Exercise
Given
z(x,y)=\arctan (\frac{x}{y})
f(u,v)=u\sin v
g(u,v)=u\cos v
Z(u,v)=z(f(u,v),g(u,v))
Calculate the derivative
Z'_u,Z'_v
Final Answer
Solution
We will use the chain rule to calculate the derivative
Z'_u=z'_f\cdot f'_u+z'_g\cdot g'_u
We have
Z(u,v)=z(f(u,v),g(u,v))
And function u is
z(x,y)=\arctan(\frac{x}{y})
We put it in the function above and get
z(f,g)=\arctan(\frac{f}{g})
Hence, the partial derivatives of z are
z'_f=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{1}{g}
z'_g=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{-f}{g^2}
Also, we calculate the derivatives of f and g
f'_u=\sin v
g'_u=\cos v
We put all the derivatives and get
Z'_u=z'_f\cdot f'_u+z'_g\cdot g'_u=
=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{1}{g}\cdot \sin v+\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{-f}{g^2}\cdot \cos v=
=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}\cdot \frac{1}{u\cos v}\cdot \sin v+\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}\cdot \frac{-u\sin v}{{(u\cos v)}^2}\cdot \cos v=
=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}(\frac{\sin v}{u\cos v}+\frac{-u\sin v}{u^2\cos^2 v}\cdot \cos v)=
=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}(\frac{\sin v}{u\cos v}-\frac{\sin v}{u\cos v})=
=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}\cdot 0=0
We will use the chain rule to calculate the partial derivatives of Z.
Z'_v=z'_f\cdot f'_v+z'_g\cdot g'_v
We got above
z'_f=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{1}{g}
z'_g=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{-f}{g^2}
Alao, we calculate the derivatives of f and g.
f'_v=u\cos v
g'_v=-u\sin v
We put the derivatives and get
Z'_v=z'_f\cdot f'_v+z'_g\cdot g'_v
=\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{1}{g}\cdot u\cos v+\frac{1}{1+{(\frac{f}{g})}^2}\cdot \frac{-f}{g^2}\cdot(-u\sin v)=
=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}\cdot \frac{1}{u\cos v}\cdot u\cos v+\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}\cdot \frac{-u\sin v}{{(u\cos v)}^2}\cdot(-u\sin v)=
=\frac{1}{1+{(\frac{u\sin v}{u\cos v})}^2}(\frac{u\cos v}{u\cos v} +\frac{-u\sin v}{{(u\cos v)}^2}\cdot(-u\sin v))=
=\frac{1}{1+\frac{\sin^2 v}{\cos^2 v}}(1+\frac{u^2\sin^2 v}{u^2\cos^2 v})=
=\frac{1}{1+\tan^2 v}(1+\tan^2 v)=
=\frac{1+\tan^2 v}{1+\tan^2 v}=1
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