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

In summary: It's not ∂y/∂x=∂f/∂x*∂y/∂f. (That's not even valid notation; you can't use ∂ for a function of just 'x'.) It's ∂y/∂x=∂f/∂g*∂g/∂x. If you have a function defined by say y=f(x), and x=g(t), then y=f(g(t)). The chain rule relates the derivative of y to the derivatives of f and g.
  • #1
jmher0403
22
0

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?
 
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  • #2
jmher0403 said:

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.
 
  • #3
Can please give me some directions as to how to start on it?
I am really confused...
 
  • #4
jmher0403 said:
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:
[tex] \frac{\partial y(x,t)}{\partial x} = \lim_{h \to 0} \frac{y(x+h,t)-y(x,t)}{h},[/tex]
substitute in the given form of y(x,t), carry out the steps, and see what you get.
 
  • #5
Ray's way is nice, but alternatively you may use the chain rule.
 
  • #6
CAF123 said:
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.
 
  • #7
Ray Vickson said:
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
 
  • #8
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?
 
  • #9
jmher0403 said:
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.
 
  • #10
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.

:(
 
  • #11
jmher0403 said:
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.
 

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

1. What is partial differentiation?

Partial differentiation is a mathematical concept used in multivariable calculus to calculate how a function changes when one variable is changed while holding all other variables constant. It involves taking the derivative of a function with respect to one of its variables, while treating all other variables as constants.

2. Why is partial differentiation important?

Partial differentiation is important because it allows us to analyze the rate of change of a function with respect to specific variables. This is essential in many scientific fields, such as physics, economics, and engineering, where variables are often interdependent and change simultaneously.

3. How is partial differentiation different from regular differentiation?

Regular differentiation, also known as total differentiation, calculates the rate of change of a function with respect to a single variable. Partial differentiation, on the other hand, calculates the rate of change of a multivariable function with respect to one of its variables while holding all other variables constant.

4. Can partial differentiation be applied to any function?

Yes, partial differentiation can be applied to any function that has multiple variables. However, the function must be continuous and differentiable, meaning that it must have a well-defined derivative at each point in its domain.

5. What are some real-world applications of partial differentiation?

Partial differentiation has numerous real-world applications, including optimization problems in economics, determining rates of chemical reactions in chemistry, and analyzing the behavior of complex systems in physics. It is also used in engineering to optimize designs and in finance to calculate risk and return on investment.

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