Verifying Solution for Partial Differentiation of a Function of x-ct

[jmher0403]

Homework Statement

y(x,t) = f(x-ct)

verify this solution satisfies equation
∂y²/∂x² = 1/c²*∂y²/∂t²

Homework Equations

The Attempt at a Solution

∂y/∂x = ∂f/∂x = 1
∂y²/∂x² = 0


∂y/∂t = ∂f/∂t = -c
∂y²/∂t² = 0

Is this the way to do it?

 

[Ray Vickson]

Homework Statement

y(x,t) = f(x-ct)

verify this solution satisfies equation
∂y²/∂x² = 1/c²*∂y²/∂t²

Homework Equations

The Attempt at a Solution

∂y/∂x = ∂f/∂x = 1
∂y²/∂x² = 0


∂y/∂t = ∂f/∂t = -c
∂y²/∂t² = 0

Is this the way to do it?

Start over: everything you did was wrong.

 

[jmher0403]

Can please give me some directions as to how to start on it?
I am really confused...

 

[Ray Vickson]

Can please give me some directions as to how to start on it?
I am really confused...

OK: to see how to get $\partial y / \partial x$, use the definition:
\frac{\partial y(x,t)}{\partial x} = \lim_{h \to 0} \frac{y(x+h,t)-y(x,t)}{h},
substitute in the given form of y(x,t), carry out the steps, and see what you get.

 

[CAF123]

Ray's way is nice, but alternatively you may use the chain rule.

 

[Ray Vickson]

Ray's way is nice, but alternatively you may use the chain rule.

I agree, but did not want to suggest that. The OP seems to have a fundamental conceptualization problem, and just having him/her apply some "rules" without thinking seemed to me to be counterproductive. I would rather have the OP grind through things from first principles.

 

[CAF123]

I agree, but did not want to suggest that. The OP seems to have a fundamental conceptualization problem, and just having him/her apply some "rules" without thinking seemed to me to be counterproductive. I would rather have the OP grind through things from first principles.

Ok, that makes sense

 

[jmher0403]

I'm getting similar answer...

∂y/∂x = lim (x+h-ct-x+ct)/h
= lim h/h
= lim 1
=1
∂y/∂t = lim (x-ct+h-x+ct)/h
= lim 1
= 1

am i on the right track?

 

[Dick]

I'm getting similar answer...

∂y/∂x = lim (x+h-ct-x+ct)/h
= lim h/h
= lim 1
=1
∂y/∂t = lim (x-ct+h-x+ct)/h
= lim 1
= 1

am i on the right track?

No. ∂y/∂x = lim (f((x+h)-ct)-f(x-ct))/h=lim (f((x-ct)+h)-f(x-ct)). It's not 1. You left out f altogether. And I would suggest if you know the chain rule (and you probably should) then use it. Then look back at how that difference quotient will let you prove it.

 

[jmher0403]

Ok..if I try doing chain rule

∂y/∂x = ∂f/∂x * ∂y/∂f

right?

I have only dealt with questions where the function is actually given.. like the terms are defined

something like y(x,t) = ax + xt + xt^2 blah blah

then I would keep one of the x and t as a constant and differentiate.

but I am really stuck as to how to this question as it only says function of x-ct.

:(

 

[Dick]

Ok..if I try doing chain rule

∂y/∂x = ∂f/∂x * ∂y/∂f

right?

I have only dealt with questions where the function is actually given.. like the terms are defined
but I am really stuck as to how to this question as it only says function of x-ct.

:(

y=f(x-ct). Define g=(x-ct). Then y=f(g). ∂y/∂x=f'(g)*∂g/∂x. f is a function of a single variable. I'm not sure you are getting this chain rule thing very well.