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Find EQ of tangent line for y=sin x at x = 6*pi

  1. Dec 7, 2011 #1
    1. The problem statement, all variables and given/known data

    Find the tangent line equation for y = sin x at x = [itex]6\,\pi[/itex].


    2. Relevant equations

    [itex]m_{tangent}\,=\,\lim_{h \to 0}\,\frac{f(c+h)\,-\,F(c)}{h}[/itex]

    [itex]y\,-\,y_1\,=\,m\,(x\,-\,x_1)[/itex]


    3. The attempt at a solution

    [itex]m_{tangent}\,=\,\lim_{h \to 0}\,\frac{f(c+h)\,-\,F(c)}{h}[/itex]

    [itex]m_{tangent}\,=\,\lim_{h \to 0}\,\frac{f(6\pi\,h)\,-\,F(6\pi)}{h}[/itex]

    [itex]m_{tangent}\,=\,\lim_{h \to 0}\,\frac{sin(6\pi\,+\,h)}{h}[/itex]

    Now use L'Hopital's Rule and take the derivative of the denominator and numerator...

    [itex]\frac{d}{dh}\,\left[\frac{sin(6\pi\,+\,h)}{h}\right]\,=\,cos(h)[/itex]

    [itex]y\,-\,y_1\,=\,m\,(x\,-\,x_1)[/itex]

    [itex]y\,-\,(0)\,=\,(cos\,h)\,(x\,-\,6\pi)[/itex]

    Since h is zero, the coefficient, m, is 1.

    So your tangent line equation is then...

    [itex]y\,=\,x\,-\,6\pi[/itex]
     
  2. jcsd
  3. Dec 7, 2011 #2

    Mentallic

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    Homework Helper

    This is one of those cases where two wrongs can make a right. You have the right answer, but the way you got there is just flat out wrong in many ways.


    These lines should instead be:

    [tex]m=\lim_{h \to 0}\frac{f(c+h)-f(c)}{h}[/tex]

    [tex]m=\lim_{h \to 0}\frac{f(6\pi+h)-f(6\pi)}{h}[/tex]

    [tex]m=\lim_{h \to 0}\frac{\sin(6\pi+h)-sin(6\pi)}{h}[/tex]

    Now, I don't understand why you would take the approach of finding the derivative by this method, and then use L'Hospital's rule. Why didn't you just take the derivative of sin(x) to begin with?
    Regardless,

    To use L'Hospital's rule correctly, you need to take the derivative of the numerator and denominator separately as you said, so, this should instead be:

    [tex]\frac{\frac{d}{dh}\left(\sin(6\pi+h)-\sin(6\pi)\right)}{\frac{d}{dh}\cdot h}=\frac{\cos(6\pi+h)-0}{1}=\cos(6\pi+h)[/tex]

    Then you apply the rules you used in a similar fashion:

    As a note, a much quicker, tidier, and easier method is to just start with the equation, and find the values you need accordingly.

    [tex]y-y_0=m(x-x_0)[/tex]

    Finding [itex]x_0,y_0,m[/itex]

    We are given that [itex]x_0=6\pi[/itex]

    And we have the function [itex]f(x)=\sin(x)[/itex] so [itex]f(x_0)=y_0=f(6\pi)=\sin(6\pi)=0[/itex]

    And finally, m is just the gradient at that point so what we are looking for is [itex]\sin'(6\pi)[/itex] taking the derivative of sin(x) gives cos(x), and [itex]\cos(6\pi)=1[/itex] and there you have it, all the values you need to plug into the point-gradient formula for a line.
     
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