MHB 7.8.11 Find amplitude, period, PS, VS. graph 2 periods

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The discussion focuses on analyzing the function y=3cos(πx-2)+5 to determine its amplitude, period, phase shift (PS), and vertical shift (VS). The amplitude is identified as 3 and the vertical shift as 5. The period is calculated using the formula T=2π/ω, resulting in a period of 2. There is a debate regarding the phase shift, with one participant suggesting φ=2 and another indicating φ should be -2, highlighting the importance of the sign in determining the direction of the wave's shift. The conversation emphasizes the need to clarify the general form of the cosine function used in the analysis.
karush
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$\tiny{\textbf{7.8.11 Campbell HS}}$
Find (A)mplitude, (P)eriod, PS, VS. graph 2 periods
$y=3\cos(\pi x-2)+5$

by observation we have A=3 and VS=5
ok assume $\omega=\pi$
so if period is $T=\dfrac{2\pi}{\omega}$ then $T=\dfrac{2\pi}{\pi}=2$
 
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karush said:
$\tiny{\textbf{7.8.11 Campbell HS}}$
Find (A)mplitude, (P)eriod, PS, VS. graph 2 periods
$y=3\cos(\pi x-2)+5$

by observation we have A=3 and VS=5...
and $\omega=\pi$ ...
Otherwise, good!

-Dan
 
ok i think $\phi =2$ then PS is $\dfrac{\phi}{\omega}=\dfrac{2}{\pi}$

really! :unsure:
 
Okay, check with your general form of the sine wave. I use
[math]y = A ~ cos( \omega x + \phi ) + y_0[/math]

Your source might be using
[math]y = A ~ cos( \omega x - \phi ) + y_0[/math]
in which case, yes, [math]\phi = 2[/math]. In this model I'm using [math]\phi = - 2[/math]. The negative sign is important because it tells which way the wave has been shifted along the x-axis.

-Dan
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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