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Solutions
Question 1.


1
1 
2w
2u 1
∂(x, y, z)

J=
=
1 2v
∂(u, v, w)
1
1
Question 2.
a) J =
∂(x,y)
∂(u,v)
µ
=
2u 1
1 2v
¶
a) Solution I. Use the formula
µ
(rule)
õ
¶
∂
∂x
∂
∂y
=
∂(x, y)
∂(u, v)
¶t !−1 µ
∂
∂u
∂
∂v
¶
where t means transposed and −1 means inverse matrices, see lectures. Then you
need to know the formula for the inverse 2 × 2 matrix (from Linear Algebra):
µ
(inv)
A :=
a b
c d
¶
, A
1
=
det A
−1
µ
d −b
−c a
¶
Using that formula and det J = 4uv − 1, we have
õ
t −1
(J )
=
∂(x, y)
∂(u, v)
¶t !−1
1
=
4uv − 1
µ
2v
−1
−1
2u
¶
and
µ
∂
∂x
∂
∂y
¶
1
=
4uv − 1
µ
2v
−1
−1
2u
¶µ
∂
∂u
∂
∂v
¶
1
=
4uv − 1
µ
∂
∂
− ∂v
2v ∂u
∂
∂
− ∂u
+ 2u ∂v
¶
a) Solution II. Let f : R2 → R be and arbitrary smooth function and we consider
the compound function
f (u, v) = f (x(u, v), y(u, v))
and we need to express fx0 and fy0 in terms of fu0 and fv0 . By the chain rule,
½
(chain)
fu0 = 2ufx0 + fy0
fv0 = fx0 + 2vfy0
µ
OR
fu0
fv0
¶
µ
=
2u 1
1 2v
¶µ
fx0
fy0
¶
Now, we need to solve the 2 × 2 linear system (chain) with respect to unknowns fx0
and fy0 (=invert the matrix J t ). To this end, multiplying first equation by 2v and
substracting the second one, we have
(4uv − 1)fx0 = 2vfu0 − fv0 , fx0 =
1
¡
¢
1
2vfx0 − fy0 .
4uv − 1
2
Analogously, multiplying the second equation by 2u and substracting the first one,
we have
(4uv − 1)fv0 = 2ufv0 − fu0 .
Thus,
µ
∂
∂x f
∂
∂y f
¶
1
=
4uv − 1
µ
∂
∂
2v ∂u
f − ∂v
f
∂
∂
− ∂u f + 2u ∂v f
¶
and, omitting f , we get the final answer.
Remark: Both solutions are, of course, is equivalent and the only difference is
the way how do we compute the inverse matrix, so the choice of the method is a
question of taste (and of how confident is you in inverting 2 × 2 matrices!!). If you
are sure that you are able to reproduce (inv) correctly (during the test or exam),
then the first way is preferable. Just memorize the (rule) and do not forget that
the Jacobi matrix is NOT always a symmetric matrix (in contrast to the Hessian)
and that you need to invert the transposed Jacobi matrix.
c) Solution I. Write the total differential of f as a function of x and y
df = fx0 dx + fy0 dy
and then substitute dx and dy by the differentials of x(u, v) and y(u, v) respectively:
df = fx0 dx+fy0 dy = fx0 (2u du+dv)+fy0 (du+2v dv) = (2ufx0 +fy0 )du+(fx0 +2vfv0 )dv.
c) Solution II. Write from the very beginning that
df = fu0 du + fv0 dv,
but after you need to express fu0 and fv0 through fx0 and fy0 using the CHAIN rule
(chain):
fu0 = 2ufx0 + fy0 , fv0 = fx0 + 2vfx0
and we end up with the same answer! The advantage of the 1st method is that you
need not to memorize the chain rule (chain), but deduce it instead!