MATHEMATICS 226, FALL 2016, PROBLEM SET 4 – SOLUTIONS
Section 12.4, Question 11: 6 marks We have
fx (x, y) =
fxx (x, y) =
and by symmetry fyy (x, y) =
x2
2x
2y
, fy (x, y) = 2
,
2
+y
x + y2
2(x2 + y 2 ) − 2x · 2x
2y 2 − 2x2
=
,
(x2 + y 2 )2
(x2 + y 2 )2
2x2 −2y 2
.
(x2 +y 2 )2
Therefore
fxx (x, y) + fyy (x, y) =
2y 2 − 2x2
2x2 − 2y 2
+
= 0.
(x2 + y 2 )2 (x2 + y 2 )2
Section 12.4, Question 17: 6 marks
We have
x2
1
x2
1
2
2
2
2
ut = − t−3/2 e−x /4t + t−1/2 e−x /4t · 2 = − 3/2 e−x /4t + 5/2 e−x /4t
2
4t
2t
4t
x
x
2
) = − 3/2 e−x /4t
2t
2t
x
x
1
x2
1
2
2
2
2
uxx = − 3/2 e−x /4t − 3/2 e−x /4t (− ) = − 3/2 e−x /4t + 5/2 e−x /4t
2t
2t
2t
2t
4t
so that ut = uxx .
ux = t−1/2 e−x
2 /4t
(−
Section 12.5, Question 17: 6 marks
We have
∂
f (t sin s, t cos s) = f1 sin s + f2 cos s
∂t
∂2
∂
f (t sin s, t cos s) = sin s f1 (t sin s, t cos s) + cos sf1
∂s∂t
∂s
∂
+ cos s f2 (t sin s, t cos s) − sin sf2
∂s
= sin s(t cos sf11 − t sin sf12 ) + cos sf1
+ cos s(t cos sf21 − t sin sf22 ) − sin sf2
= t sin t cos t(f11 − f22 ) + t(cos2 s − sin2 s)f12 + cos sf1 − sin sf2 .
with all derivatives evaluated at (t sin s, t cos s).At the last step we used that f12 = f21 .
1
Section 12.5, Question 19: 6 marks
We have
∂
f (y 2 , xy, −x2 ) = yf2 − 2xf3
∂x
∂2
∂ f (y 2 , xy, −x2 ) =
yf2 (y 2 , xy, −x2 ) − 2xf3 (y 2 , xy, −x2 )
∂y∂x
∂y
= f2 + 2y 2 f21 + xyf22 − 4xyf31 − 2x2 f32
with all derivatives evaluated at (y 2 , xy, −x2 ).
Section 12.5, Question 23: 8 marks We have
∂z
∂z
∂z ∂z
∂z
∂z
= es cos t
+ es sin t ,
= −es sin t
+ es cos t
∂s
∂x
∂y ∂t
∂x
∂y
∂z
∂z
∂ 2z
s
s
=
e
cos
t
+
e
sin
t
∂s2
∂x
∂y
2
∂
z
∂ 2z s
s
s
+ e cos t e cos t 2 + e sin t
∂x
∂y∂x
2
∂ 2z ∂ z
s
s
s
+ e sin t 2
+ e sin t e cos t
∂x∂y
∂y
∂z
∂z
∂ 2z
= −es cos t
− es sin t
2
∂t
∂x
∂y
2
∂ z
∂ 2z s
s
s
− e sin t − e sin t 2 + e cos t
∂x
∂y∂x
2
∂ z
∂ 2z + es cos t − es sin t
+ es cos t 2
∂x∂y
∂y
∂ 2z ∂ 2z ∂ 2z ∂ 2z ∂ 2z ∂ 2z
2s
2
2
2
2
+
=
e
(cos
t
+
sin
t)
+
=
(x
+
y
)
+
∂s2
∂t2
∂x2 ∂y 2
∂x2 ∂y 2
Section 12.6, Question 7: 6 marks
The differential is dz = 2xe3y dx + 3x2 e3y dy. At (3, 0), we have z = 9, so that at (3.05, −0.02)
we have
z ≈ 9 + dz = 9 + 6dx + 27dy = 9 + 6(0.05) + 27(−0.02) = 8.76
2
Section 12.6, Question 11: 6 marks
Let x, y, z denote the edges of the box, then we have |dx| ≤ 0.01x, |dy| ≤ 0.01y, |dz| ≤ 0.01z.
(a) V = xyz so dV = yzdx+xzdy +xydz, |dV | ≤ yz(0.01x)+xz(0.01y)+xy(0.01z) = 0.03V .
The maximum percentage error is 3%.
(b) A = xy so dA = ydx + xdy, |dA| ≤ y(0.01x) + x(0.01y) = 0.02A. The maximum
percentage error is 2%.
(c) D2 = x2 + y 2 + z 2 so 2DdD = 2xdx + 2ydy + 2zdz, |2DdD| ≤ 2x(0.01x) + 2y(0.01y) +
2z(0.01z) = 0.02D2 and |dD| ≤ 0.01D. The maximum percentage error is 1%.
Note: Instead of the inequalities |dx| ≤ 0.01x etc., the solution manual simply plugs in the
maximal error dx = 0.01x. Strictly speaking, this requires an additional justification, but
since several solved examples in the textbook are done similarly, I will accept such solutions
as correct and give full credit for them.
Section 12.6, Question 19: 6 marks
2x
z
y
4 1 2
Df =
, Df (2, 2, 1) =
− ln z 2y −x/z
0 4 −2
 


−0.02
dx
6
4 1 2 
0.01 
f (1.98, 2.01, 1.03) ≈ f (2, 2, 1) + Df (2, 2, 1) dy  =
+
4
0 4 −2
dz
0.03
6
−0.01
5.99
=
+
=
4
−0.02
3.98
3