I know it varies. So my problem.

I also have a problem:My child uses around 23 nappys per week.

Which model’ll be cheaper thinking in a correlative period of two years?

I have 5 different models in the supermarket (a, b, c, d, e). Each one differs 0,01 from the others (the price is for each nappy, not for the whole box). The content of the different boxes are also different: 32, 32, 34, 40, 42.

Assuming that my child will have a bigger shit in an exponential number of 0.002 grams per month and that i’ll forget to feed her at least 1 night per week…

LOL this is real example of applied math

What’s a “nappy”?

Diaper, man. Diaper.

doesn’t this formula help? cos(a-b)=cos(a)cos(b) +sin(a)sin(b)

not really

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

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.

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

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.

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