7.8.99 find PS, VS Period, graph

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Discussion Overview

The discussion revolves around finding the amplitude, period, phase shift (PS), and vertical shift (VS) of the function \(y=\cos\left(x+\dfrac{\pi}{2}\right)\). Participants explore these concepts in the context of trigonometric functions, specifically focusing on their graphical representations and properties.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculates the amplitude as \(|A|\), the period as \(T=\dfrac{2\pi}{\omega}=\pi\), and suggests a phase shift of \(\dfrac{\pi}{4}\), but expresses uncertainty about the vertical shift.
  • Another participant states the amplitude is 1, the period is \(T=2\pi\), and identifies the phase shift as \(\dfrac{\pi}{2}\) to the left, asserting there is no vertical shift.
  • Several participants question the value of \(\omega\) and express uncertainty about its determination, with one suggesting it might be 2 or 4.
  • One participant clarifies that the general cosine equation can vary in notation, identifying \(y_0\) as the vertical shift and discussing the terms used for amplitude, angular frequency, and phase shift.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the values of the amplitude, period, phase shift, and vertical shift, with multiple competing views presented regarding these parameters.

Contextual Notes

There is ambiguity regarding the definitions and values of \(\omega\) and the vertical shift, as well as the notation used for the cosine function, which may depend on different educational contexts.

karush
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$\tiny\textbf{7.8.a09 Radford HS}$
Find amplitude, period, PS, VS. then graph.
$y=\cos\left(x+\dfrac{\pi}{2}\right)$For the graphs of $y=A\sin(\omega x - \phi)$ or $y=A\cos(\omega x - \phi),\omega>0$
Amplitude $=|A|$
Period $T=\dfrac{2\pi}{\omega}=\dfrac{2\pi}{2}=\pi$
PS $=\dfrac{\phi}{\omega}=\dfrac{\pi}{4}$

well so far
I don't know what the greek letter is for VS or Vertical Shift? which is usually D
 
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$y=\cos\left(x + \dfrac{\pi}{2}\right)$

amplitude = 1

period, $T = 2\pi$

phase shift = $\dfrac{\pi}{2}$ left

no vertical shift

fyi, $\cos\left(x+\dfrac{\pi}{2}\right) = -\sin{x}$
 
[math]y = A ~ sin( \omega x + \phi ) + y_0[/math]

What was your [math]\omega[/math] again?

-Dan
 
topsquark said:
[math]y = A ~ sin( \omega x + \phi ) + y_0[/math]

What was your [math]\omega[/math] again?

-Dan

$y=\cos\left(x+\dfrac{\pi}{2}\right)$
well thot it was 2 maybe 4? it was kinda :unsure:
 
karush said:
$y=\cos\left(x+\dfrac{\pi}{2}\right)$
well thot it was 2 maybe 4? it was kinda :unsure:

try 1
 
karush said:
$y=\cos\left(x+\dfrac{\pi}{2}\right)$
well thot it was 2 maybe 4? it was kinda :unsure:
[math]y = cos \left ( x + \dfrac{ \pi }{2} \right )[/math]

[math]y = A ~ cos( \omega x + \phi ) + y_0[/math]

What is the coefficient of x in your cosine argument??

Geez, dude! You are better than that!

-Dan
 
topsquark said:
[math]y = cos \left ( x + \dfrac{ \pi }{2} \right )[/math]

[math]y = A ~ cos( \omega x + \phi ) + y_0[/math]

What is the coefficient of x in your cosine argument??

Geez, dude! You are better than that!

-Dan
$y = \cos \left( 1 \left( x + \dfrac{ \pi }{2} \right )\right )$

are you using $y_0$ as VS
 
karush said:
$y = \cos \left( 1 \left( x + \dfrac{ \pi }{2} \right )\right )$

are you using $y_0$ as VS
Yes. There really is no standard way of writing the general cosine equation. It varies from class to class and text to text. (In fact I learned it as sine in College.)

A - wave amplitude
[math]\omega[/math] - angular frequency
[math]\phi[/math] - phase angle, or phase shift as you are calling it
[math]y_0[/math] - vertical displacement, or vertical shift as you are calling it. Some would also call this "h."

-Dan
 

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