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

  • Thread starter seniorhs9
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  • #1
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Homework Statement



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]

Homework 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]

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.
 
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Answers and Replies

  • #2
SammyS
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...
I know....

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

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

The Attempt at a Solution



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

I'm not seeing the connection...

Thank you.
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.
 
  • #3
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Hi SammyS. Thanks for your answer.

But I don't understand your answer either

"If y = cx is not tangent to the graph of y = sin(x) "near the second hump", then the graph of y = cx will intersect the graph of y = sin(x) in either 7 places, or in 3 places "
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].
 
  • #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.
 
  • #5
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Hi SammyS. Thanks.
 

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