y 3  1 .

For example,  2 x  3 y  5   3 x  y    y  3
2
2
  0 gives the straight line
2 x  3 y  5 along with the isolated point 1,3 .
Question 3
Use Autograph to investigate x k  kxy  y k  1 for other values of k .
Can you explain why you get two parallel lines in the case k  2 ?
© MEI 2007
A calculus approach
If the graph of x3  3xy  y 3  1 is a straight line, then differentiating
dy
x3  3xy  y 3  1 with respect to x should lead to
 1 .
dx
This requires the technique of implicit differentiation:
x3  3 xy  y 3  1  3 x 2  3 x
dy
dy
 3y  3y2
0 
dx
dx
When x  1 and y  1 we have the unhelpful result
dy
x2  y

x  y2
dx
dy 0
 . There’s
dx 0
something very strange about that point!
What about the other points on the ‘curve’?
For example, at  0,1 ,
22   1
dy
dy
02  1
2,
1
,
and
at



 1 ,



1


2
dx
0  12
dx
2   1
and so on.
dy
 1 then we would know that the curve is a straight
dx
line. (How would any isolated points such as  1, 1 affect this?)
If we could show that
x2  y
dy
 1 and the only fact
 1 , we need to show that 
x  y2
dx
we have at our disposal is x3  3xy  y 3  1 . I can’t see how to do this; please
let me know if you can.
To show that
d2 y
d2 y

everywhere.
0
 0 implies
dx 2
dx 2
that the gradient never changes, and since we know that the gradient is -1 at,
for example,  2, 1 , and the curve passes through  0,1 , then the curve must
An alternative approach is to show that
be the line x  y  1 .
Question 4
By differentiating 3x 2  3x
x  y
dy
 dy 
 y   
dx
2
 dx 
3
3
x  3xy  y  1 .
2
x
2
dy
dy
 3 y  3 y2
 0 with respect to x , show that
dx
dx
d
2
y
dx 2
and that the left hand side is 0 for the curve
© MEI 2007
In 3 dimensions
There are lots of ways of starting with x3  3xy  y 3  1 and introducing z s in
order to make an equation in three dimensions. One natural way is
x3  3xyz  y 3  z 3 . This has the benefit that every term contains 3 variables
multiplied together so is ‘homogeneous’.
Surprisingly (?) Autograph gives a plane!
If you sliced through this plane parallel to the x  y plane cutting the z  axis at
z  1 , then you would get the graph of x  y  1 shown on the first page above.
How about looking at
x3  3xyz  y 3  1
instead of
x3  3xyz  y 3  z 3 ?
As you can see, this
gives something more
interesting and well
worth exploring.
© MEI 2007
Answers
Question 1
Substitute y  1  x in x3  3 xy  y 3  1 . What does this show?
x3  3x 1  x   1  x   x3  3x  3x 2  1  3 x  3 x 2  x3  1 . This shows that
3
every point on the line y  1  x satisfies x3  3xy  y 3  1 but not vice versa; i.e.
there may be other points off the line satisfying x3  3xy  y 3  1 .
 x  y
Similarly, x  y  1 
3
 1  x3  3xy  x  y   y 3  1 and the final
equation is the same as x3  3 xy  y 3  1 if x  y  1 .
Question 2
Can you find any other points besides  1, 1 that satisfy x3  3xy  y 3  1 but
not x  y  1 ?
There are no other points as is shown in the section ‘An algebraic approach’.
Question 3
Use Autograph to investigate x k  kxy  y k  1 for other values of k .
Can you explain why you get two parallel lines in the case k  2 ?
x 2  2 xy  y 2  1 is the same as  x  y   1 and this gives the two lines
2
x  y  1 and x  y  1 .
For even values of k  2 the
graph looks something like this:
2
whereas for odd values of k  3
the graph looks something like this:
2
y
y
1
1
x
x
–1
1
–1
2
1
2
–1
–1
–2
–2
Both are symmetrical in the line y  x . This can be proved by interchanging x
and y in the equation; the equation remains unchanged.
There are lots of other aspects that can be explored. See the Chapter 7 of the
MEI FP2 textbook ‘Investigation of Curves’ for some ideas.
© MEI 2007
Question 4
By differentiating 3x 2  3x
x  y
dy
 dy 
x   y   
dx
2
 dx 
3
3
x  3xy  y  1 .
2
Differentiating 3x 2  3x
2
dy
dy
 3 y  3 y2
 0 with respect to x show that
dx
dx
d
2
y
dx 2
and that the left hand side is 0 for the curve
dy
dy
 3y  3 y2
0:
dx
dx
2
2
2
2


dy
d 2 y dy
dy
 dy  
 dy  
2 d y
2 d y








3 2x   x 2   y
2
y
0
2
x
y
x
y





  
  


dx
dx
dx
dx 2
dx
dx 2
 dx  
 dx  


Substitute

2
dy
x2  y
and multiply throughout by  x  y 2  :

2
dx
x y
2 x  x  y 2    x 2  y  x  y 2   y  x 2  y 
2
2

   x  y2 
d2 y
dx 2
3
x  y 
y  xy  
2 3
Expand the brackets and simplify: 3x y  xy  x
2
Factorise: xy  3 xy  y  x  1
3
It follows that
3
2
x  y 

2 3
2
4
d2 y
dx 2
4
2
 xy   0 
d2 y
dx 2
x  y 

2 3
2
d2 y
dx 2
d2 y
0.
dx 2
dy
x2  y

with respect to x :
Alternatively, you could differentiate
dx
x  y2
 x  y   2 x  ddyx    x
2
dy
x y

dx
x  y2
2
Substitute

2
d y

dx 2
2
x  y 
dy 

 y  1  2 y 
dx 

2 2
dy
x2  y
and simplify:

dx
x  y2
d2 y
6 x 2 y 2  2 xy  2 xy 4  2 x 4 y


3
dx 2
 x  y2 

2 xy
x  y 
2 3
 3xy  1  y
© MEI 2007
3
 x3   
2 xy
x  y 
2 3
0
You can use Autograph to plot
dy
x2  y

:
dx
x  y2
If you now click on any point of the grid you will be given the
graph of the function that passes through that point and satisfies
dy
x2  y
the differential equation

. For example:
dx
x  y2
© MEI 2007