# A second opinion for a tricky problem?

1. Oct 10, 2007

### mlazos

1. The problem statement, all variables and given/known data
$$z=e^{i(\kappa x-\omega t)$$

$$\theta=x^2+t$$ and $$\phi=x-t$$

find $$\frac{\partial z}{\partial \theta}$$ and $$\frac{\partial z}{\partial \phi}$$ in terms of x and t only.

2. Relevant equations

3. The attempt at a solution
for the first partial I thought to solve the $$\theta=x^2+t$$ for t and to substitute to the z and then to use the chain rule. So to do this

$$t=\theta-x^2$$ and so $$z=e^{i(\kappa x-\omega (\theta-x^2))}$$

then in order to find $$\frac{\partial z}{\partial \theta}$$ I will do this $$\frac{\partial z}{\partial \theta}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial \theta}$$

Now $$\frac{\partial z}{\partial x}=e^{i(\kappa x-\omega (\theta-x^2))}(i(\kappa x-\omega (\theta-x^2)))'=e^{i(\kappa x-\omega (\theta-x^2))}(i\kappa - 2\omega x)$$

the $$\frac{\partial x}{\partial \theta}=\frac{1}{2x}$$

so $$\frac{\partial z}{\partial \theta}=\frac{\partial z}{\partial x} \frac{\partial x}{\partial \theta}=\frac{e^{i(\kappa x-\omega (\theta-x^2))}(i\kappa - 2\omega x)}{2x}$$

Now we substitute again the t and we get

$$\frac{\partial z}{\partial \theta}=\frac{e^{i(\kappa x-\omega t)}(i\kappa - 2\omega x)}{2x}$$

For the second partial I follow the same way

$$x=\phi + t$$ I substitute to z and I get $$z=e^{i(\kappa (\phi + t) -\omega t)}$$

and then $$\frac{\partial z}{\partial \phi}=\frac{\partial z}{\partial t}\frac{\partial t}{\partial \phi}$$

We find that $$\frac{\partial t}{\partial \phi}=-1$$ and

$$\frac{\partial z}{\partial t}=e^{i(\kappa (\phi + t) -\omega t)}(i(\kappa (\phi + t) -\omega t))'=e^{i(\kappa (\phi + t) -\omega t)}(i \kappa - \omega)$$

so $$\frac{\partial z}{\partial \phi}=\frac{\partial z}{\partial t}\frac{\partial t}{\partial \phi}=-e^{i(\kappa (\phi + t) -\omega t)}(i \kappa - \omega)$$

I am not sure if this is the solution, i need a second opinion. I am a mathematician and I need someone else to tell me if im wrong and where. Thank you for your time. Please read carefully before to answer. Thank you

2. Oct 11, 2007

### HallsofIvy

$$\frac{\partial z}{\partial \phi}\ne\frac{\partial z}{\partial t}\frac{\partial t}{\partial \phi}$$
$$\frac{\partial z}{\partial \phi}=\frac{\partial z}{\partial t}\frac{\partial t}{\partial \phi}+ \frac{\partial z}{\partial x}\frac{\partial x}{\partial \phi}$$
and the same for $\theta$.