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Math Help - Parametric solving

  1. #1
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    Question Parametric solving

    Hi again;

    I have been given the following questions:

    Code:
    Find the equation of the (i) tangent and (ii) normal lines to the curve given parametrically
    by x = t2 + t, y = t2 − t at the point corresponding to t = 0.
    now this is what I have done so far:

    dx/dt = 2t
    dy/dt = 2t
    THEREFORE
    dy/dx = 2t/2t

    But am not sure if I am correct, or if I am what do I do next?

    Thanks
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  2. #2
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    \frac{dy}{dt}=2t-1

    \frac{dx}{dt}=2t+1

    When t=0, dy/dx = -1

    \frac{dy}{dx}=m=\frac{2t-1}{2t+1}

    (x,y)=(0,0)

    Now use y-y_{1}=m(x-x_{1})

    y-0 = -1(x-0)

    y=-x

    Now, it's easy to find the normal.
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  3. #3
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    Quote Originally Posted by galactus View Post

    Now use y-y_{1}=m(x-x_{1})
    Is that like y=mx+c?
    could you please explain that

    ALSO

    so the normal would have gradient -1/m
    so the normal would be -1/-1
    giving y = 1x+c?
    Last edited by taurus; October 23rd 2007 at 03:11 PM.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by taurus View Post
    Is that like y=mx+c?
    could you please explain that

    ALSO

    so the normal would have gradient -1/m
    so the normal would be -1/-1
    giving y = 1x+c?
    the formula he used is called the point-slope formula. it is very commonly used, you should note it

    yes, the normal's gradient is the negative inverse of the gradient of the tangent line
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  5. #5
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    for normal in my case i got m = 1
    therefore y = 1x+c => y = x+c

    How do i get c then? In the answers i have its just x?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by taurus View Post
    for normal in my case i got m = 1
    therefore y = 1x+c => y = x+c

    How do i get c then? In the answers i have its just x?
    the line goes through the origin. you can use the point slope form as galalctus showed you, or you could just realize that if a line passes through the origin, the y-intercept is zero
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  7. #7
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    Question

    Quote Originally Posted by galactus View Post
    \frac{dy}{dt}=2t-1

    \frac{dx}{dt}=2t+1

    When t=0, dy/dx = -1

    \frac{dy}{dx}=m=\frac{2t-1}{2t+1}

    (x,y)=(0,0)

    Now use y-y_{1}=m(x-x_{1})

    y-0 = -1(x-0)

    y=-x

    Now, it's easy to find the normal.
    How do you know what values of x and y to use in this point slope equation?
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