Finding the horizontal shift of a function

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Homework Help Overview

The discussion revolves around finding the horizontal shift of a function, particularly in the context of trigonometric functions and their parameters. Participants are exploring the implications of using specific points and values in calculations, as well as the geometry involved in the problem.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the use of the inverse cosine function and its implications on the values obtained. There are questions about the correct points to use in calculations and the geometry that underlies the problem. Some participants express confusion regarding the relationship between the measurements and the function parameters.

Discussion Status

The discussion is active, with participants sharing their calculations and questioning the assumptions made in the original post. Some have suggested that a mistake may have been made regarding the points used in the calculations, while others are exploring the implications of these potential errors.

Contextual Notes

Participants note discrepancies between the expected and actual values used in the calculations, particularly regarding the point (0, 5) versus (0, 4.586). There is an ongoing examination of the geometry involved and how it affects the calculations of the function parameters.

Einstein44
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Homework Statement
I was trying to find the horizontal shift of the function, as shown in the picture attached below. It clearly states, that this was found through simultaneous eqn's, but I am unsure how this is done.
Relevant Equations
##y=1.5 cos(\frac{2\pi }{23}x+C)+5.4##
Screenshot 2021-12-22 at 15.24.40.png

I've never actually done this, so I was wondering if someone could show me how this is done. One way I tried was by simply using ##cos^{-1}## in order to cancel the cosine, but that gave me a different value, so I assume this is not how you are supposed to do this.

--> I know I am supposed to know this, but here I am... :)
 
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Yes, he plugged in the x,y values from (0,5) and used ##\cos^{-1}##.
I have no idea what the "From my measurements" line is talking about. It might have something about simultaneous equations, but I can't guess.
I also can't guess why the value of C is that particular solution. There may be some geometry involved that is not shown in your post.
 
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FactChecker said:
Yes, he plugged in the x,y values from (0,5) and used ##\cos^{-1}##.
I have no idea what the "From my measurements" line is talking about. It might have something about simultaneous equations, but I can't guess.
I also can't guess why the value of C is that particular solution. There may be some geometry involved that is not shown in your post.
Hmmm... This is the website, in case you want to have a look: https://www.ukessays.com/essays/mat...area-of-complex-three-dimensional-objects.php
I can't figure out what kind of geometry could have been used here, as it seems to be a pretty simple scenario to me (and no indication on the website whatsoever), but by calculating this using ##\cos^{-1}## gave me a different value... Have you tried calculating it?

Edit: I believe the "from my measurments" in this case refers to the measurments he took in order to find the rest of the function (a,b,d).
 
Last edited:
The work on the website is all about the geometry of a vase. That probably determines which of the set of solutions for C are valid in that situation..
 
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Einstein44 said:
One way I tried was by simply using ##cos^{-1}## in order to cancel the cosine, but that gave me a different value, so I assume this is not how you are supposed to do this.
The closest thing I got to their answer was ##\arccos(-0.4/1.5)-2\pi = 1.8407291226283 - 2\pi = -4.44245618455128##, so I don't know what they did.
 
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FactChecker said:
The closest thing I got to their answer was ##\arccos(-0.4/1.5)-2\pi = 1.8407291226283 - 2\pi = -4.44245618455128##, so I don't know what they did.

No idea either!

1640238664046.png


?
 
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FactChecker said:
The closest thing I got to their answer was ##\arccos(-0.4/1.5)-2\pi = 1.8407291226283 - 2\pi = -4.44245618455128##, so I don't know what they did.
Alright, I guess this will remain a mystery. Thanks for the contribution.
 
The author has made a mistake. Simple as that!

In the Post #3 link, they say they are using the point (0, 5). But if you look carefully at the diagram
https://images.ukessays.com/180520/3/0651715.022.jpg
the correct point is clearly shown to be (0, 4.586), and this is in fact what has been used to calculate C (not (0,5)).

Easy to confirm: evaluate y using x=0 and the calculated value of C = -4.13893405 and we get:
##y = 1.5 cos(\frac{2\pi }{23}0-4.13893405)+5.4 = 4.5861937##
as required.
 
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Steve4Physics said:
The author has made a mistake. Simple as that!

In the Post #3 link, they say they are using the point (0, 5). But if you look carefully at the diagram
https://images.ukessays.com/180520/3/0651715.022.jpg
the correct point is clearly shown to be (0, 4.586), and this is in fact what has been used to calculate C (not (0,5)).

Easy to confirm: evaluate y using x=0 and the calculated value of C = -4.13893405 and we get:
##y = 1.5 cos(\frac{2\pi }{23}0-4.13893405)+5.4 = 4.5861937##
as required.
True, that makes sense.
 
  • #10
Steve4Physics said:
The author has made a mistake. Simple as that!

In the Post #3 link, they say they are using the point (0, 5). But if you look carefully at the diagram
https://images.ukessays.com/180520/3/0651715.022.jpg
the correct point is clearly shown to be (0, 4.586), and this is in fact what has been used to calculate C (not (0,5)).

Easy to confirm: evaluate y using x=0 and the calculated value of C = -4.13893405 and we get:
##y = 1.5 cos(\frac{2\pi }{23}0-4.13893405)+5.4 = 4.5861937##
as required.
I thought about that too. But the y-ordinate 4.586 is not a measured value. It came from using the (wrongly) calculated sine graph parameters!.
 
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neilparker62 said:
I thought about that too. But the y-ordinate 4.586 is not a measured value. It came from using the (wrongly) calculated sine graph parameters!.
Aha, you are right! I rushed in without reading the text carefully. Apologies to all. But let me try to redeem myself...

In the text of the Post #3 link
https://www.ukessays.com/essays/mat...area-of-complex-three-dimensional-objects.php
the author says:
“This means that either the point of the base (25, 4.05) or the top point (0, 5) [must be used to find C]”.

In fact the author has used the point (25, 4.05), not (0, 5) to calculate C.

##C = cos^{-1}(\frac{(y-5.4)}{1.5}) - \frac{2π}{23}x##

##=cos^{-1}(\frac{(4.05-5.4)}{1.5}) - \frac{2π}{23}25 = -4.1389834##
 
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