Results 1 to 3 of 3

Math Help - Tangency

  1. #1
    Member
    Joined
    Sep 2009
    Posts
    80

    Tangency

    Another problem I'm stuck on.

    "If k >/= 1 (greater than or equal to 1), the graphs of y=sin(x) and y=ke^-x intersect for x >/= 0. Find the smallest value of k for which the graphs are tangent. What are the coordinates of the point of tangency?"

    Help, please?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,719
    Thanks
    634
    Hello, Rumor!

    I think I have a solution.
    Please check my reasoning and my work.


    If k \geq 1, the graphs of: y\:=\:\sin x and y\:=\:ke^{-x} intersect for x \geq 0.
    Find the smallest value of k for which the graphs are tangent.
    What are the coordinates of the point of tangency?
    We have: . \begin{array}{ccc} f(x) &=& \sin x \\ g(x) &=& ke^{-x} \end{array}

    The graphs intersect at the point of tangency.
    . . Hence: . f(x) \:=\:g(x) \quad\Rightarrow\quad \sin x \:=\:ke^{-x} \quad\Rightarrow\quad e^x\sin x \:=\:k [1]

    Their slopes are equal at the point of tangency.
    . . Hence: . f'(x) = g'(x) \quad\Rightarrow\quad \cos x \:=\:-ke^{-x} \quad\Rightarrow\quad e^x\cos x \:=\:-k [2]


    Divide [1] by [2]: . \frac{e^x\sin x}{e^x\cos x} \:=\:\frac{k}{\text{-}k} \quad\Rightarrow\quad \tan x \:=\:-1 \quad\Rightarrow \quad x \:=\:\frac{3\pi}{4}


    We have: . \begin{array}{ccccccc}f\left(\frac{3\pi}{4}\right) &=& \sin\frac{3\pi}{4} &=& \frac{1}{\sqrt{2}} \\ \\[-3mm]<br />
g\left(\frac{3\pi}{4}\right) &=& ke^{-\frac{3\pi}{4}}\end{array}


    Since g\left(\tfrac{3\pi}{4}\right) \,=\,f\left(\tfrac{3\pi}<br />
{4}\right), we have: . ke^{-\frac{3\pi}{4}} \:=\:\frac{1}{\sqrt{2}}

    Therefore: . \boxed{k \:=\:\frac{e^{\frac{3\pi}{4}}}{\sqrt{2}}}

    And the point of tangency is: . \left(\frac{3\pi}{4},\:\frac{1}{\sqrt{2}}\right)

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2009
    Posts
    80
    Ah, of course!

    Your work looks sound to me. I don't know why I couldn't remember to set them equal. Thank you very much!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Tangency to a curve
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 3rd 2009, 07:47 AM
  2. Calculus Tangency
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 29th 2009, 01:12 AM
  3. Tangency Question
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 9th 2008, 05:54 AM
  4. tangency
    Posted in the Calculus Forum
    Replies: 3
    Last Post: March 19th 2008, 12:33 PM
  5. Points of tangency.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 2nd 2008, 09:33 PM

Search Tags


/mathhelpforum @mathhelpforum