Winter 2012
Math 255
Problem Set 6
Solutions
15.3.20
Find the first partial derivatives of the function f (s, t) =
fs (s, t) =
t2 (t2 − s2 )
,
(s2 + t2 )2
ft (s, t) =
st2
.
s2 + t2
2s3 t
.
(s2 + t2 )2
Z
15.3.24
Find the first partial derivatives of the function f (x, y) =
x
cos(t2 )dt.
y
fx (x, y) = cos(x2 ),
fy (x, y) = − cos(y 2 ).
15.3.42 Use implicit differentiation to find ∂z/∂x and ∂z/∂y for yz =
ln(x + z). (Visit TEC Visual
http://www.brookscole.com/math_d/templates/student_resources/visuals_modules_5e/visuals/visual_14_3/
to see what this surface looks like.)
Note that x and y are independent variables, and z is a function of x
and y. By differentiating the equation implicitly with respect to x treating
y as a constant, we obtain
1
∂z
∂z
y
=
1+
.
∂x
x+z
∂x
1
Therefore, we obtain
∂z
1
=
.
∂x
(x + z)y − 1
By differentiating the equation implicitly with respect to y treating x as a
constant, we obtain
∂z
1 ∂z
z+y
=
.
∂y
x + z ∂y
Therefore, we obtain
(x + z)z
∂z
=
.
∂y
1 − (x + z)y
15.3.44
URL.)
Same as above but for sin(xyz) = x + 2y + 3z. (Visit the above
By differentiating the equation implicitly with respect to x treating y as
a constant, we obtain
∂z
∂z
cos(xyz) yz + xy
=1+3 .
∂x
∂x
Therefore, we obtain
∂z
1 − yz cos(xyz)
=
.
∂x
xy cos(xyz) − 3
By differentiating the equation implicitly with respect to y treating x as a
constant, we obtain
∂z
∂z
cos(xyz) xz + xy
=2+3 .
∂y
∂y
Therefore, we obtain
∂z
2 − xz cos(xyz)
=
.
∂y
xy cos(xyz) − 3
2
15.3.60
Find partial derivatives frss and frst for f (r, s, t) = r ln(rs2 t3 ).
We obtain
fr = ln(rs2 t3 ) +
then
rs2 t3
= ln(rs2 t3 ) + 1,
rs2 t3
2
frs = .
s
Therefore,
frss = −
2
,
s2
frst = 0.
15.3.68 Determine whether each of the following functions is a solution of
Laplace’s equation uxx + uyy = 0.
(a) u = x2 + y 2 ,
(b) u = x2 − y 2 ,
(c) u = x3 + 3xy 2 ,
p
(d) u = ln x2 + y 2 ,
(e) u = sin x cosh y + cos x sinh y,
(f) u = e−x cos y − e−y cos x.
(a) not solution, (b) solution , (c) not solution, (d) solution, (e) solution,
(f) solution.
15.3.71 If f and g are twice differentiable functions of a single variable,
show that the function
u(x, t) = f (x + at) + g(x − at)
is a solution of the wave equation utt = a2 uxx .
3
The second derivatives are calculated as
2
2
2 d f (u) 2 d g(u) utt = a
+a
,
du2 u=x+at
du2 u=x−at
uxx
d2 f (u) d2 g(u) =
+
.
du2 u=x+at
du2 u=x−at
Therefore, we have utt = a2 uxx .
15.3.80
The wind-chill index is modeled by the function
W = 13.12 + 0.6215T − 11.37v 0.16 + 0.3965T v 0.16
where T is the temperature (◦ C) and v is the wind speed (km/h). When
T = −15◦ C and v = 30 km/h, by how much would you expect the apparent
temperature to drop if the actual temperature decreases by 1◦ C? What if
the wind speed increases by 1 km/h?
We have
∂W
= 0.6215 + 0.3965v 0.16 ,
∂T
∂W
= 0.16(−11.37 + 0.3965T )v −0.84 .
∂v
If T decreases by 1◦ C, the apparent temperature drops 0.6215+0.3965(300.16 ) =
1.304754 · · · = 1.3◦ C. If v increases by 1 km/h, the apparent temperature
drops |0.16(−11.37 + 0.3965(−15))30−0.84 | = | − 0.1591561 · · · | = 0.16◦ C.
15.4.2 Find an equation of the tangent plane to the surface z = 9x2 + y 2 +
6x − 3y + 5 at the point (1, 2, 18).
Let us define f (x, y) = 9x2 + y 2 + 6x − 3y + 5. The tangent plane is
expressed as
z − 18 = fx (1, 2)(x − 1) + fy (1, 2)(y − 2).
