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Math Help - [SOLVED] tangent/normal lines

  1. #1
    Member >_<SHY_GUY>_<'s Avatar
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    Question [SOLVED] tangent/normal lines

    1. fint the lines that are tangent and normal to the curve y=x^3 at (1,1)

    im guessing that i have to use the definition of a derivative, and substitute f(x) and x in the equation with one, and solve it?

    what about the normal line?
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  2. #2
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    1. fint the lines that are tangent and normal to the curve y=x^3 at (1,1)?
    The tangent line has slope y'(1); the normal line has slope \frac{-1}{y'(1)}.
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    1. fint the lines that are tangent and normal to the curve y=x^3 at (1,1)

    im guessing that i have to use the definition of a derivative, and substitute f(x) and x in the equation with one, and solve it?

    what about the normal line?
    Correct, take the first derivative of y=x^3 and solve f'(1)

    The normal is always perpendicular to the tangent and so the product of their gradients must be -1: m_1m_2 = -1 where m_1 = f'(1) = the gradient at that point and m_2 = \frac{-1}{m_1}

    In both cases use the equation of a straight line formula:

    y-y_1 = m(x-x_1) where (x_1,y_1) is a coordinate. In your case this is (1,1)
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    Member >_<SHY_GUY>_<'s Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    Correct, take the first derivative of y=x^3 and solve f'(1)

    The normal is always perpendicular to the tangent and so the product of their gradients must be -1: m_1m_2 = -1 where m_1 = f'(1) = the gradient at that point and m_2 = \frac{-1}{m_1}

    In both cases use the equation of a straight line formula:

    y-y_1 = m(x-x_1) where (x_1,y_1) is a coordinate. In your case this is (1,1)
    ok, by finding the derivative, i got 3x^2
    and substituting 1 gives me 3.

    i know that the slope of the tangent line is 3 and the normal is -1/3.
    so i use y=mx+b to find the formula correct?

    Thanks!
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    Quote Originally Posted by >_<SHY_GUY>_< View Post
    ok, by finding the derivative, i got 3x^2
    and substituting 1 gives me 3.

    i know that the slope of the tangent line is 3 and the normal is -1/3.
    so i use y=mx+b to find the formula correct?

    Thanks!
    Correct. You can use either y = mx+b or the formula I used above, both work
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