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Differential equations proof

by th3chemist
Tags: differential, equations, proof
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th3chemist
#1
Nov8-12, 09:50 AM
P: 66
1. The problem statement, all variables and given/known data
Show that Acos(ωt) + Bsin(ωt) can be written in the form r*sin(ωt - θ). Determine r and θ in terms of A and B. If Rcos(ωt - ε) = r*sin(ωt - θ), deermine the relationship among R, r, ε and θ.

2. Relevant equations

Trig identities.

3. The attempt at a solution

I'm really confused on this question. What I first tried to do is to use the sum-difference forumla on r*sin(ωt - θ) = r*sin(ωt)cos(θ) - r*cos(ωt)sin(θ).
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LCKurtz
#2
Nov8-12, 11:57 AM
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Quote Quote by th3chemist View Post
1. The problem statement, all variables and given/known data
Show that Acos(ωt) + Bsin(ωt) can be written in the form r*sin(ωt - θ). Determine r and θ in terms of A and B. If Rcos(ωt - ε) = r*sin(ωt - θ), deermine the relationship among R, r, ε and θ.

2. Relevant equations

Trig identities.

3. The attempt at a solution

I'm really confused on this question. What I first tried to do is to use the sum-difference forumla on r*sin(ωt - θ) = r*sin(ωt)cos(θ) - r*cos(ωt)sin(θ).
Start by writing your expression like this$$
\sqrt{A^2+B^2}\left( \frac A {\sqrt{A^2+B^2}}\cos(\omega t)+ \frac B {\sqrt{A^2+B^2}}
\sin(\omega t) \right)$$That quantity in the large parentheses looks like an addition formula. Think about a right triangle with legs ##A## and ##B##.
th3chemist
#3
Nov8-12, 12:04 PM
P: 66
Is that from the Pythagorean theorem?
I'm still unsure how to set this problem up :(

LCKurtz
#4
Nov8-12, 12:09 PM
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Differential equations proof

Quote Quote by th3chemist View Post
Is that from the Pythagorean theorem?
I'm still unsure how to set this problem up :(
Did you draw the triangle I suggested? Do those fractions look like sines and cosines of some angle? Can you make the expression in large parentheses look like an addition formula?
th3chemist
#5
Nov8-12, 12:14 PM
P: 66
It would be cos(theta) and sin(theta)
LCKurtz
#6
Nov8-12, 12:37 PM
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Quote Quote by th3chemist View Post
It would be cos(theta) and sin(theta)
So.......?
th3chemist
#7
Nov8-12, 12:41 PM
P: 66
Quote Quote by LCKurtz View Post
So.......?
If I use the sum-difference identity I would get cosine term not the sine :(
LCKurtz
#8
Nov8-12, 12:59 PM
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Isn't ##\displaystyle{\frac A {\sqrt{A^2+B^2}}}## the cosine of some angle in your triangle?
th3chemist
#9
Nov8-12, 01:09 PM
P: 66
Quote Quote by LCKurtz View Post
Isn't ##\displaystyle{\frac A {\sqrt{A^2+B^2}}}## the cosine of some angle in your triangle?
yeah so you get cos(θ)cos(ωt) + sin(θ)(sin(ωt))
LCKurtz
#10
Nov8-12, 01:36 PM
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Quote Quote by th3chemist View Post
yeah so you get cos(θ)cos(ωt) + sin(θ)(sin(ωt))
This gives you the form ##r\cos(\omega t -\theta)## (don't forget there is that square root out front). Now getting it in a sine form is what I think last sentence in your problem is about.
th3chemist
#11
Nov8-12, 01:46 PM
P: 66
Quote Quote by LCKurtz View Post
This gives you the form ##r\cos(\omega t -\theta)## (don't forget there is that square root out front). Now getting it in a sine form is what I think last sentence in your problem is about.
But I did not show the first part. I'm quite sure though that if you have rcos(θ) and rsin(θ) they're constants.
th3chemist
#12
Nov8-12, 01:59 PM
P: 66
Here is something else I did. I got r*sin(wt)cos(theta) - r*cos(wt)sin(theta)
Then I set rewrote as r*cos(theta) + (-rsin(theta)cos(wt)

I can then get A = r*cos(theta) and B = (-r*sin(theta))
LCKurtz
#13
Nov8-12, 02:45 PM
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Quote Quote by th3chemist View Post
But I did not show the first part.
Haven't we shown$$
A\cos(\omega t)+B\sin(\omega t) = \sqrt{A^2+B^2}\left( \frac A {\sqrt{A^2+B^2}}\cos(\omega t)+ \frac B {\sqrt{A^2+B^2}}
\sin(\omega t) \right) = \sqrt{A^2+B^2}(\cos\theta\cos\omega t + \sin\theta \sin(\omega t) = \sqrt{A^2+B^2}\cos(\omega t - \theta)$$To get it to a sine function I think you have to work the last sentence in the original problem.
th3chemist
#14
Nov8-12, 03:39 PM
P: 66
Quote Quote by LCKurtz View Post
Haven't we shown$$
A\cos(\omega t)+B\sin(\omega t) = \sqrt{A^2+B^2}\left( \frac A {\sqrt{A^2+B^2}}\cos(\omega t)+ \frac B {\sqrt{A^2+B^2}}
\sin(\omega t) \right) = \sqrt{A^2+B^2}(\cos\theta\cos\omega t + \sin\theta \sin(\omega t) = \sqrt{A^2+B^2}\cos(\omega t - \theta)$$To get it to a sine function I think you have to work the last sentence in the original problem.
What can I do though? Use the identity on both sides? The first line states to write it in the sin form. We did it in cos :(.
LCKurtz
#15
Nov8-12, 06:03 PM
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Quote Quote by LCKurtz View Post
Haven't we shown$$
A\cos(\omega t)+B\sin(\omega t) = \sqrt{A^2+B^2}\left( \frac A {\sqrt{A^2+B^2}}\cos(\omega t)+ \frac B {\sqrt{A^2+B^2}}
\sin(\omega t) \right) = \sqrt{A^2+B^2}(\cos\theta\cos\omega t + \sin\theta \sin(\omega t) = \sqrt{A^2+B^2}\cos(\omega t - \theta)$$To get it to a sine function I think you have to work the last sentence in the original problem.
Quote Quote by th3chemist View Post
What can I do though? Use the identity on both sides? The first line states to write it in the sin form. We did it in cos :(.
Yes, we have the cosine. Did you read my previous post, copied above? That is part of your problem, to figure out how to change the cosine to a sine. I have helped you with half of your problem. Now you work on finishing it.
th3chemist
#16
Nov8-12, 06:34 PM
P: 66
Is A = r*cos(theta) and B = (-r*sin(theta)) correct for the first part then? To get theta and r in terms of A and B?
th3chemist
#17
Nov8-12, 07:38 PM
P: 66
Hold on, I have a question. How can you assume that will give you cosine instead of sine? If I switch, A and B I would get sine.
LCKurtz
#18
Nov8-12, 08:16 PM
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Quote Quote by th3chemist View Post
Is A = r*cos(theta) and B = (-r*sin(theta)) correct for the first part then? To get theta and r in terms of A and B?
You have A and B in terms of r and theta. You want r and theta in terms of A and B.


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