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Derivative of Cosine with unit vector

  • #1

Homework Statement


Take ∂2E/∂t2 E(r,t)=E0cos((k(u^·r−ct)+φ) in which u^ is a unit vector.

Homework Equations


d/dx(cosx)=-sinx

The Attempt at a Solution


I had calc 3 four years ago and can't for the life of me remember how to differentiate the unit vector. I came up with -c2u2^cos(k(u^r-ct)+φ), but I feel like I'm not doing that right.
 
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Answers and Replies

  • #2
RPinPA
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I think in this equation ##\hat u## is a constant in time. It's the direction of motion of this wave. And ##r## is held constant when doing the partial with respect to ##t##. So you don't need to differentiate it, just treat ##\hat u \cdot r## as a constant, which it is.
 
  • #3
I think in this equation ##\hat u## is a constant in time. It's the direction of motion of this wave. And ##r## is held constant when doing the partial with respect to ##t##. So you don't need to differentiate it, just treat ##\hat u \cdot r## as a constant, which it is.
Ah I made a mistake. I meant to write ∂2E/∂r2
 
  • #4
RPinPA
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How is the unit vector defined? Unit vector in what direction?
 
  • #5
vela
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Ah I made a mistake. I meant to write ∂2E/∂r2
You're using the symbol r to stand for different things, so it's not clear what you're trying to do.

There's the vector ##\vec r##, which is apparently what appears in the argument of the cosine. It wouldn't make sense to write ##r\cdot \hat u## since ##r=|\vec r|## isn't a vector. On the other hand, you say you're trying to calculate ##\partial^2 E/\partial r^2##. Is ##r## here the magnitude of ##\vec{r}##?
 

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