Since fx (x, y) = 18x + 6 and fy (x, y) = 2y − 3, the equation is obtained as
24x + y − z − 8 = 0.
4
15.4.6 Find an equation of the tangent plane to the surface z = ex
the point (1, −1, 1).
2 −y 2
Let us define f (x, y) = ex
2 −y 2
at
. The tangent plane is expressed as
z − 1 = fx (1, −1)(x − 1) + fy (1, −1)(y + 1).
Since fx (x, y) = 2xex
tained as
2 −y 2
and fy (x, y) = −2yex
2 −y 2
, the equation is ob-
2x + 2y − z + 1 = 0.
15.4.24
Find the differential of the function v = y cos xy.
dv =
∂v
∂v
dx +
dy = −y 2 sin (xy)dx + (cos xy − xy sin xy) dy.
∂x
∂y
15.4.38 Four positive numbers, each less than 50, are rounded to the first
decimal place and then multiplied together. Use differentials to estimate the
maximum possible error in the computed product that might result from the
rounding.
Let p, q, r, and s be the four positive numbers, and w = pqrs be the
multiplied number. The differential is calculated as
dw =
∂w
∂w
∂w
∂w
dp+
dq +
dr +
ds = qrs dp+prs dq +pqs dr +pqr ds. (1)
∂p
∂q
∂r
∂s
Let p0 , q0 , r0 , and s0 be the rounded numbers, which are less than 50. To
estimate the error, we use dp = dq = dr = ds = 0.05 since we round up or
down depending on whether the second decimal digit is greater or less than
0.05. We use this together with p = p0 , q = q0 , r = r0 , and s = s0 . The
maximum possible error is estimated as
dw = (q0 r0 s0 + p0 r0 s0 + p0 q0 s0 + p0 q0 r0 )0.05 ≤ 4(503 )0.05 = 25000.
5
(The following solution with dp = dq = dr = ds = 0.1 is also fine.)
Let p, q, r, and s be the four positive numbers, and w = pqrs be the
multiplied number. The differential is calculated as
dw =
∂w
∂w
∂w
∂w
dp +
dq +
dr +
ds = qrs dp + prs dq + pqs dr + pqr ds.
∂p
∂q
∂r
∂s
Let p0 , q0 , r0 , and s0 be the rounded numbers, which are less than 50. To
estimate the error, we can use dp = dq = dr = ds = 0.1 together with p = p0 ,
q = q0 , r = r0 , and s = s0 . The maximum possible error is estimated as
dw = (q0 r0 s0 + p0 r0 s0 + p0 q0 s0 + p0 q0 r0 )0.1 ≤ 4(503 )0.1 = 50000.
15.4.41 Prove that if f is a function of two variables that is differentiable
at (a, b), then f is continuous at (a, b).
Hint: Show that
lim
f (a + ∆x, b + ∆y) = f (a, b)
(∆x,∆y)→(0,0)
Since f is differentiable at (a, b), we have
f (a + ∆x, b + ∆y) = f (a, b) + fx (a, b)∆x + fy (a, b)∆y + ε1 ∆x + ε2 ∆y,
where ε1 and ε2 → 0 as (∆x, ∆y) → (0, 0). Hence,
lim
f (a + ∆x, b + ∆y) = f (a, b).
(∆x,∆y)→(0,0)
That is, f is continuous at (a, b).
15.4.42
6
(a)
The function
xy
2 + y2
x
f (x, y) =

0


if (x, y) 6= (0, 0)
if (x, y) = (0, 0)
is graphed in the figure. Using the result of Prob. 41, show that fx (0, 0) and
fy (0, 0) both exist but f is not differentiable at (0, 0).
Let us calculate fx (0, 0) and fy (0, 0) according to the definition.
f (h, 0) − f (0, 0)
= 0,
h→0
h
f (0, h) − f (0, 0)
= 0.
h→0
h
fx (0, 0) = lim
fy (0, 0) = lim
When ∆x = ∆y = ∆,
∆2
1
= 6= f (0, 0).
2
2
∆→0 ∆ + ∆
2
lim f (∆, ∆) = lim
∆→0
Therefore, f (x, y) is not continuous at (0, 0). This means that, if we write
f (∆x , ∆y ) as
f (∆x , ∆y ) = f (0, 0) + fx (0, 0)∆x + fy (0, 0)∆y + ε1 ∆1 + ε2 ∆2 = ε1 ∆1 + ε2 ∆2 ,
ε1 and ε2 do not go to zero when ∆x and ∆y go to zero as ∆x = ∆y = ∆ → 0.
