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Thread: 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 $\displaystyle \cos^2 \theta + \sin^2 \theta = 1$.
    Now using this we can square things and add them
    $\displaystyle x^2 + y^2 = a^2 \cos^2 \theta + b^2 \sin^2 \theta$
    However, this doesn't get rid of the $\displaystyle \sin \theta$ or the $\displaystyle \cos \theta$. 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 $\displaystyle \frac{a}{b}$ squared it and then added them. The following is the result
    $\displaystyle x^2 + \frac{a^2}{b^2}y^2 = a^2 \cos^2 \theta + \frac{a^2b^2}{b^2} \sin^2 \theta = a^2$. Finally, the solution to a) is
    $\displaystyle x^2 + \frac{a^2}{b^2}y^2 = a^2$.

    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 $\displaystyle <\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
    $\displaystyle \left(\dfrac{\partial f}{\partial x},\dfrac{\partial f}{\partial y}\right)$
    In your case $\displaystyle f = x^2 + \frac{a^2}{b^2}y^2 - a^2$
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