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Math Help - Parametric Differentiation help....

  1. #1
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    Parametric Differentiation help....

    Need help having trouble with this question....
    x=2cos\theta, y=\frac{1}{\sqrt2}sin\theta
    a) Find gradient at the point where theta=pi
    b) Hence find the cartesian co-ordinates
    c) Find equations of the tangent and normal at this point.
    My workings
    a)
    \frac{dx}{d\theta}=-2sin\theta , \frac{dy}{d\theta}=\frac{1}{\sqrt2}cos\theta
    \frac{d\theta}{dx}=-\frac{1}{2sin\theta}
    \frac{dy}{dx}=\frac{dy}{d\theta} x \frac{d\theta}{dx}
    \frac{dy}{dx}=\frac{1}{\sqrt2}cos\theta x -\frac{1}{2sin\theta}
    \frac{dy}{dx}=-\frac{\frac{1}{\sqrt2}cos\theta}{2sin\theta}
    \frac{dy}{dx}=-\frac{\frac{1}{\sqrt2}}{2tan\theta}
    \frac{dy}{dx}=-\frac{\frac{1}{\sqrt2}}{2tan\pi}
    as tan pi = 0, i will get an undefined answer is so how can i do the following parts? :s can someone please help me out

    And yes Theta=Pi heres the original question, Question 3 here [/URL]
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  2. #2
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    e^(i*pi)'s Avatar
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    Your working looks fine and I get the same answer. It's probably just undefined
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  3. #3
    MHF Contributor Calculus26's Avatar
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    at (-2,0)

    you have a vertical tangent and a horizontal normal
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  4. #4
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    how did you get that co-ordinate of -2,0 just need help on b and c if my a is correct. thanks.
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  5. #5
    MHF Contributor Calculus26's Avatar
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    x= 2cos(pi) y = 1/(2^1/2)sin(pi)
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  6. #6
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    How do i get equations of tangent+normal ? :s im doing this

    (y-0)=m(x-(-2)) but what do i put in for m infinity? =/ never had a question like this before t.t
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  7. #7
    MHF Contributor Calculus26's Avatar
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    remember your old algebra days --the eqn of a vertical line is x = c

    in this case x = -2
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  8. #8
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    x=-2, and y=-2 but is there any working out needed ? i know its really basic but how would i know that the tangent + normal were going to be horizontal and vertical?
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Kevlar View Post
    x=-2, and y=-2 but is there any working out needed ? i know its really basic but how would i know that the tangent + normal were going to be horizontal and vertical?
    the slope was undefined, this means the tangent is vertical. the normal is perpendicular to the tangent, so it had to be horizontal
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  10. #10
    MHF Contributor Calculus26's Avatar
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    y = 0
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