Therefore, f is not differentiable at (0, 0).
7
(b)
Explain why fx and fy are not continuous at (0, 0).
A function g(x, y) is continuous at (0, 0) if
lim
g(x, y) = g(0, 0).
(x,y)→(0,0)
If
lim
lim g(x, y) 6= lim
lim g(x, y) ,
∆x →0 ∆y →0
∆y →0 ∆x →0
then lim(x,y)→(0,0) g(x, y) doesn’t exist, and g(x, y) is not continuous. Note
that
x(x2 − y 2 )
y(y 2 − x2 )
,
f
(x,
y)
=
.
fx (x, y) = 2
y
(x + y 2 )2
(x2 + y 2 )2
Let g(x, y) be fx (x, y) or fy (x, y) and consider two limits lim∆x→0 lim∆y→0
and lim∆y→0 lim∆x→0 .
h
i2
∆y (∆y)2 − (∆x)2
h
i = 0.
lim lim fx (∆x, ∆y) = lim lim
∆x→0 ∆y→0
∆x→0 ∆y→0
(∆x)2 + (∆y)2
h
i
∆y (∆y)2 − (∆x)2
(∆y)3
lim lim fx (∆x, ∆y) = lim lim h
→ ∞.
i2 = lim
∆y→0 ∆x→0
∆y→0 ∆x→0
∆y→0 (∆y)4
(∆x)2 + (∆y)2
h
i
∆x (∆x)2 − (∆y)2
(∆x)3
lim lim fy (∆x, ∆y) = lim lim h
→ ∞.
i2 = lim
∆x→0 (∆x)4
∆x→0 ∆y→0
∆x→0 ∆y→0
(∆x)2 + (∆y)2
h
i
∆x (∆x)2 − (∆y)2
lim lim fy (∆x, ∆y) = lim lim h
i2 = 0.
∆y→0 ∆x→0
∆y→0 ∆x→0
(∆x)2 + (∆y)2
Thus, fx and fy are not continuous at (0, 0).
8
15.5.6 Use the Chain Rule to find dw/dt for w = xy + yz 2 , x = et ,
y = et sin t, z = et cos t.
dw
dt
∂w dx ∂w dy ∂w dz
+
+
∂x dt
∂y dt
∂z dt
t
2 t
= ye + (x + z )e (sin t + cos t) + 2yzet (cos t − sin t)
=
= yet + (x + 2yz + z 2 )et cos t + (x − 2yz + z 2 )et sin t.
15.5.28
Use
∂F/∂x
Fx
dy
=−
=−
to find dy/dx for
dx
∂F/∂y
Fy
y 5 + x2 y 3 = 1 + yex
2
2
Let F (x, y) be F (x, y) = y 5 + x2 y 3 − 1 − yex . Then we have F (x, y) = 0.
2 − ex2
−2xy
y
dy
∂F/∂x
=−
= 4
.
dx
∂F/∂y
5y + 3x2 y 2 − ex2
15.5.56 Suppose that the equation F (x, y, z) = 0 implicitly defines each
of the three variables x, y, and z as functions of the order two: z = f (x, y),
y = g(x, z), x = h(y, z). If F is differentiable and Fx , Fy , and Fz are all
nonzero, show that
∂z ∂x ∂y
= −1
∂x ∂y ∂z
By the implicit function theorem, we obtain
Fy
∂z ∂x ∂y
Fx
Fz
= −
−
−
= −1.
∂x ∂y ∂z
Fz
Fx
Fy
9
(Solution without using the implicit function theorem)
We have the differentials as
∂x
∂x
dy +
dz,
∂y
∂z
∂y
∂y
dx +
dz,
∂x
∂z
∂z
∂z
dx +
dy.
∂x
∂y
dx =
dy =
dz =
(2)
(3)
(4)
Equations (2), (3), and (4) should hold for any dx, dy, and dz. In particular,
when dy = 0 in Eqs. (2) and (4), we obtain
∂x
∂z
∂z ∂x
∂z ∂x
dy +
dz +
dy =
dz.
dz =
∂x ∂y
∂z
∂y
∂x ∂z
Thus, we have
1=
∂z ∂x
.
∂x ∂z
(5)
By Eqs. (2) and (3), we obtain
∂y
∂x
∂x ∂y
dx +
dz +
dz.
dx =
∂y ∂x
∂z
∂z
When dx = 0, we have
0=
∂x ∂y ∂x
+
.
∂y ∂z
∂z
Using Eqs. (5) and (6), we obtain
∂z ∂x ∂y
= −1.
∂x ∂y ∂z
10
(6)