MATH 253 WORKSHEET 13
THE CHAIN RULE
(1) Dene z as a function of x, y as the solution to 2x + 3y − 4z − exyz−1 = 0.
∂z
∂z
(a) Find ∂x
and ∂y
.
∂z
∂z
− yzexyz−1 − xyexyz−1 ∂x
= 0 and solve
Solution: We dierentiate the equation to get 2 − 4 ∂x
for zx to get
∂z
Similarly, 3 − 4 ∂x
∂z
2 − yzexyz−1
.
=
∂x
4 + xyexyz−1
∂z
− xzexyz−1 − xyexyz−1 ∂y
= 0 and hence
∂z
3 − xzexyz−1
.
=
∂y
4 + xyexyz−1
Discussion: Let F (x, y, z) = 2x+3y−4z−exyz−1 , and let z = z(x, y) be the function implicitely
dened by F = 0. Then the two-variable composite function (x, y) 7→ F (x, y, z(x, y)) is the
constant zero (that's how z(x, y) is dened!). Its derivatives are therefore zero. But we can also
calculate them using the chain rule:
∂F
∂F ∂z
∂F (x, y, z(x, y))
=
+
.
∂x
∂x
∂z ∂x
Since both methods for calculating the derivative must give the same answer (zero), we can
solve for zx :
Fx (x, y, z(x, y))
.
Fz (x, y, z(x, y))
(b) Find the plane tangent to this surface at (1, 1, 1).
Solution: We verify that (1, 1, 1) is on the surface: 2 · 1+3 · 1−4 · 1−e1 · 1 · 1−1 = 2+3−4−e0 = 0.
∂z
· 1 · e1 · 1 · 1−1 = 1 and ∂z (1, 1) = 3−1 · 1 · e1 · 1 · 1−1 = 2 . It
Now at (1, 1, 1) we have ∂x
(1, 1) = 2−1
5
∂y
5
4+1 · 1 · e1 · 1 · 1−1
4+1 · 1 · e1 · 1 · 1−1
zx = −
follows that the plane has the equation
z−1=
or
2
1
(x − 1) + (y − 1)
5
5
5z − x − 2y = 2 .
35
(c) Find an approximate solution to 35 + 72 − 4z − e 36z−1 = 0.
Solution: We recognize the equation as F 56 , 76 , z = 0, that is as the equation dening z 56 , 67 .
We can approximate this value by a linear approximation about z (1, 1) (which we already know
from part (b)). The linear approximation is
z(x, y) ≈ 1 +
Plugging in x =
z
5 7
,
6 6
5
6
and y =
≈1+
1
5
1
2
(x − 1) + (y − 1) .
5
5
7
6
gives
5
2 7
1
2
31
−1 +
−1 =1−
+
=
,
6
5 6
30 30
30
(2) Suppose that w = x2 + yz − ln (1 + z), that x = st, that y = s + t and that z = st . Find
Solution: We have
∂w
∂s
and
∂w ∂x ∂w ∂y ∂w ∂z
1
1
s
t
1
∂w
=
+
+
= (2x)t + z · 1 + y −
= 2st2 + + s + t −
.
∂s
∂x ∂s
∂y ∂s
∂z ∂s
1+z t
t
s+t t
Date
: 4/10/2013.
1
∂w
∂t
.
Similarly,
s
∂w
1
s
t
s
.
= (2x)s + z · 1 + y −
− 2 = 2s2 t + − s + t −
∂s
1+z
t
t
s + t t2
.
(3) Suppose that z is a function of x, y and that x, y are functions of r, θ according to x = r cos θ,
2
2
∂z
∂z
y = r sin θ. Express ∂z
+ r12 ∂z
in terms of ∂x
and ∂y
.
∂r
∂θ
∂z ∂x
∂z ∂y
∂z
∂z
∂z ∂x
∂z ∂y
∂z
sin θ and ∂z
Solution: We have ∂r = ∂x ∂r + ∂y ∂r = ∂x cos θ + ∂y
∂θ = ∂x ∂θ + ∂y ∂θ =
∂z
∂z
∂x (−r sin θ) + ∂y (r cos θ). It follows that
∂z
∂r
2
1
+ 2
r
∂z
∂θ
2
2
1
2
[r (zy cos θ − zx sin θ)]
r2
=
(zx cos θ + zy sin θ) +
=
=
zx2 cos2 θ + zy2 sin2 θ + 2zx zy cos θ sin θ + zx2 sin2 θ + zy2 cos2 θ − 2zx zy cos θ sin θ
zx2 cos2 θ + sin2 θ + zy2 cos2 θ + sin2 θ
=
zx2 + zy2 .
(4) You are driving at a constant speed on a road that keeps a xed compass direction as it goes over a
2
2
hill. Say you position at time t is (1 − t, t), and hill is described by z = e−x −y . How fast is your
elevation changing at time t? When is your elevation maximal? What is it then?
2
2
2
2
∂y
∂z
−x2 −y 2
Solution: Since ∂x
(−1)−2ye−x −y (1) = 2 (x − y) e−x −y ,
∂t = −1, ∂t = 1 we have ∂t = −2xe
that is
2
∂z
= 2 (1 − 2t) e−(1+2t −2t) .
∂t
This vanishes when x − y = 0 that is when 1 − 2t = 0 or t =
1
1
√
e− 22 − 22 = 1/ e.
2
1
2
, at which point the elevation is