Exercise
Given the differentiable function (with parameters a,b)
w=f(x+at, y+bt)
Prove the equation
w'_t=aw'_x+bw'_y
Proof
Define
u=x+at
v=y+bt
We get the function
w=f(u,v)
And the internal functions
u(x,t)=x+at
v(y,t)=y+bt
We will use the chain rule to calculate the partial derivatives of w.
w'_t=f'_u\cdot u'_t+f'_v\cdot v'_t=
=f'_u\cdot a+f'_v\cdot b
w'_x=f'_u\cdot u'_x+f'_v\cdot v'_x=
=f'_u\cdot 1+f'_v\cdot 0=
=f'_u
w'_y=f'_u\cdot u'_y+f'_v\cdot v'_y=
=f'_u\cdot 0+f'_v\cdot 1=
=f'_v
We got the equations
w'_t=f'_u\cdot a+f'_v\cdot b
w'_x=f'_u
w'_y=f'_v
Hence, we get
w'_t=w'_x\cdot a+w'_y\cdot b
w'_t=aw'_x+bw'_y
As required.
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