Recent content by kiwi101

I get it! Thank you so much for bearing with me and explaining this to me step by step!

Is the second partial derivative going to be: ∂2U/∂t = f"(t-R√με) *(1) ∂2U/∂R = f"(t-R√με) * (με)

Can you tell me what the answer for U(r,t) = t2 + R2 would be? Unless I see a visual I am going to remain confused because I know the answer is very obvious. I am just confused.

This is how I interpreted your example x=t2 + R2 ∂U(t,R)/∂t = f'(x) * ∂x/∂t = cos(x) * (2t) = cos(t2 + R2) * 2t

Actually if x = t-R√με & U = f(x) ∂U/ ∂t = df/dx * ∂x/∂t = 1 * 1 = 1

Thanks the example helped So according to this ∂U(t,R)/∂t = t -R√με * (1)

What I am confused about is what are u deriving f'(g(t,R) in terms of? So for example the answer to ∂U(t,R)/∂t = f'g(t,R) * 1 I don't understand where we are going with this. I am sorry for testing your patience but I am confused and I want to learn this

I know how to solve via chain rule but there is no given value of t & R so... ∂U/∂f = ∂U/∂t * dt/dU + ∂U/∂R * dR/dU = (1) * (?) + (√με) * (?)

∂U/∂t = 1 ∂U/∂R = √με So that is the partial derivative of U(t,R) I can't take the second derivative of this because it is constants.

oh okay thank you so much guys!

That means that the function is constant regardless of time at all points of space

Well isn't a twice differentiable function one that has a second derivative? Could you elaborate or demonstrate another example about linking partial derivatives to the ordinary ones

I think it means f"(t-R√με)

By differentiating a general function you mean differentiating t-R√με in terms of t?

Homework Statement Prove by direct substitution that any twice differentiable function of (t-R\sqrt{με}) or of (t+R\sqrt{με}) is a solution of the homogeneous wave equation. Homework Equations Homogeneous wave equation = ∂2U/ ∂R2 - με ∂2U/∂t2 = 0 The Attempt at a Solution Could you...