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Math Help - relations

  1. #1
    Newbie mathproblem's Avatar
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    Angry relations

    without using a graphing calculator
    ---------------------------------
    How would you find the coordinates of all points(x,y) of the tangent lines of a relation such as (x^2)*y - y^3 = 8 (btw this graph looks like 3 parabolas oppening, one to the right, left and one downwards) where the tangent line is horzontal ?

    Would using implicit differentiation work getting y'=(-2xy)/((x^2) -3y^2) then equating it to zero(slope) ?
    --------------------------------------------------------------------
    Consider x^3 + y^3 - 9xy = 0 , dy/dx=y'=(9y -3x^2) / (3y^2 - 9x)

    how can you isolate the y on one side in the form y = x ?

    without using a graphing calculator how would you go about finding the points where the tangent lines to the graph of x^3 + y^3 - 9xy = 0 are horizontal ?

    how would you find all the coordinates when y'= (9y -3x^2) / (3y^2 - 9x)= 0 like solving for them algebraicaly
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  2. #2
    MHF Contributor
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    Quote Originally Posted by mathproblem
    without using a graphing calculator
    ---------------------------------
    How would you find the coordinates of all points(x,y) of the tangent lines of a relation such as (x^2)*y - y^3 = 8 (btw this graph looks like 3 parabolas oppening, one to the right, left and one downwards) where the tangent line is horzontal ?

    Would using implicit differentiation work getting y'=(-2xy)/((x^2) -3y^2) then equating it to zero(slope) ?
    --------------------------------------------------------------------
    Consider x^3 + y^3 - 9xy = 0 ,

    how can you isolate the y on one side in the form y = x ?

    without using a graphing calculator how would you go about finding the points where the tangent lines to the graph of x^3 + y^3 - 9xy = 0 are horizontal ?

    how would you find all the coordinates when y'= (9y -3x^2) / (3y^2 - 9x)= 0 like solving for them algebraicaly
    I like your avatar (Pythagorean theorem, eh?) so let me give one way to answer your questions.

    If the tangent line is horizontal, that means its slope is zero, and y' is zero.

    (x^2)*y -y^3 = 8 --------(i)
    You diffferentiated it implicitly with respect to x, and you arrived at (which is correct):
    y' = (-2xy)/((x^2) -3y^2)
    Which is the same as,
    y' = (2xy)/(3y^2 -x^2)
    Set y' to zero,
    0 = (2xy)/(3y^2 -x^2)
    Multiply both sides by (3y^2 -x^2) /2,
    0 = xy
    Which means x=0, or y=0, or both x and y are zeros.

    When x=0,
    (x^2)*y -y^3 = 8 --------(i)
    0 -y^3 = 8
    y^3 =-8
    y = -2
    So, (0,-2) is a point of tangency for a horizontal tangent line.

    When y=0,
    (x^2)*y -y^3 = 8 --------(i)
    0 -0 = 8
    No way.

    I found only one such point of tangency then.

    --------------------------
    x^3 + y^3 - 9xy = 0 -------(ii)
    You got,
    dy/dx=y'=(9y -3x^2) / (3y^2 - 9x) ----which is correct also.
    That will simplify to:
    y' = (3y -x^2) /(y^2 -3x)

    For horizontal tangent lines,
    y' = 0
    0 = (3y -x^2) /(y^2 -3x)
    0 = 3y -x^2
    x^2 = 3y
    y = (x^2) /3 -----***
    Substitute that into (ii),
    x^3 +[(x^2) /3]^3 - 9x[(x^2) /3] = 0
    x^3 +(x^6)/27 - 3x^3 = 0
    (x^6)/27 -2x^3 = 0
    Clear the fraction, multiplyboth sides by 27,
    x^6 -54x^3 = 0
    (x^3)(x^3 -54) = 0

    x^3 = 0
    x = 0 -------***

    x^3 -54 = 0
    x = cuberoot(54) = 3cbrt(2) ------***

    When x=0,
    x^3 + y^3 - 9xy = 0 -------(ii)
    0 +y^3 -0 = 0
    y = 0
    So, the origin (0,0) is a point of tangency.

