SOLUTION TO PROBLEMS 37, 38- SECTION 14.6
Problem 37: Suppose that x > 0, y > 0 and that z can be expressed
either as a function of Cartesian coordinates (x, y) or as a function of
polar coordinates (r, θ), so that z = f (x, y) = g(r, θ). (Recall that x =
r cos θ, y = r sin θ, r = x2 + y 2 , and, for x > 0, y > 0, θ = arctan(y/x))
(a) Use the chain rule to find ∂z/∂r and ∂z/∂θ in terms of ∂z/∂x and
∂z/∂y.
(b) Solve the equations you have just written down for ∂z/∂x and
∂z/∂y in terms of ∂z/∂r and ∂z/∂θ.
(c) Show that the expressions you get in part (b) are the same as you
would get by using the chain rule to find ∂z/∂x and ∂z/∂y in
terms of ∂z/∂r and ∂z/∂θ.
Solution. (a) As z depends on x and y which in turn both depend on
r and θ, the chain rule gives us
∂z
∂z ∂x ∂z ∂y
=
+
∂r
∂x ∂r ∂y ∂r
∂z
∂z ∂x ∂z ∂y
=
+
∂θ
∂x ∂θ ∂y ∂θ
As x = r cos θ, y = r sin θ, we obtain
∂z
∂z
∂z
(1)
=
cos θ +
sin θ
∂r
∂x
∂y
∂z
∂z
∂z
(2)
=
(−r sin θ) +
(r cos θ)
∂θ
∂x
∂y
(b) Multiplying (1) by r sin θ and (2) by cos θ and adding the two
gives
r sin θ
∂z
∂z
∂z
+ cos θ
=
(r sin2 θ + r cos2 θ)
∂r
∂θ
∂y
As sin2 θ + cos2 θ = 1, we have
∂z
∂z cos θ ∂z
= sin θ
+
∂y
∂r
r ∂θ
In a similar manner, one can obtain
∂z
∂z sin θ ∂z
= cos θ
−
∂x
∂r
r ∂θ
1
2
SOLUTION TO PROBLEMS 37, 38- SECTION 14.6
(c) As z depends on r, θ which depend on x and y, the chain rule
gives
∂z
∂z ∂r ∂z ∂θ
=
+
∂x
∂r ∂x ∂θ ∂x
∂z
∂z ∂r ∂z ∂θ
=
+
∂y
∂r ∂y ∂θ ∂y
p
As r = x2 + y 2 ,
2x
1
∂r/∂x = p
= cos θ
2 x2 + y 2
(As we are evaluating ∂r/∂x, we cannot have a dependence on
x in the right hand side.) Similarly, ∂r/∂y = sin θ. As θ =
arctan(y/x), we have
−y
− sin θ
1
(−y/x2 ) = 2
=
.
∂θ/∂x =
2
2
1 + (y/x)
x +y
r
Similarly, ∂θ/∂y = cos θ/r. Putting everything together, we obtain the same expressions as in (b).
Problem 38. Show that
2 2 2
2
∂z
∂z
∂z
1 ∂z
+
=
+ 2
.
∂x
∂y
∂r
r
∂θ
Solution. From (a), the R.H.S is
2
2
∂z
∂z
∂z
∂z
2
cos θ + 2
cos θ sin θ +
sin2 θ
∂x
∂x
∂y
∂y
2
2
∂z
∂z
∂z
∂z
+
sin2 θ − 2
cos θ sin θ +
cos2 θ
∂x
∂x
∂y
∂y
As cos2 θ + sin2 θ = 1, this gives R.H.S =
2 2
∂z
∂z
+
,
∂x
∂y
which equals the L.H.S as needed.