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Math Help - Derivatives: normal and tangent line

  1. #1
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    Derivatives: normal and tangent line

    Find the equations for the tangent line and normal line to the circle at the indicated points. (4,3)(-3,4)
    x^2 + y^2 = 25

    Which coordinate set do I use to find both lines? Because I get a different answer for the normal line when I use (4,3) than when I use (-3,4)

    The derivative solved implicitly is -x/y. I do not know where to go from here.

    Thanks for your help!
    Last edited by trigcalc4; November 20th 2008 at 02:56 PM.
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  2. #2
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    Quote Originally Posted by trigcalc4 View Post
    Find the equations for the tangent line and normal line to the circle at the indicated points. (4,3)(-3,4)
    x^2 + y^2 = 25

    The derivative solved implicitly is -x/y. I do not know where to go from here.

    Thanks for your help!
    Remember that the derivative of a function gives the slope at any point (x,y). Plugging in the ordered pairs will yield the slope of the tangent line at the indicated points above. Then, since you have a slope, and you have a point, you can then write the equation for the tangent lines to the function at those points. Do this for (4,3) and (-3.4).

    After you have found the equations for the two tangent lines, you can then find each of the lines normal to the points by remembering that a line normal to the tangent line will have the opposite reciprocal slope of the tangent line. We know the slope of the tangent line is -x/y, so that means that the slope of the line normal to the curve at any given point (x,y) is y/x (opposite reciprocal). Once again, to write the equation of the line normal, you have the ordered pairs and a slope..then you can find the y-intercept to find the equation of those lines normal at each of the points (4,3) and (-3,4).

    Hope this helps.
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