Finding the horizontal shift of a function

In summary: This means that either the point of the base (25, 4.05) or the top point (0, 5) [must be used to find C]The author has used the point (25, 4.05), not (0, 5) to calculate C.
  • #1
Einstein44
125
31
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... :)
 
Physics news on Phys.org
  • #2
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.
 
  • Informative
Likes Einstein44
  • #3
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:
  • #4
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..
 
  • Like
Likes Einstein44
  • #5
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.
 
  • Like
Likes Einstein44
  • #6
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


?
 
  • Like
Likes Einstein44 and FactChecker
  • #7
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.
 
  • #8
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.
 
  • Like
  • Informative
Likes FactChecker and Einstein44
  • #9
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!.
 
  • Like
Likes Steve4Physics
  • #11
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##
 
  • Like
Likes Einstein44

1. What is the horizontal shift of a function?

The horizontal shift of a function is the amount by which the graph of the function is moved horizontally from its original position. It is also known as the horizontal translation.

2. How do you find the horizontal shift of a function?

To find the horizontal shift of a function, you need to determine the difference between the x-coordinate of the original function and the x-coordinate of the shifted function. This difference will give you the amount of shift in the x-direction.

3. What is the formula for finding the horizontal shift of a function?

The formula for finding the horizontal shift of a function is given by:
Horizontal shift = x-coordinate of shifted function - x-coordinate of original function.

4. Can the horizontal shift of a function be negative?

Yes, the horizontal shift of a function can be negative. This means that the graph of the function has shifted to the left. A positive horizontal shift indicates a shift to the right.

5. How does the horizontal shift affect the graph of a function?

The horizontal shift affects the graph of a function by moving the entire graph to the left or right. This means that all the points on the graph will be shifted by the same amount in the x-direction. The shape of the graph remains the same, but its position changes.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
3
Views
1K
  • Precalculus Mathematics Homework Help
Replies
18
Views
1K
  • Precalculus Mathematics Homework Help
Replies
9
Views
1K
Replies
3
Views
793
Replies
4
Views
661
  • Precalculus Mathematics Homework Help
Replies
15
Views
613
  • Atomic and Condensed Matter
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
934
  • Precalculus Mathematics Homework Help
Replies
8
Views
766
  • Precalculus Mathematics Homework Help
Replies
11
Views
492
Back
Top