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Math Help - Calculating the acute angle between OX and the tangent to a curve.

  1. #1
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    Calculating the acute angle between OX and the tangent to a curve.

    I don't know where to begin with this question. It's in the revision section in my textbook and I've had a look back through the book and I can't find anything helpful. I don't even know what OX is supposed to mean. Here it is:

    Calculate, to the nearest degree, the acute angle between OX and the tangent to the curve y = {x}^{4} - {2x}^{3} + {3x}^{2} - 4x + 1 at the point where
    x = 2.

    According to the back of the book the answer is 86 degrees, but I would like to know how to get there.
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  2. #2
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    Quote Originally Posted by MAX1592 View Post
    I don't even know what OX is supposed to mean. Here it is: Calculate, to the nearest degree, the acute angle between OX and the tangent to the curve y = {x}^{4} - {2x}^{3} + {3x}^{2} - 4x + 1 at the point where x = 2. According to the back of the book the answer is 86 degrees, but I would like to know how to get there.
    In this case OX stands for the x-axis.
    The slope of a line is the tangent of the angle the line makes with the x-axis in the positive direction.
    So you need to find \arctan\left(y^{\,\prime}(2)\right).
    BTW: You have written the question wrong.
    y=x^4-2x^3+3x^2-4x+1.
    The answer is about 86.424^o
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