Answer: You’re almost there. First, a miscalculation: your ∂v∂x has the wrong sign. In particular: we indeed have u=12log(x2+y2),v=tan−1(y/x). It follows that ∂u∂x=xx2+y2,∂u∂y=yx2+y2, which you calculated correctly. However, ∂v∂x=−yx2⋅11+(y/x)2=−yx2+y2,∂v∂y=1x⋅11+(y/x)2=xx2+y2. Now, in order to see that JJ′=1, write both Jacobians as a function of the same set of variables. That is, either as a function of x and y or as a function of u and v. If we write J′ as a function of u and v, then we find J′=x2+y2(x2+y2)2=1×2+y2=1[eucosv]2+[eusinv]2=1e2u. In greater detail: we have v(x,y)=tan−1(y/x). We find the derivative of this using the chain rule. Because ddttan−1(t)=11+t2, we have ∂v∂x=11+(y/x)2⋅∂∂x[(y/x)]. To calculate the partial derivative of y/x with respect to x, we use the power rule. In particular: f(x,y)=y/x=y⋅x−1⟹∂f∂x=−yx−2=−y⋅1×2=−yx2. Reply
Answer:
You’re almost there. First, a miscalculation: your ∂v∂x has the wrong sign. In particular: we indeed have
u=12log(x2+y2),v=tan−1(y/x).
It follows that
∂u∂x=xx2+y2,∂u∂y=yx2+y2,
which you calculated correctly. However,
∂v∂x=−yx2⋅11+(y/x)2=−yx2+y2,∂v∂y=1x⋅11+(y/x)2=xx2+y2.
Now, in order to see that JJ′=1, write both Jacobians as a function of the same set of variables. That is, either as a function of x and y or as a function of u and v.
If we write J′ as a function of u and v, then we find
J′=x2+y2(x2+y2)2=1×2+y2=1[eucosv]2+[eusinv]2=1e2u.
In greater detail: we have v(x,y)=tan−1(y/x). We find the derivative of this using the chain rule. Because ddttan−1(t)=11+t2, we have
∂v∂x=11+(y/x)2⋅∂∂x[(y/x)].
To calculate the partial derivative of y/x with respect to x, we use the power rule. In particular:
f(x,y)=y/x=y⋅x−1⟹∂f∂x=−yx−2=−y⋅1×2=−yx2.