1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Folium of Descartes - very interesting

  1. Sep 24, 2007 #1
    This is my problem:

    1) Suppose that the function y = f(x) satisfies the equation x^3+y^3 = 3xy and is differentiable.
    Find a formula for the derivative y' showing it dependent on x and y.

    Well, I really don't know what to do but here goes....

    Is it y' = f'(x) = x^3+y^3 = 3xy

    Then i have to isolate y (for example y^3) and then derive?

    That gives me y = 3sqrt(3xy-x3) and I should derive that formula right ?

    If i'm right about this I'm still stuck in problem two which is:

    2= Show that for t is not −1 and t is not 0 the line y = tx intersects the folium of Descartes in exactly one point different from (0, 0), and find that point.

    I'm clueless now.

    Anyone ?
  2. jcsd
  3. Sep 24, 2007 #2
    well as i understood it you only need to take the derivataive wrt to x of the equation:
    x^3+y^3=3xy and then isolate the y' from the other terms.
  4. Sep 24, 2007 #3
    what does wrt to x mean? (English is not my first language)
  5. Sep 24, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    This makes no sensef'(x) is NOT equal to x^3+ y^3!!!

    You could but you don't have to. Use "implicit differentiation" instead.

    If x^3+ y^3= 3xy, you have d(x^3+ y^3)/dx= d(3xy)/dx. Since you do not know y as a function of x, use the chain rule: d(y^3)/dx= 3y^2 dy/dx.

    You have two simultaneous equations: x^3+ y^3= 3xy and y= tx. Putting the second equation into the first, x^3+ t^4x^3= 3t x^2. Obviously x= 0 satisfies that. What other solutions (x as a function of t) does that have?
  6. Sep 24, 2007 #5


    User Avatar
    Science Advisor
    Homework Helper

    with respect to
  7. Sep 24, 2007 #6
    OK! Thanks for the advice!
    I used "implicit differ" and found out that y' = -x^2/y^2 (The first time I use this method)

    But in the second problem I know that x=0 satisfies the equation. But there is on other number that also satisfies that equation, but I don't know how to solve the other point for x.
  8. Sep 24, 2007 #7


    User Avatar
    Homework Helper

    There is a problem here: that is not the correct result for y' (or dy/dx).

    The result y = 3sqrt(3xy-x3) is also not a correct derivation from the original expression for the folium, x^3+y^3 = 3xy . (This would have to involve *cube* roots and is a mess to rewrite as y = f(x). That is why we suggest implicit differentiation.)

    HallsofIvy suggested that you start with your original expression

    x^3+y^3 = 3xy

    and then take the derivative of each side with respect to x:

    d/dx [(x^3+ y^3)] = d/dx (3xy) .

    You will need to use the Chain Rule; since we don't really know how to write y as a function of x, we just differentiate it, knowing that it has a relation to x. HofI gave you one equation for dealing with the (y^3) term:

    d/dx (y^3) = (3y^2) dy/dx = 3(y^2)·y' .

    You will also need to use the Product Rule and Chain Rule to find d/dx (3xy). You will then have an equation with a (dy/dx) or y' factor on each side. You will need to rearrange the equation so that you can solve for y'. What do you find?

    By the way, it is a good idea to show your work, so the readers here can see how you are getting your results.

    HofI also suggested substituting y with tx (the equation given in the problem) in your equation for the folium, giving

    x^3 + (tx)^3 = 3·x·(tx) or

    (x^3) + (t^3)(x^3) = 3t·(x^2)
    [typo in earlier post corrected]

    Since we already found x = 0 as a solution, we now want to look at the solution for x when x is *not* zero. How does your result explain why the problem said we cannot have t = 0 or t = -1 ?
  9. Sep 24, 2007 #8
    Ok, before I type this I must say big thanks for taking your time helping me! It is very helpful.

    I've got from part 1:

    d/dx(x^3+y^3) = d/dx(3xy) (I will say from now d/dx = y')

    Used the product rule on the right side and got (3x)'(y)+(3x)(y)' = 3y + 3xy'

    From the left side I got 3x^2+3y^2y'

    That does 3x^2+3y^2y' = 3y + 3xy'

    I'm not good at this but here goes....

    I take the 3xy' from right to left and combine that with 3y^2y':

    (3y^2+3x)y' = 3y-3x^2

    and then isolate y':

    y' = (3y-3x^2)/(3y^2+3x)

    take the 3's and the ^ thingies (don't know the english word for ^)

    y' = (1-x)/(y-1)

    Is this right ????

    And from the part 2

    Since we already found x = 0 as a solution, we now want to look at the solution for x when x is *not* zero. How does your result explain why the problem said we cannot have t = 0 or t = -1 ?

    Well if t = 0 we have x^3 = 0 and since x has to be something else but zero this doesn't add up.

    If t = -1 we have x^3 - x^3 = -3(x^2)

    Obviously doesn't fit for x *not* zero.

    So t can be every number but -1 and 0 but what about x?
    Last edited: Sep 24, 2007
  10. Sep 24, 2007 #9


    User Avatar
    Homework Helper

    You're OK to here.

    Watch your signs: that should be (3y^2 - 3x)y' , so you'll have

    y' = (3y - 3x^2)/(3y^2 - 3x)

    The " ^ " is called a caret or a circumflex. In mathematical use, we'd read that as "to the power of"; because of that, some people call this the "uppen" symbol... ^_^

    You can cancel out the 3's, but the powers of x and y can't just be removed. Your result will be

    y' = [ y - (x^2) ]/[ (y^2) - x ] ;

    that's as simple as it's going to get. But it also answers the question as to why dy/dx depends on both x and y.

    You still need to solve this:

    (x^3) + (t^3)(x^3) = 3t·(x^2)

    for x (as a function of t). If x is not zero, what does this reduce to?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Folium of Descartes - very interesting
  1. Folium of Descartes (Replies: 1)