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!

Tangent lines to Parametric Equation

  1. Mar 15, 2010 #1
    HALP! I have already killed a forest trying to work this one out on paper.

    1. The problem statement, all variables and given/known data

    a. Use Implicit differentiation to find the equations of the horizontal tangent lines to the parametric equation: yx^3-2x^2y^2+xy^3=192.

    b. Use Implicit differentiation to find the equations of the horizontal tangent lines to the parametric equation: yx^3-2x^2y^2+xy^3=192.

    2. Relevant equations

    Differentiation and I used the open-source program Graph and believe that the equations of the horizontal tangent lines should be y=6 and y=-6. This is just a good approximation.

    3. The attempt at a solution

    Ok, since my work spanned about 5 pages of paper in my best attempt, I will merely list the sequence of steps and my answer for each sequence.

    I used implicit differentiation to determine:
    dy/dx = (4xy^2-3x^2y+y^3)/(x^3-4x^2y+3xy^2)

    I set the derivative equal to zero since that is the slope that my 2 tangent lines will have.

    (4xy^2-3x^2y+y^3)/(x^3-4x^2y+3xy^2) = 0

    However, I believe I only need to set the numerator equal to 0, so:

    4xy^2-3x^2y+y^3 = 0

    I think this needs to be solved using some clever substitution (like the last problem I posted) and so I tried to eliminate all y-variables by expressing them in terms of x, so I can then solve for x.

    From here I determine y^3=3x^2y-4xy^2
    and y^2=3x^2-4xy

    I used these values to substitute into the original equation and start over by taking the implicit derivative of the original equation, but now with my substitutions in place. So, now for dy/dx, I get:

    dy/dx=(72x-2xy-99y)/(28x)

    Again I set this equal to 0 (the slope of my tangent lines).

    (72x-2xy-99y)/(28x)=0

    Solving for y:

    y=(72x)/(2x+99)

    I now use this value of y to substitute into the original equation along with my values of y^3 and y^2. I attempt to solve for x and get to here:

    (2016x^4)/(2x+99)-18x^4=192

    I am stuck with the algebra here and I don't have any confidence in the answer this provides. Do I need to do the differentiation again instead of just solving for x?

    I did try to differentiate and got: (1287/4)=x^2+57x
    But this didn't seem to help either....

    I feel like I am on the right track and probably just screwed up my arithmetic. Is my approach the right one? If I screwed up the math somewhere, I would appreciate it if someone could just point it out. If I am correct in my approach, when I go to solve the equations of the vertical tangent lines, I just need to set the denominator to 0 and do it that way, right?
     
  2. jcsd
  3. Mar 15, 2010 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    That doesn't look quite right....you'd better post your calculation fr this step.
     
  4. Mar 15, 2010 #3
    Well, If my first derivative is wrong, then all subsequent work will be wrong. So starting from scratch, here is my work for the derivative of the original parametric equation y(x^3)-2(x^2)(y^2)+x(y^3)=192:

    dy/dx = y'(x)
    dy/dx [y(x^3)-2(x^2)(y^2)+x(y^3)=192]
    d/dx y(x^3) - 2 [d/dx (x^2)(y^2)] + d/dx x(y^3) = d/dx 192
    3(x^2)y+(x^3)y'(x)-2[2(x^2)(y)y'(x)+2x(y^2)]+3x(y^2)y'(x)+y^3=0
    3(x^2)y+(x^3)y'(x)-4(x^2)(y)y'(x)-4x(y^2)+3x(y^2)y'(x)+y^3=0
    (x^3)y'(x)-4(x^2)(y)y'(x)+3x(y^2)y'(x)=4x(y^2)-3(x^2)y-y^3
    y'(x)[(x^3)-4(x^2)(y)+3x(y^2)]=4x(y^2)-3(x^2)y-y^3
    y'(x)=[4x(y^2)-3(x^2)y-y^3]/[(x^3)-4(x^2)(y)+3x(y^2)]

    Here is my result for the implicit differentiation of the parametric equation. I have tried simplifying with no success. What am I overlooking?

    Thanks for the help and sorry that my notation changes midway in the work....I usually do all work in Leibniz notation, but figured prime notation would be easier to show here.
     
  5. Mar 15, 2010 #4

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    That's better. Your original post had +y^3 instead of -y^3.

    Now you can simplify this...start by factoring the numerator and denominator separately. For example, the numerator factors as

    [tex]4xy^2-3x^2y-y^3=-y(y^2-4xy+3x^2)=-y(y-x)(y-3x)[/tex]
     
  6. Mar 15, 2010 #5
    Yeah, I caught that mistake as I went back through my work...I idn't even realize that I had done it wrong so many times until I did it right 2X! There was only one guy in my Calculus class that was able to get this so he helped me out. I will post the solution once I have it typed up. The ace up your sleeve to solve this is "completing the square".
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook