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Math Help - Please Help Derivatives... last problem of the day...

  1. #1
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    Please Help Derivatives... last problem of the day...

    Find the equation of the tangent and normal line to the graph of the equation at the indicated point.

    y = \frac{4}{x^2};(2,1)

    the answer from the back of my book is:
    tangent: y=-x+3; normal y = x - 1
    how could this be?
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  2. #2
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find the equation of the tangent and normal line to the graph of the equation at the indicated point.

    y = \frac{4}{x^2};(2,1)

    the answer from the back of my book is:
    tangent: y=-x+3; normal y = x - 1
    how could this be?
    Thus,
    y'=-\frac{8}{x^3}
    x=2
    Thus, (the derivative at that point),
    y'(2)=-\frac{8}{8}=-1
    Using slope-intercept formula,
    y-y_0=-1(x-x_0)
    (x_0,y_0)=(2,1)
    Thus,
    y-1=-x-2
    Thus,
    y=-x-1
    This is equation of tangent line.

    Thus you are correct in saying the book is wrong.
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  3. #3
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    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    Find the equation of the tangent and normal line to the graph of the equation at the indicated point.

    y = \frac{4}{x^2};(2,1)

    the answer from the back of my book is:
    tangent: y=-x+3; normal y = x - 1
    how could this be?
    The tangent line will be of the form y = mx + b.

    We can get the slope from the derivative:
    y' = -\frac{8}{x^3}

    At x = 2 this becomes: y' = -\frac{8}{8} = -1

    So we need the intercept. At x = 2 we know that y = 1, so
    1 = -1 \cdot 2 + b

    b = 3

    So the tangent line is y = -x + 3

    The line normal to this will have a slope of -\frac{1}{-1} = 1, so the form of the normal line will be y = x + b' and passes through (2, 1), so
    1 = 2 + b'

    b' = -1

    So the normal line is y = x - 1.

    (Sorry about the graph. The lines don't look perpendicular because the x and y scales are different. I haven't figured out how to fix this yet.)

    -Dan
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