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Math Help - tangent and normal

  1. #1
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    tangent and normal

    Hi i wonder what the wrong is in my calculation is :
    deside the tangent and the normal to the curve f(x) = ln(x + sqrt(1 + x^2) in the pont (0,0).

    f(x)= 1/(sqrt(x^2 + 1))
    f(0) = 1/(sqrt(0^2 + 1)) = 1
    y = kx + m, y = 0, x = 0, k = 1
    m = y -kx = 1 - 1 * 0 = 0

    but I must have done wrong somewere, for the right answer is tangen, y= x, and the normal, y = -x.

    help someone :/
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  2. #2
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    Quote Originally Posted by paulaa View Post
    Hi i wonder what the wrong is in my calculation is :
    deside the tangent and the normal to the curve f(x) = ln(x + sqrt(1 + x^2) in the pont (0,0).

    f(x)= 1/(sqrt(x^2 + 1))
    f(0) = 1/(sqrt(0^2 + 1)) = 1
    y = kx + m, y = 0, x = 0, k = 1
    m = y -kx = 1 - 1 * 0 = 0

    but I must have done wrong somewere, for the right answer is tangen, y= x, and the normal, y = -x.

    help someone :/
    1. This is definitely the wrong forum!

    2. You solved the question perfectly - honestly!

    You got: k = 1 and m = 0. Therefore the equation of the tangent is:

    y = 1 \cdot x + 0~\implies~\boxed{y=x}

    3. The normal has the slope k_n=-\frac11 and passes through the tangent point T(0, 0).
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