I Deriving expression as a function of another variable

AI Thread Summary
To derive an expression as a function of another variable, such as f in the equation B(r) = Bsin(A f r + θ), one can manipulate the equation by first rewriting it as g(r) = Bsin(Afr + θ). B is confirmed to be a constant, and the process involves dividing both sides by B and applying the inverse sine function. While some participants suggested using Laplace transforms, it was clarified that the problem is simpler and does not directly relate to calculus. The discussion highlights the importance of understanding trigonometric functions and their inverses in this context. Overall, the thread emphasizes a straightforward approach to isolating f in the given expression.
kenef
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I am attempting to understand how I can derive a new expression as a function of another variable in this existing expression. This is a college undergraduate course but I can't recall doing this much before, maybe in high school. I have taken all the way up to ODE's and PDE's.

The confusion that begins to arise is when attempting to manipulate this expression due to its trigonometric function.

With that being said, if one has an equation such that:

B(r) = Bsin(A f r + θ)

How can one solve to express this as a function of let's say f?
Any insight towards mathematical properties would be greatly appreciated!
 
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mfb said:
What does "Bsin(A f r + θ)" mean? Is B a constant here? On the left side it appears to be a function.

You'll need inverse trigonometric functions.
Yes B is a constant, let's say the original amount.

I was thinking that perhaps one might need to use Laplace transform
 
If ##x = \sin(y)##, then ##y = \sin^{-1}(x)##. Or to use an alternative notation, ##y = \arcsin(x)##. But surely you've seen this already, if you've studied ODEs and PDEs.
 
kenef said:
B(r) = Bsin(A f r + θ)

kenef said:
I was thinking that perhaps one might need to use Laplace transform
Laplace transforms? No, this is much simpler than that.

Let's rewrite the equation above as g(r) = Bsin(Afr + θ) so that B doesn't have two different roles.
1. Divide both sides by B
2. Take the inverse sine of both sides
3. Etc.
 
Thread moved as it doesn't appear to have anything to do directly with Calculus...
 
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Mark44 said:
Thread moved as it doesn't appear to have anything to do directly with Calculus...
Thank you, it was originally under General Math, and someone moved it to Calculus. Its not really related to "Calculus" as you have said. I will take your suggestion for the solution, thank you!
 
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