Review Problems for Exam 1:
Problem 1 Find the differential of fx, y  y arctanxy.
solution)

1
Note that arctan u  1u
2 .
f
y2

,
x
1  x2y2
f
xy
 arctanxy 
.
y
1  x2y2
So,
df  f x dx  f y dy

y2
xy
dx  arctanxy 
1  x2y2
1  x2y2
dy.
Problem 2 Evaluate the Jacobian of the mapping u  sin x cos y, v  cos x sin y. 
solution)
u
x
u
y
v
x
v
y

cos x cos y  sin x sin y
 sin x sin y cos x cos y
 cos 2 x cos 2 y  sin 2 x sin 2 y.
z
z
 ax y
 0.
Problem 3 If z  fax 2  by 2 , show that by x
solution)
Let t  ax 2  by 2 . Then by the chain rule,
z  z t  z 2ax
t x
t
x
z  z t   z 2by.
t y
t
y
Hence,
by z  ax z  2abxy z  2abxy z  0.
y
t
t
x
Problem 4 Go over the problem 7 on page 100.
solution) see the homework solution.
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Problem 5 Functions u and v are defined implicitly as functions of x, y and z by the following
equations
x 2  yv  zu  6
xyz  uv  8.
u
x
u
y
u
z
Find ,
and .
solution)
Taking differentials, we get
2xdx  vdy  ydv  udz  zdu  0
yzdx  xzdy  xydz  vdu  udv  0.
Rewriting the above,
zdu  ydv  2xdx  vdy  udz
vdu  udv  yzdx  xzdy  xydz.
By making use of Cramer’s rule to solve for du, we have
du 
2xdx  vdy  udz
y
yzdx  xzdy  xydz
u
z y
v
2x y
u
dx
v
y
dy
u y
yz u
xz u
xy u



uz  vy
uz  vy
uz  vy
2
2
2
2xu  y z
uv  xyz
u  xy
 uz  vy dx  uz  vy dy  uz  vy dz.
dz
Thus,
2
u  2xu  y z ,
uz  vy
x
u  uv  xyz ,
uz  vy
y
2
2
u  u  xy .
uz  vy
z
2
Problem 6 Consider the mapping given by u  ux, y, z, v  vx, y, z, w  wx, y, z. The
x,y,z
u,v,w
Jacobian x,y,z of the mapping is 4. Find the Jacobian u,v,w of the inverse mapping
x  xu, v, w, y  yu, v, w, z  zu, v, w.
solution) Using the identity
x, y, z

u, v, w
1
u,v,w
x,y,z
,
we can easily get
x, y, z
 1.
4
u, v, w
Problem 7 Let x  r cos , y  r sin . Compute

x
and

y
.
solution)
dx  cos dr  r sin d,
dy  sin dr  r cos d.
Solving for dr, we get
cos  dx
d 
sin  dy
cos  r sin 
sin 

 sin dx  cos dy
.
r
r cos 
Hence,
   sin  ,
r
x
  cos  .
r
y
3
Problem 8 Consider the surface x 2  3y  4z  7. Find a) the unit normal to the surface, and
b) the equation of the tangent plane at 3, 2, 1.
solution)
a) Let Fx, y, z  x 2  3y  4z. The gradient at 3, 2, 1 is
Fx, y, z  2x, 3, 4  F3, 2, 1  6, 3, 4.
Hence the unit normal to the surface Fx, y, z  7 at 3, 2, 1 is
6, 3, 4
6, 3, 4
n 

.
941
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b) The equation of the tangent plane at 3, 2, 1 is
6x  3  3y  2  4z  1  0
6x  3y  4z  16.
Problem 9 Let z  x 2  xy  y. Find the directional derivative of z along the curve
y  x 2  x  1 at x, y  2, 1.
solution) Parametirc representation of the curve y  x 2  x  1 is
r t  t, t 2  t  1.
Note that x, y  2, 1 at t  2.
Now,


r t  1, 2t  1  r 2  1, 3,
and zx, y  2x  y, x  1  z2, 1  3, 1.
Hence the directional derivative of z along the curve r t at t  2 is

z2, 1 
r 2

r 2
 3, 1 
1, 3
 0.
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Problem 10 Find  such that fx, y, z  x 2  y 2  z 2 is harmonic.
solution)
f
2f
 2x 
 2,
x
x 2
f
2f
 2y 
 2,
y
y 2
2f
f
 2x 
 2.
z
z 2
Note that we say that f is harmonic if  2 f  0. Hence,
2f
2f
2f


2f 
x 2
y 2
z 2
 2  2  2  0
   2.
Problem 11 Let fx, y  yx 2  y 2  2 . Use the Laplacian formula in polar coordinates to
compute  2 f.
2f
2f
f
(Hint:  2 f  r 2  r12  2  1r r . )
solution)
In polar coordinates, f  r sin r 2  2  r 3 sin .
Now,
f
f
2f
 3r 4 sin  
 
 12r 5 sin ,
r
r r
r 2
f
f
2f
 r 3 cos  
 
 r 3 sin .
2
 


So,
2
2f
1  f  1 f

2f 
r r
r 2
r 2  2
 12r 5 sin   r 5 sin   3r 5 sin 
 8r 5 sin .
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