    When x = cbrt(54),
    x^3 + y^3 - 9xy = 0 -------(ii)
    54 +y^3 - 9y[cbrt(54)] = 0
    y^3 -(34.01786835)y +54 = 0 ------(iii)
    Zeez, I will leave that to you to get the 3 roots of (iii). It's late in the evening here now.
    [By "rule of signs" (?), there are 2 positive roots and one negative root.]

    Therefore, there should be 4 points of tangency for horizontal tangents.

    --------
    Believe it or not, I don't know how to use any graphing calculator, any Internet "grapher", any graphing application like MathCad or what you call them.
    But I can sketch any graph the old-fashion way---some points, critical points, asymptotes, known characteristics, etc.

    So, since I did not sketch the graphs of your two curves, I cannot say 100% sure that the points of tangency are really points of tangency. Algebraically, or if y'=0, they are correct, though.
    Last edited by ticbol; December 13th 2005 at 06:32 AM.
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by mathproblem
    without using a graphing calculator
    ---------------------------------
    How would you find the coordinates of all points(x,y) of the tangent lines of a relation such as (x^2)*y - y^3 = 8 (btw this graph looks like 3 parabolas oppening, one to the right, left and one downwards) where the tangent line is horzontal ?

    Would using implicit differentiation work getting y'=(-2xy)/((x^2) -3y^2) then equating it to zero(slope) ?
    --------------------------------------------------------------------
    Consider x^3 + y^3 - 9xy = 0 , dy/dx=y'=(9y -3x^2) / (3y^2 - 9x)

    how can you isolate the y on one side in the form y = x ?

    without using a graphing calculator how would you go about finding the points where the tangent lines to the graph of x^3 + y^3 - 9xy = 0 are horizontal ?

    how would you find all the coordinates when y'= (9y -3x^2) / (3y^2 - 9x)= 0 like solving for them algebraicaly
    Equation of curve is:

    x^3 + y^3 - 9xy = 0 .

    Assuming what you have done is correct, you are interested in points where:

    \frac{9y-3x^2}{3 y^2-9x}=0

    which implies:

    9y-3x^2=0, or x^2=3y.

    Substitute this back into the original equation (Opps substituted back
    into wrong equation corrected now):

    x^6-54x^3=0,

    or:

    x^3(x^3-54)=0

    The values of x you require are the real roots of the 6-tic,
    and you can find the values of y by substituting these real
    x's back into the original equation and then solving for y.


    RonL
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  4. #4
    Newbie mathproblem's Avatar
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    Quote Originally Posted by ticbol
    Multiply both sides by (3y^2 -x^2) /2,
    0 = xy
    Which means x=0, or y=0, or both x and y are zeros.

    When y=0,
    (x^2)*y -y^3 = 8 --------(i)
    0 -0 = 8
    No way.

    I found only one such point of tangency then.
    Is y=0 then considered to be an extraneous solution of (i) ?
    Last edited by mathproblem; December 13th 2005 at 01:31 PM.
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  5. #5
    MHF Contributor
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    Quote Originally Posted by mathproblem
    Is y=0 then considered to be an extraneous solution of (i) ?
    No.
    y=0 is not a solution of (i).

    >>>>>Originally Posted by ticbol
    Multiply both sides by (3y^2 -x^2) /2,
    0 = xy
    Which means x=0, or y=0, or both x and y are zeros.

    When y=0,
    (x^2)*y -y^3 = 8 --------(i)
    0 -0 = 8
    No way. <<<<<

    As you can see, if y=0, then (i) is going to the moon. (To hide, I believe. Because if y=0, then (i) does not make sense. So if y=0, then (i) is embarrassed to appear here, I think. Maybe on the moon, 0 minus 0 = 8.)

    y=0, as shown before, is a possible solution to the first derivative of (i) with respect to x, if this first derivative is equated to 0.
    Or, y=0 is a possible solution to y'=0-------to the dy/dx of (i), if this dy/dx is zero.
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