Hi,
I am trying to graph the following function with a Casio 9850 GB+.
y = 6.5sin(pi/6) (t-C) + D
However, I only get a straight line when i do this...any body able to help me through this?
John
How To Graph This General Sine Function?
Started by
yoonglai2001
, Apr 09 2005 05:42 AM
9 replies to this topic
#1
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Posted 09 April 2005 - 05:42 AM
#2
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Posted 09 April 2005 - 09:03 AM
6.5sin(pi/6) is a constant, t-C too unless you select "type"->"param" in graph menu. What do you want to draw?
In the case you mean "sin(Xpi/6)" (and just forgot the X in your post) you probably have switched your calc to DEG instead of RAD.
In the case you mean "sin(Xpi/6)" (and just forgot the X in your post) you probably have switched your calc to DEG instead of RAD.
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Posted 09 April 2005 - 01:34 PM
6.5sin(pi/6) is a constant, t-C too unless you select "type"->"param" in graph menu. What do you want to draw?
In the case you mean "sin(Xpi/6)" (and just forgot the X in your post) you probably have switched your calc to DEG instead of RAD.
Hi
No, I didn't forget the X...there is none..the calculator is set to rad, and the t would be X, i guess...on the calculator i typed in Y1 =6.5sin(pi/6)(x-C) +20.5, where x would be the independent variable. However I didn't set param..i though that was just for circles, not trigonometric functions..
what should i do?
John
#4
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Posted 09 April 2005 - 01:51 PM
No if you use X you needn't to set param (it's when you use T only where you can specify a function then each for X and Y ordinates dependent on T for non unique x->y mappings, for example xt=sin T, yt=cos T would give a circle but you can draw other things too)
NOTE: there's a difference between sin 3X and sin (3)(X) on that calc
What are the view window settings and what's the value of C? Btw. do you expect to see a sine curve? Then you have to set X within the brackets belonging to sin, e.g. sin((pi/6)/(x-C)), else X is not used for the argument.6.5sin(pi/6)(x-C) +20.5
NOTE: there's a difference between sin 3X and sin (3)(X) on that calc
#5
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Posted 09 April 2005 - 08:54 PM
With that expression you should see a straight line that looks like the general expression Y=A*X+B in your case Y=A*(X-C)+B, the only difference is that the second is shifted to the left/right depending on the value of C
#6
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Posted 10 April 2005 - 12:54 AM
My task was to model temperatures using the general sine function y=AsinB(x-C)+D.
the question went like this:
Below is a table which shows the mean monthly maximum temperature (degrees centigrade) for a city in Greece.
Month Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec
Temp 15 14 15 18 21 25 27 26 24 20 18 16
Here my reasoning:
I have to get a model in the form of the general sine function of
y=AsinB(x-C)+10
The amplitude was max(27)-min(14)/2 for 6.5, which is A
Since the period is twelve months, setting 2pi/B=12 gives us a B value of pi/6
Now the principal axis is max(27)+min(14)/2 which gives us 20.5, and a model of y=6.5sin(pi/6)(t-C) +20.5 where t is the month.
At t=4.9 and 9.9, the curve touches the principal axis, y=20.5. This could give us a C value of 9.9 or 4.9. How do I know which one to use,
i.e y=6.5sin(pi/2)(x-4.9)+20.5 or y=6.5sin(pi/6)(x-9.9)+20.5?
The textbook book gives y=7.5sin(pi/2)(x-9.9)+20.5 as the answer...I assume they must be wrong (typo) with the A value (7.5) , but why not 4.9?
Very confused here....maybe the book is wrong...
Now I know this is not a math forum, but I put the problem in to show you what I was trying to do. The C value determines the horizontal translation of the sine curve, but I see two possible C values from my reasoning. The problem then, was how to use technology, i.e. The Casio 9850 GBPlus to graph the different models. I kept getting the straight line. I do see how it could look like a linear function the way I have input it.
So you can see, there are actually two problems here. One, the reason why 4.9 could not be used as a C value for a model, and two, the technique to graph two sine curves with different C values, following that general model. The X in the equation would obviously correspond to the month..i.e. jan =1, May =5, etc.
I am still listening...just confused. It seems technology could really help here in determining the bet model.
John
the question went like this:
Below is a table which shows the mean monthly maximum temperature (degrees centigrade) for a city in Greece.
Month Jan Feb Mar Apr May Jun Jul Aug Sept Oct Nov Dec
Temp 15 14 15 18 21 25 27 26 24 20 18 16
Here my reasoning:
I have to get a model in the form of the general sine function of
y=AsinB(x-C)+10
The amplitude was max(27)-min(14)/2 for 6.5, which is A
Since the period is twelve months, setting 2pi/B=12 gives us a B value of pi/6
Now the principal axis is max(27)+min(14)/2 which gives us 20.5, and a model of y=6.5sin(pi/6)(t-C) +20.5 where t is the month.
At t=4.9 and 9.9, the curve touches the principal axis, y=20.5. This could give us a C value of 9.9 or 4.9. How do I know which one to use,
i.e y=6.5sin(pi/2)(x-4.9)+20.5 or y=6.5sin(pi/6)(x-9.9)+20.5?
