Calculus 1

The curve with the equation y^2=5x^4-x^2 is called a kampyle of Eudoxus. Find an equation of the tangent line to this curve at the point (-1,2).

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  1. y^2=5x^4-x^2

    2y dy/dx = 20x^3 - 2x
    dy/dx = (20x^3 - 2x)/(2y)
    = (10x^3 - x)/y
    at (-1,2)
    dy/dx = (-10+1)/2
    = -9/2

    so the tangent equation is 9x + 2y = c
    with (-1,2) lying on it, so
    -9 + 4 = c = -5

    the tangent equation is 9x + 2y = -5

    check:
    http://www.wolframalpha.com/input/?i=y%5E2%3D5x%5E4-x%5E2+%2C+9x+%2B+2y+%3D+-5+%2C+from+-2+to+0

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  2. nope that is incorrect
    the tangent equation is -9/2x-5/2

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  3. First of all -9/2x-5/2 is NOT an equation

    if you meant y = -(9/2)x - 5/2
    then by golly, that is what my equation is

    9x + 2y = -5
    2y = -9x - 5
    divide each term by 2

    y = (-9/2)x - 5/2

    My answer is correct.

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  4. well the online homework system didn't take it as a complete equation because the y= part was already outside of the answer box

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  5. well, there you go, how about that

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  6. It worked for mine, thanks Reiny.

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