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Math Help - Cartesian Equation and Parametric Equation

  1. #1
    Zel
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    Question Cartesian Equation and Parametric Equation

    Hey,
    I need some help with the following questions:
    a) Write down the Cartesian equation of the curve C represented by the parametric equations x=acos(theta) and y=bsin(theta), a>b>0
    b) Find an expression for the gradient of the tangent to C
    c) Hence show that the equation of the tangent can be written in the form:
    bxcos(theta) + aysin(theta) = ab
    I would really appreciate it
    Thanks in advance
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  2. #2
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    Remember that \cos^2 \theta + \sin^2 \theta = 1.
    Now using this we can square things and add them
    x^2 + y^2 = a^2 \cos^2 \theta + b^2 \sin^2 \theta
    However, this doesn't get rid of the \sin \theta or the \cos \theta. We need both of these terms to have the same number in front of them, but they don't right now. So we could have squared the x term then multiplied the y term by \frac{a}{b} squared it and then added them. The following is the result
    x^2 + \frac{a^2}{b^2}y^2 = a^2 \cos^2 \theta + \frac{a^2b^2}{b^2} \sin^2 \theta = a^2. Finally, the solution to a) is
    x^2 + \frac{a^2}{b^2}y^2 = a^2.

    For the last two, I am unsure how you find the gradient of a function at your level. What has your instructor defined as the gradient and is there a formula?
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  3. #3
    Zel
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    lvleph, thanks for part a.
    For part b, I assume they want us to find the derivative of the equation, which I assume would be
    2x + 2y (dy/dx) = 0 ... Would this be correct, or would I need to include the a and b values as well?
    Thanks again
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  4. #4
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    The gradient of a function f(x,y,z) should be  <\frac{df}{dx}, \frac{df}{dy}, \frac{df}{dz} > .
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  5. #5
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    For some reason I thought this was in the pre-calc or trig section. Sorry for that.
    The gradient is always a vector and as Math Major pointed out it should be
    \left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}\right)
    In your case  f = x^2 + \frac{a^2}{b^2}y^2 - a^2
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