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Meaning of y = c if f(x) = cg(x)

  1. Oct 31, 2011 #1
    1. The problem statement, all variables and given/known data

    Hi. I post part A too but I'm having trouble only with Part B.

    Question.[PLAIN]http://img100.imageshack.us/img100/8297/20104ivwhytangenttoseco.png [Broken]

    2. Relevant equations

    I know....

    [itex]y = sin x [/itex] and [itex] y = cx[/itex] intersects when ...

    [itex]sin x = cx[/itex] so [itex] \frac{sin x}{x} = c[/itex]

    3. The attempt at a solution

    I just don't understand why [itex] \frac{sin x}{x} = c[/itex]
    means that [itex] y = c [/itex] has to be tangent to
    [itex] y = \frac{sin x}{x}[/itex] at the second hump?

    I'm not seeing the connection...

    Thank you.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Oct 31, 2011 #2

    SammyS

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    Look at the second graph. If y = c x is not tangent to the graph of y = sin(x) "near the second hump", then the graph of y = c x will intersect the graph of y = sin(x) in either 7 places, or in 3 places.
     
  4. Oct 31, 2011 #3
    Hi SammyS. Thanks for your answer.

    But I don't understand your answer either

    Generally, I have trouble relating intersection of [itex] y = sin x [/itex] and [itex] y = cx [/itex] to graph of [itex] y = \frac{sin x}{x} [/itex] and [itex] y = c[/itex]. But I understand [itex] sin x = cx [/itex] means [itex] \frac{sin x}{x} = c [/itex].
     
  5. Oct 31, 2011 #4

    SammyS

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    To solve y = c x , and y = sin(x) simultaneously, you have:

    c x = sin(x) .

    That's equivalent to [itex]\displaystyle c=\frac{\sin(x)}{x}\,.[/itex]

    One way to solve this equation is graphically, i.e., where does the graph of y = c intercept the graph of [itex]\displaystyle y=\frac{\sin(x)}{x}\,?[/itex] I think the question to be asking is, "Should these two graphs be tangent at the point of intersection?"

    If y = c is tangent to [itex]\displaystyle y=\frac{\sin(x)}{x}\,[/itex] near [itex]\displaystyle x=\frac{5\pi}{2}\,,[/itex] then it must be true that on either side of the point of tangency, [itex]\displaystyle c>\frac{\sin(x)}{x}\,.[/itex] It can be shown that this inequality holds.

    Therefore, the slope of [itex]\displaystyle y=\frac{\sin(x)}{x}\,[/itex] is zero at the same value of x at which the the graph of y = c x is tangent to the graph of [itex]\displaystyle y=\sin(x)\,[/itex] near the second hump. Use this value of x to solve for c.
     
  6. Nov 4, 2011 #5
    Hi SammyS. Thanks.
     
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