Is the amplitude is the maximum displacement in the y direction? So I have to find the derivative?
That I do understand, but I found many different equations for Φ and I don't know which one is the correct one. For example, here it says that ##Φ=NBAcosθ## , on this website ##Φ=NBA## and...
Homework Statement
[/B]
A radio transmitter radiates isotropically at the frequency of 90.8 MHz. The peak magnetic field at
a receiver, 9km from the transmitter, is ##9x10^{-10}T##. Calculate the maximum amplitude of the induced emf in a 12 turn coil with area A = ##90cm^{2}## at the receiver...
Now I got it! :partytime:
##-2R(R^2+Rr+Rr+r^2)+R^3+3R^2r+3Rr^2+r^3=0##
##-2R^3-2R^2r+(-2R^2r)-2Rr^2+R^3+3R^2r+3Rr^2+r^3=0##
##-R^3-R^2r+Rr^2+r^3=0##
and then if I substitute R=r, I get the same as you did: ##-R^3-R^3+R^3+R^3=0##
Thank you!
But now I still have to find the absolute maximum, or...
Well, the numerator was ##-2RV^2##, so I rewrote it as ##-2R(-2)V^2##.
But I just realized that this was nonsense because it would only work if the numerator was ##-2(R+V^2)##.
I'll try it again, give me a few minutes.
But if I substitute R+r with S, then I have three variables. And I don't understand where the ##\frac{P}{V^2}## is coming from. If we have a P there, then there are four variables (S, V, P and R).
Ok, that makes sense to me now. If I do it, it looks like this:
##\frac{-2R(-2)V^2}{(R+r)^3}+\frac{V^2}{(R+r)^2}=0##
divide by ##V^2## :
##\frac{4R}{(R+r)^3}+\frac{1}{(R+r)^2}=0##
common denominator:
##\frac{4R(R+r)^2}{(R+r)^5}+\frac{(R+r)^3}{(R+r)^5}=0##
multiply by common denominator...
I understand now how to find the derivative but I don't understand how to simplify to ##-R^3-R^2r+Rr^2+r^3 =0##
How did you get R^2r or Rr^2? I tried using the binomial theorem for (R+r)^2 and (R+r)^3 but that made everything even more confusing.
What I ended up with is basically this...
Thank you so much for your help!
I thought that the derivative might be wrong. Thanks for the idea to convert the equation into a product. It's so much easier, I'm always going to do that from now on.
Anyways, I tried it on my own, but I got something different than you have. I'm not sure if my...