What’s a “nappy”?Diaper, man. Diaper.

The problem is impossible to solve without knowing the value of the exponent that the amount of “doo doos” are multiplied by. Which leads me to say, its your experience. Even calculating the solution for you won’t change that. I am withdrawing myself from this, as it is not going to be helped no matter how I pursue it. And it won’t benefit me much. In fact, a problem like this will only result in more unnecessary experiences if I were to pursue it indefinitely. Maybe someone else can help you.

Piece of cake!This is wrong… :)) ( the situation not the solution )I = a1² + a2² + 2a1a2cos( p2-p1 )

What do you mean, Enabled? Does that spell something gross? I mean, why do you think that equation is wrong?

- Bought between 10 and 49 items
- Exclusive Author
- Has been a member for 5-6 years
- Sold between 100 and 1 000 dollars

igisoftware, i think we did learn most of them.

I dont know serbian but my parents do and i can also use google translator, so ok you can explain it in serbian if you wish.

Thanks

Before I attempted to solve your problem, which I failed to solve correctly, I trusted that… no I was too gullible in believing that simply approaching a problem with the intent to solve it does not prevent me from facing the consequences.

I really learned something from this. I learned how screwed up things can get if I try to solve your problems. Next time you ask something like this on a public forum, to solve your problems, at least warn others who might think about attempting at it, that the whole situation was screwed up from the beginning. Such a problem requires the exact answer. But I learned that my understanding was not enough to prevent my world from being rewritten in such a way that one familiar road I travel ends up in an area where it never should have ended up. That road I traveled tonight took me much longer than it normally would to reach an exit. On top of that, the exit point was on the opposite side of the town. It was not like that before. Now, I don’t know what I can do to fix it, to correct it, to solve it correctly. Now I am the one who needs help. And it isn’t for some 5 mark reward of recognition. I was fortunate I even made it home, considering two things. A “labyrinth-like” situation that changed what I was familiar with, and something that resembled a ship that appeared to me as a triangle, with lights that shone brightly. Standing up for myself has gotten difficult. Sometimes I consider wishing I’d not agreed to accept talking to anyone. I shouldn’t have tried to solve this.

Enabled, it would have been nice if you’d elaborated. But I know you weren’t obligated to. Still, it would have helped.

- Italy
- Sold between 10 000 and 50 000 dollars
- Has been a member for 4-5 years
- Microlancer Beta Tester
- Beta Tester
- Repeatedly Helped protect Envato Marketplaces against copyright violations
- Exclusive Author
- Author had a Free File of the Month
- Bought between 10 and 49 items

this thread reminds me of how bad I am in mathematics. Probably I should tell my boss here at NASA that I don’t really understand the equations they pretend from me to integrate in the new space ship logical unit.

Farting on myself is one of the things I can rely on in order to realize that something is still going how it should. And that, too, is becoming difficult. However humiliating that might seem, I’m telling the truth.

So… repeated in a corrected way;

y1 and y2 are 2D-vectors, given by

y1 = a1.cos(wt-p1)

y2 = a2.cos(wt-p2)

where a1 and a2 are the lengths of the vectors and wt,p1 and p2 are angles (with respect to some axis).

Show that the squared-length I = |y1+y2|^{2 of the sum of the two vectors is given by
I = a12 + a22 +2.a1.a2.cos(p2-p1)}

For the solution

I = |y1+y2|2 = |y1|^{2 + |y2|}2 + 2.y1 times y2

= |y1|^{2 + |y2|}2 + 2.|y1||y2|.cos( angle between y1 and y2 )

= a12 + a22 + 2.a1.a2.cos( (wt-p1)-(wt-p2) )

= a12 + a22 + 2.a1.a2.cos( p2-p1 )

where |y| means the absolute value/length of vector y and y1*y2 means inner product between vectors y1 and y2

I wonder what your professor thinks of this thread?

I also wondered about creating a dud account to play a trick on you, but then I discovered that your home town of Prishtina has two almost identically named universities and it would be even more difficult to search and make a guess at your professor’s name. wicked I know, but it seems I have been foiled! Lol

So… repeated in a corrected way;y1 and y2 are 2D-vectors, given by

y1 = a1.cos(wt-p1)

y2 = a2.cos(wt-p2)

where a1 and a2 are the lengths of the vectors and wt,p1 and p2 are angles (with respect to some axis).

Show that the squared-length I = |y1+y2|^{2 of the sum of the two vectors is given by I = a12 + a22 +2.a1.a2.cos(p2-p1)}For the solution

I = |y1+y2|2 = |y1|

where |y| means the absolute value/length of vector y and y1*y2 means inner product between vectors y1 and y2^{2 + |y2|}2 + 2.y1 times y2

= |y1|^{2 + |y2|}2 + 2.|y1||y2|.cos( angle between y1 and y2 )

= a12 + a22 + 2.a1.a2.cos( (wt-p1)-(wt-p2) )

= a12 + a22 + 2.a1.a2.cos( p2-p1 )

Your solution is partially correct, but explanation is not so good

y1(t) and y2(t) are not vectors they are functions and you can’t apply this formula on functions only on vectors.

He said this is physics problem, so they probably talk about waves, signals or something similar represented with wave function y(t) = A.cos(wt – p)

where A is amplitude, w is frequency, and p is phase.

so y1 and y2 are not vectors, they are cosine waves, where a1 and a2 are amplitude of each wave, w is frequency, t is current time and p1 and p2 are phases(how much each wave lates in time).

This two cosine waves have same frequency, so their sum will be also wave with same frequency but different phase and different amplitude.

I(t) = y1(t) + y2(t) = Iamplitude * cos(wt – p3)

Solution:

let’s stop time at the moment where t=0 we will have:

wt=0;

y1(t) = a1.cos(-p1)

y2(t)=a2.cos(-p2)

now we can draw vectors in phasor diagram

so there are 3 vectors with different lengths. Now we can use cosine law or the method that patrickjansen wrote, to calculate length of Iamplitude.

if we move begining of the a1 vector at the and of a2 we will get triangle with sides a1, a2 and Iamplitude. We can use cosine law to calculate length of Iamplitude.

so we have now:

Iamplitude ^ 2 = a1 ^ 2 + a2 ^ 2 – 2.a1.a2.cos(180-p2+p1)

where 180-p2-p1 is angle between a1 and a2 in formed triangle.

knowing that cos(180-x)=-cos(x)

Iamplitude ^ 2 = a1 ^ 2 + a2 ^ 2 + 2.a1.a2.cos(p2-p1)

and that’s it

- Bought between 10 and 49 items
- Exclusive Author
- Has been a member for 5-6 years
- Sold between 100 and 1 000 dollars

Hey igisoftware, thats great explanation … I know understand it, and why he told us that we can never solve it

Thanks everyone for trying to help me.