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Homework Help: 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


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    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


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    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


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    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


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    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


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    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?
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