1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

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

    Yes!

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Equation of Tangent to the Line
  1. Tangent line equation (Replies: 9)

  2. Equation of Tangent Line (Replies: 14)

Loading...