A second opinion for a tricky problem?

[mlazos]

Homework Statement

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.

Homework Equations

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 I am wrong and where. Thank you for your time. Please read carefully before to answer. Thank you

 

[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.