Mathematical question i physics

[nhrock3]

$$\vec{E}=f_1(z-ct)\hat{x}+f_2(z-ct)\hat{y}+0\hat{z}$$$$f_1(u)$$ and $$f_2(u)$$ are functions of "u"
u=z-ct
i have the formula
$$\nabla \times \vec{E}=-\frac{db}{dt }$$
$$\nabla\times\vec{E}=|\begin{array}{ccc} \hat{x} & \hat{y} & \hat{z}\\ \frac{{d}}{dx} & {\frac{{d}}{dy}} & \frac{{d}}{dz}|=\\ f_{1}(z-ct) & f_{2}(z-ct) & 0\end{array}\hat{-x}\frac{{df_{2}}}{dz}-\hat{y}\frac{{df_{1}}}{dz}+\hat{z0}=-\frac{{d\overrightarrow{B}}}{dt}$$

i want to find B
so i need to integrate the left side by t
in order to get B
but f_1 f_2 are function of u
how to make a result
?


and then i need to show that $$B\bullet E=0$$
i can't get 0

 

[tiny-tim]

hi nhrock3!

(write "tex" not "TEX"; and write "itex" rather than "tex", and it won't keep starting a new line )

$$\vec{E}=f_1(z-ct)\hat{x}+f_2(z-ct)\hat{y}+0\hat{z}$$
$f_1(u)$ and $f_2(u)$ are functions of "u"
u=z-ct
i have the formula
$$\nabla \times \vec{E}=-\frac{db}{dt }$$
$$\nabla\times\vec{E}=|\begin{array}{ccc} \hat{x} & \hat{y} & \hat{z}\\ \frac{{d}}{dx} & {\frac{{d}}{dy}} & \frac{{d}}{dz}|=\\ f_{1}(z-ct) & f_{2}(z-ct) & 0\end{array}\hat{-x}\frac{{df_{2}}}{dz}-\hat{y}\frac{{df_{1}}}{dz}+\hat{z0}=-\frac{{d\overrightarrow{B}}}{dt}$$

i want to find B
so i need to integrate the left side by t
in order to get B
but f_1 f_2 are function of u
how to make a result
?


and then i need to show that $$B\bullet E=0$$
i can't get 0

you notation is very confused …

perhaps because you are? ​

f₁ is a function of only one variable, and has only one derivative, call it f₁' …

(z - ct) is a function of two variables, and has two derivatives …

applying the chain rule:

∂f₁(z - ct)/∂z = f₁'(z - ct); ∂f₁(z - ct)/∂t = -cf₁'(z - ct) …​

carry on from there