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Question about Solve
08-25-2015, 07:16 AM (This post was last modified: 08-25-2015 07:22 AM by salvomic.)
Post: #1
Question about Solve
hi,
an Italian student asked to me what's the best way to solve these equations in Solve:

E1: A*(2*π*10)*COS(P)+11*A*SIN(P) = 0
E2: 11*A*COS(P)-A*(2*π*10)*SIN(P) = 23
where A is amplitude, P phase, the rest 2*PI*frequency (here 10 Hz)

I suggest only to set the Prime in DEG mode (he is to get something like 0.36 and 80°, he said).
I get different values with different invite suggestion for A and P (but they are values MOD 360, I think), however also about 0 and about 99.93 (giving A=0.001); giving for A 2 and for P 0 I get about 0 and about 80 also...
For examples the student got -20350160.0698 whit his tries:
-20350160.0698 MOD 360 is about 280 (about 279.93), that's 360-80...
Whit my values (about -0.36 and 99.93) the equations are consistent but a negative amplitude hasn't a physic sense.

But the guy is searching a general good solution if he doesn't know what invite values to give...
Suggestions?

Salvo

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08-25-2015, 10:27 AM
Post: #2
RE: Question about Solve
when i remember correct, Setting the angular mode to deg isn't a good aproach. This should be best in radians.

regards

Wolfgang
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08-25-2015, 10:29 AM
Post: #3
RE: Question about Solve
(08-25-2015 10:27 AM)ww63 Wrote:  when i remember correct, Setting the angular mode to deg isn't a good aproach. This should be best in radians.

regards

Wolfgang

hi Wolfgang, you're right, but in this problem he must calculate phase in DEG (sin, cos), as usually speaking about electricity...

Salvo

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08-25-2015, 12:44 PM
Post: #4
RE: Question about Solve
Hi Salvo,

You are both right. If you plot the equations in the advance graphing app (changing with P as X and A as Y) you will see there is an infinite number of solution:

   

and here is one solution zoomed in:

   

Road
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08-25-2015, 12:47 PM
Post: #5
RE: Question about Solve
(08-25-2015 12:44 PM)roadrunner Wrote:  Hi Salvo,

You are both right. If you plot the equations in the advance graphing app (changing with P as X and A as Y) you will see there is an infinite number of solution:
...
Road

in fact, Road!
this is the actual solution: an infinite number of solutions.
Thank you, I'll say about this to the student.

Salvo

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08-25-2015, 02:00 PM (This post was last modified: 08-25-2015 02:03 PM by DrD.)
Post: #6
RE: Question about Solve
(08-25-2015 07:16 AM)salvomic Wrote:  hi,
an Italian student asked to me what's the best way to solve these equations in Solve:

E1: A*(2*π*10)*COS(P)+11*A*SIN(P) = 0
E2: 11*A*COS(P)-A*(2*π*10)*SIN(P) = 23
where A is amplitude, P phase, the rest 2*PI*frequency (here 10 Hz)

I suggest only to set the Prime in DEG mode (he is to get something like 0.36 and 80°, he said).

Here is the solution he was seeking (Advanced Graphing App with X=P, Y=A):
   
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08-25-2015, 03:43 PM
Post: #7
RE: Question about Solve
(08-25-2015 02:00 PM)DrD Wrote:  Here is the solution he was seeking (Advanced Graphing App with X=P, Y=A):

thank you Dale,
this is the solution he was seeking, however, as Road wrote, it shouldn't be unique...

Salvo

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08-25-2015, 07:50 PM
Post: #8
RE: Question about Solve
There was no mention of boundary conditions in the problem, but you did mention an electrical context. The important solution he was after occurs in the first 90°, and is periodic forevermore. Since the phase is constant, after the first cycle there wouldn't be much point in reporting periodic values for subsequent cycles. For this problem, perhaps that's why he was seeking that particular result(?)
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08-25-2015, 07:52 PM
Post: #9
RE: Question about Solve
(08-25-2015 07:50 PM)DrD Wrote:  There was no mention of boundary conditions in the problem, but you did mention an electrical context. The important solution he was after occurs in the first 90°, and is periodic forevermore. Since the phase is constant, after the first cycle there wouldn't be much point in reporting periodic values for subsequent cycles. For this problem, perhaps that's why he was seeking that particular result(?)

yes, you're right. So, 80° is the right solution, I think he would to find only this one.

Salvo

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