Results 1 to 2 of 2

Math Help - A new puzzle?

  1. #1
    Banned
    Joined
    Oct 2009
    Posts
    769

    A new puzzle?

    Some days ago, I did a puzzle using exp(x) and ln(x) and it occurred to me that these functions can be used to start off a different type of puzzle.

    Let x = 1. Then you have the points (1,e) and (1,0). Now draw tangents to the curves at those points. The tangent lines (most likely - I haven't formally figured it out this far) would intersect at a point. Figure out what this point is. Now let x = 2 to get the points (2, exp(2)) and (2, ln(2)). Now draw tangents to those two points and figure out what the point of intersection is for the next tangent lines. Let x take on various different values and derive more points of intersection.

    Can you produce a graph showing the locus of intersection points? As an extra bonus question, can you derive an equation for these locus points? (to take it a step further, you may want to explore what happens letting y = x +1 or using parabolas for variety sakes).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    5
    Awards
    2
    This was a fun problem. Here's my answer. For the function e^{x}, we write the equation of the tangent line to the point (x,e^{x}) as y_{1}(x_{1}). Similarly, for the function \ln(x), we write the equation of the tangent line as y_{2}(x_{2}). Now, given that (e^{x})'=e^{x} and (\ln(x))'=1/x, we can immediately write down the equations of the tangent lines:

    y_{1}-e^{x}=e^{x}(x_{1}-x), and
    y_{2}-\ln(x)=\dfrac{1}{x}(x_{1}-x).

    We want to set these equal. Obviously, this will have to happen when x_{1}=x_{2}, so re-define \xi=x_{1}=x_{2}, and set

    e^{x}+e^{x}(\xi-x)=\ln(x)+\dfrac{1}{x}(\xi-x).

    Solving for \xi-x yields

    \xi-x=\dfrac{x(e^{x}-\ln(x))}{1-xe^{x}}.

    Plugging this back into either y_{1} or y_{2} (it shouldn't matter, although I did it in both as a check, and they both come out the same), you get

    y=\dfrac{e^{x}(1-x\ln(x))}{1-xe^{x}}.

    This is valid when xe^{x}\not=1.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Set Puzzle
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: April 6th 2008, 08:46 AM
  2. new puzzle
    Posted in the Math Challenge Problems Forum
    Replies: 1
    Last Post: October 10th 2007, 05:10 PM
  3. Puzzle
    Posted in the Math Challenge Problems Forum
    Replies: 3
    Last Post: December 22nd 2006, 04:38 PM
  4. A puzzle
    Posted in the Math Challenge Problems Forum
    Replies: 4
    Last Post: December 9th 2006, 08:40 PM
  5. more puzzle
    Posted in the Math Challenge Problems Forum
    Replies: 15
    Last Post: October 11th 2006, 09:26 PM

Search Tags


/mathhelpforum @mathhelpforum