Derivative of Cosine with unit vector

In summary, the conversation discusses differentiating the unit vector in the equation ∂2E/∂t2 E(r,t)=E0cos((k(u^·r−ct)+φ), with the goal of calculating ∂2E/∂r2. It is determined that in this equation, the unit vector is a constant in time and the vector r is held constant when taking the partial derivative with respect to t. It is also mentioned that there is some confusion as to what r represents in the equation and whether it is the magnitude of the vector or something else.
  • #1
Blue Kangaroo
41
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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  • #2
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
RPinPA said:
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
How is the unit vector defined? Unit vector in what direction?
 
  • #5
Blue Kangaroo said:
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}##?
 

What is the derivative of cosine with a unit vector?

The derivative of cosine with a unit vector is a vector that represents the slope of the cosine curve at a specific point. It is calculated by multiplying the unit vector by the negative sine of the angle at that point.

How is the derivative of cosine with a unit vector related to the unit circle?

The derivative of cosine with a unit vector is closely related to the unit circle because the unit vector represents the direction from the origin to a point on the circle, and the derivative represents the direction of the tangent line at that point.

Why is the derivative of cosine with a unit vector important in physics?

The derivative of cosine with a unit vector is important in physics because it is used to calculate the velocity and acceleration of an object moving in circular motion. It is also used in other areas of physics, such as electromagnetism and quantum mechanics.

Can the derivative of cosine with a unit vector be negative?

Yes, the derivative of cosine with a unit vector can be negative. This indicates that the slope of the cosine curve at that point is decreasing, or that the object is moving in a clockwise direction on the unit circle.

How can the derivative of cosine with a unit vector be used to find the rate of change of a function?

The derivative of cosine with a unit vector can be used to find the rate of change of a function by multiplying it by the magnitude of the unit vector and the derivative of the function with respect to the angle. This gives the instantaneous rate of change at a specific point on the function.

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