Partial derivatives and change of variables

In summary, the student attempted to solve a homework equation but was confused. He re-arranged the variables and then had trouble applying the partial derivatives. He was finally able to solve the equation by integrating it once with respect to r and then again with respect to s.
  • #1
miew
27
0

Homework Statement



Sorry I tried to use Latex but it didn't work out :/

Make the change of variables r = x + vt and s = x vt in the wave equation
partial^2y/partialx^2-(1/v^2)(partial^2y/partialt^2)=0

Homework Equations


partialy/patialx=(partialy/partialr)(partialr/partialx)+(partialy/partials)(partials/partialx)
partialy/partialy=(partialy/partialr)(partialr/partialt)+(partialy/partials)(partials/partialy)

partialr/partialx=1
partials/partialx=1
partialr/partialt=v
partials/partialt=-v


The Attempt at a Solution


I am a little bit confused about what to do. This is what I did:

Rearranging:
partial^2y/partialr^2(1-1/v)+partial^2y/partials^2(1+1/v)=0

I feel this is completely wrong, but I can't think of another way of doing it.
 
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  • #2
Start from the basics:

[tex]
\begin{array}{ccc}
\frac{\partial}{\partial x} & = & \frac{\partial r}{\partial x} \frac{\partial}{\partial r}+\frac{\partial s}{\partial x} \frac{\partial}{\partial s} \\
\frac{\partial}{\partial t} & = & \frac{\partial r}{\partial t} \frac{\partial}{\partial r}+\frac{\partial s}{\partial t} \frac{\partial}{\partial s}
\end{array}
[/tex]

Computing these yields:

[tex]
\begin{array}{ccc}
\frac{\partial}{\partial x} & = & \frac{\partial}{\partial r}+\frac{\partial}{\partial s} \\
\frac{\partial}{\partial t} & = & v\frac{\partial}{\partial r}-v\frac{\partial}{\partial s}
\end{array}
[/tex]

Then just apply these again to get the second order derivatives.
 
  • #3
Yes, that is what i did and I got:

partial^2y/partialx^2=partial^2y/partialr^2+partial^2y/partials^2
partial^2y/partialt^2=vpartial^2y/partialr^2-v(partial^2y/partials^2)

Is that right? And then I just plugged that in the initial equation.
 
  • #4
Just use:
[tex]
\frac{\partial^{2}}{\partial x^{2}}=\left(\frac{\partial}{\partial r}+\frac{\partial }{\partial s}\right)\left(\frac{\partial}{\partial r}+\frac{\partial }{\partial s}\right)
[/tex]
 
  • #5
Could you explain that ? And why what I did is wrong ? Can't I just take the partial derivative of partialy/partialx ?
 
  • #6
Where dies y come into it, I thought the original variables were x and t which you were transforming into r and s?
Okay, you should know that:
[tex]
\frac{\partial}{\partial x}=\left(\frac{\partial}{\partial r}+\frac{\partial }{\partial s}\right)
[/tex]

So just apply this operator twice.
 
  • #7
Y comes from the original equation,

partial^2y/partialx^2-(1/v^2)(partial^2y/partialt^2)=0

I did what you said and I got

4partial^2y/(partialr * partials)

Is that right ?
 
  • #8
That looks very good. set that equal to zero and you have your original equation which you have now reduced to:
[tex]
\frac{\partial^{2}y}{\partial r\partial s}=0
[/tex]
 
  • #9
Thank you very much !

How can I solve that equation ?
 
  • #10
integrate once and what do you get?
 
  • #11
do i get y=rs? I am confused...
 
  • #12
Integrate once w.r.t. r say and you obtain:
[tex]
\frac{\partial y}{\partial s}=f(s)
[/tex]
Now what would you do?
 
  • #13
now Integrate with respect to s and I get: f(s)f(r)??
 
  • #14
No, you're forgetting your basic integration:
[tex]
y=\int f(s)ds+g(r)
[/tex]
The integral of f(s) w.r.t. s is just another function, call it h(s) amd trhat is your solution.
 
  • #15
Oh okay ! thank you so much ! you were really helpful :)
 

1. What is a partial derivative?

A partial derivative is a mathematical concept that measures the rate of change of a function with respect to one of its variables, while holding all other variables constant. It is denoted by ∂ (pronounced "del") and is often used in multivariable calculus.

2. How is a partial derivative calculated?

To calculate a partial derivative, you take the derivative of the function with respect to the variable in question, treating all other variables as constants. This can be done using the standard rules of differentiation, such as the power rule or the chain rule.

3. What is the difference between a partial derivative and a total derivative?

A partial derivative measures the rate of change of a function with respect to one variable, while a total derivative measures the overall rate of change of a function with respect to all of its variables. In other words, a partial derivative only considers the effect of one variable on the function, while a total derivative takes into account the effects of all variables.

4. What is change of variables in partial derivatives?

Change of variables is a technique used to simplify the calculation of partial derivatives. It involves substituting a new variable or set of variables in place of the original variables in a function, making the partial derivatives easier to calculate.

5. Why are partial derivatives important?

Partial derivatives are important in many areas of mathematics, physics, and engineering. They are used to model and analyze systems with multiple variables, such as in optimization problems or in the study of fluid dynamics. They also play a key role in the development of multivariable calculus and advanced mathematical concepts.

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