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Equation of Tangent to the Line

  1. Feb 11, 2009 #1
    3.) Give the equation of the line tangent to the curve at the given point.

    a.) (y)(tan^-1 x) = x*y at (sqrt(3),0)

    b.) ln y = x^2 +(2)*e^x at (0, e^2)

    Please help me with this problem, I am stuck.

    Here is some of my work process:

    i have no idea how to solve for y in a.)

    but for b.) I took the expontential of both sides

    then i got y= e^(x^2) + e^(2*e^x)

    do I find the derivate of y to get the slope

    but how do i plug in the values of the pt.

    do i use the y- yl= m(x-x1) formula to find the tangent to the curve?

    please help me with these problems, I'm clueless for part a, and i have a bit of an understanding on how to do part b.)
  2. jcsd
  3. Feb 11, 2009 #2


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    Science Advisor

    Put the x value of the point into your formula for the derivative!


    You may have miscopied part (a). If y is not 0, you can divide through by it and get tan-1(x)= x which means y is not a function of x. I think that there are two points, one positive and the other negative, for which tan-1(x)= x and so the equation is true for all y. Of course, it is true for all x if y= 0 and so its graph is those three lines. If that is really is what (a) says, then since [itex]tan^{-1}(\sqrt{3})\ne \sqrt{3}[/itex], the curve in a neighbor hood of the point [itex](\sqrt{3},0)[/itex] is the line y= 0 and so its tangent is the line y= 0.
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