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Two function questions

  • 23-05-2010 03:09PM
    #1
    Registered Users, Registered Users 2 Posts: 3,745 ✭✭✭


    Hey! I've two questions relating to functions; any help would be appreciated. :)

    Question 1 - Polar Coordinates

    Consider the curve [LATEX]r=1-\cos{\theta}[/LATEX]. What is the equation in rectilinear coordinates?

    So I think you must solve for x in terms of y. Two equations relating rectilinear to polar coordinates are:

    [LATEX]x=r\cos{\theta}[/LATEX] and [LATEX]y=r\sin{\theta}[/LATEX]

    I've been crunching this for a while but I can't eliminate both r and theta. Any advise? Or is my method wrong?

    Question 2 - Implicit differenciation

    I think I've this one solved, I'd just like to verify my method.

    Find [LATEX]\frac{dy}{dx}[/LATEX] by implicitly differentiating

    [LATEX]\displaystyle \cos{(x-y)}=\frac{x}{1+y^2}[/LATEX]

    So differenciating both sides. Chain rule on the left; quotient on the right.

    [LATEX]\displaystyle -\sin{(x-y)}\frac{d}{dx}(x-y)=\frac{(1)(1+y^2)-(x)(\frac{d}{dx}(1+y^2))}{(1+y^2)^2}[/LATEX]

    Aside,

    [LATEX]\displaystyle \frac{d}{dx}(x-y)=1-\frac{dy}{dx}[/LATEX]

    and

    [LATEX]\displaystyle \frac{d}{dx}(1+y^2)=0+2y^1(\frac{dy}{dx})[/LATEX]

    Sub back in and solve for dy/dx. Is this ok?

    Thanks again! :)


Comments

  • Registered Users, Registered Users 2 Posts: 1,595 ✭✭✭MathsManiac


    For the first one, try using r=sqrt(x^2+y^2) and theta = arctan(y/x).

    That should allow you to get yout equation in terms of x and y. Then see if it can be simplified at all.

    Second one looks ok to me.

    (First one is a cardioid, by the way:
    http://www.wolframalpha.com/input/?i=r%3D1-cos%28theta%29)


  • Registered Users, Registered Users 2 Posts: 3,745 ✭✭✭Eliot Rosewater


    Thanks MathsManiac; that's great! :)


  • Registered Users, Registered Users 2 Posts: 338 ✭✭ray giraffe


    Re Question 1

    Use cos(theta) = x/r first, then replace all instances of r with sqrt(x^2+y^2).

    Square roots can be removed if desired.


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