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