The textbook book gives y=7.5sin(pi/2)(x-9.9)+20.5 as the answer...I assume they must be wrong (typo) with the A value (7.5) , but why not 4.9?
Very confused here....maybe the book is wrong...
Now I know this is not a math forum, but I put the problem in to show you what I was trying to do. The C value determines the horizontal translation of the sine curve, but I see two possible C values from my reasoning. The problem then, was how to use technology, i.e. The Casio 9850 GBPlus to graph the different models. I kept getting the straight line. I do see how it could look like a linear function the way I have input it.
So you can see, there are actually two problems here. One, the reason why 4.9 could not be used as a C value for a model, and two, the technique to graph two sine curves with different C values, following that general model. The X in the equation would obviously correspond to the month..i.e. jan =1, May =5, etc.
I am still listening...just confused. It seems technology could really help here in determining the bet model.
John
No if you use X you needn't to set param (it's when you use T only where you can specify a function then each for X and Y ordinates dependent on T for non unique x->y mappings, for example xt=sin T, yt=cos T would give a circle but you can draw other things too)
What are the view window settings and what's the value of C? Btw. do you expect to see a sine curve? Then you have to set X within the brackets belonging to sin, e.g. sin((pi/6)/(x-C)), else X is not used for the argument.
#7
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Posted 10 April 2005 - 07:12 AM
go to the stat menu of your calc.
In the first List type the numbers from 1 - 12 (Months)
In the second type the temperatures.
Then Select GRPH (F1) --> SET and set Graph1 (Topmost enrtry) to
Graph-Type: Scatter (Scat)
X-List: List1
Y-List: List2
Frequency: List 1
Marktype: (doesn't matter).
Type EXE.
Now select graph again and select Graph1 (the one you just set)
Now the points will be set into the window.
Now select CALC (Don't know the F Key on FX Calcs) and select SIN.
Now the sine function is determined. Select copy and copy the function to
Graph Menu.
Then Select Draw to see if it looks right (goes through most points or at least nearly through them).
Then Press Menu, Go to Graph Menu and write down the function shown there. This should be a sine function.
From your values I got the Equation:
6.145*sin (0.575X-2.694)+20.39
This should be pretty exact (got in RAD Mode: Setup -> Angle -> RAD).
Tell me if it was the right solution.
Cu huhn
In the first List type the numbers from 1 - 12 (Months)
In the second type the temperatures.
Then Select GRPH (F1) --> SET and set Graph1 (Topmost enrtry) to
Graph-Type: Scatter (Scat)
X-List: List1
Y-List: List2
Frequency: List 1
Marktype: (doesn't matter).
Type EXE.
Now select graph again and select Graph1 (the one you just set)
Now the points will be set into the window.
Now select CALC (Don't know the F Key on FX Calcs) and select SIN.
Now the sine function is determined. Select copy and copy the function to
Graph Menu.
Then Select Draw to see if it looks right (goes through most points or at least nearly through them).
Then Press Menu, Go to Graph Menu and write down the function shown there. This should be a sine function.
From your values I got the Equation:
6.145*sin (0.575X-2.694)+20.39
This should be pretty exact (got in RAD Mode: Setup -> Angle -> RAD).
Tell me if it was the right solution.
Cu huhn
#8
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Posted 11 April 2005 - 08:48 AM
Yes, I got exactly that as well, using your technique. Quite nice, but I see the form is y=Asin(Bx-C) + D, not AsinB(x-C) +D as in my textbook, for a general sine function. How do I resolve that?
John
John
go to the stat menu of your calc.
In the first List type the numbers from 1 - 12 (Months)
In the second type the temperatures.
Then Select GRPH (F1) --> SET and set Graph1 (Topmost enrtry) to
Graph-Type: Scatter (Scat)
X-List: List1
Y-List: List2
Frequency: List 1
Marktype: (doesn't matter).
Type EXE.
Now select graph again and select Graph1 (the one you just set)
Now the points will be set into the window.
Now select CALC (Don't know the F Key on FX Calcs) and select SIN.
Now the sine function is determined. Select copy and copy the function to
Graph Menu.
Then Select Draw to see if it looks right (goes through most points or at least nearly through them).
Then Press Menu, Go to Graph Menu and write down the function shown there. This should be a sine function.
From your values I got the Equation:
6.145*sin (0.575X-2.694)+20.39
This should be pretty exact (got in RAD Mode: Setup -> Angle -> RAD).
Tell me if it was the right solution.
Cu huhn
#9
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Posted 11 April 2005 - 09:14 AM
y=Asin(Bx-C) + D -> AsinB(x-C) +D
divide C by B and you have the new C and can write the B in front of the
brackets. (What class are you man
)
divide C by B and you have the new C and can write the B in front of the
brackets. (What class are you man
)
#10
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Posted 11 April 2005 - 12:49 PM
Hi,
Yeah, I worked that out just after I sent the post. This is grade 11 math...
thanks again
John
Yeah, I worked that out just after I sent the post. This is grade 11 math...
thanks again
John
y=Asin(Bx-C) + D -> AsinB(x-C) +D
divide C by B and you have the new C and can write the B in front of the
brackets. (What class are you man
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