Exam 1 - MATH 321: Solutions
Problem 1. (a) Can the function f (x, y) =
uous by suitably defining it at (0, 0)?
(b) The same question as in (a) for f (x, y) =
xy
x2 +y 2
be made contin-
sin(x2 +y 2 )
.
x2 +y 2
Solution. (a) The partial limits of f along the paths x = 0 and
y = 0 are equal to 0, while the partial limit of f along the path y = x
is equal to 1/2. Therefore, lim(x,y)→(0,0) f (x, y) does not exist and the
function cannot be suitably defined at (0, 0) to become continuous at
this point.
(b) We have
sin(x2 + y 2 )
sin t
= 1.
= lim+
2
2
(x,y)→(0,0)
t→0
x +y
t
lim
Thus defining f (0, 0) = 1 makes f continuous at (0, 0). Since f is also
continuous at all other points of R2 , it becomes continuous everywhere.
2
2
Problem 2. Let f (x, y) = xey − yex .
(a) Find the equation for the tangent plane to the graph of f at
(1, 2).
(b) What point on the surface z = x2 −y 2 has a tangent plane parallel
to the plane found in part (a)?
Solution. (a) The tangent plane equation at (x0 , y0 ) is z = f (x0 , y0 )+
+ ∂f
(x0 , y0 )(y − y0 ). We have f (1, 2) = e4 − 2e,
∂y
2 2
2xyex )
= e4 − 4e, ∂f
(1, 2) = (2xyey −
∂y
(x,y)=(1,2)
x2 4
e )
= 4e − e. Therefore, the tangent plane equation is
∂f
(x0 , y0 )(x − x0 )
∂x
2
∂f
(1, 2) = (ey −
∂x
(x,y)=(1,2)
z = e4 − 2e + (e4 − 4e)(x − 1) + (4e4 − e)(y − 2), or (e4 − 4e)(x −
1) + (4e4 − e)(y − 2) − z + e4 − 2e = 0.
(b) A point (x1 , y1 , z1 ) on the surface z = x2 − y 2 must satisfy the
condition that the gradient of the function F (x, y, z) = x2 − y 2 − z at
this point is parallel to the normal vector to the plane of part (a), i.e.,
(2x, −2y, −1) = t(e4 − 4e, 4e4 − e, −1). This is possible only for t = 1,
2 2
x1 = 21 e4 −2e, y1 = 2e4 − 12 e, z1 = x21 −y12 = 12 e4 −2e − 2e4 − 12 e =
15 2
e
4
−
15 8
e.
4
Problem 3. Determine the equation of the tangent line to the given
path at the specified value of t.
1
2
(a) (sin 3t, cos 3t, 2t5/2 ); t = 1.
(b) (cos2 t, 3t − t3 , t); t = 0.
−c (t) in R3 at t = t is
Solution. The tangent line to the given path →
0
→
−
→
−
→
−
0
given by ` (u) = c (t0 )+ c (t0 )u, with a free parameter −∞ < u < ∞.
−c (1) = (sin 3, cos 3, 2), →
−c 0 (1) = (3 cos 3t, −3 sin 3t, 5t3/2 ) =
(a) We have →
t=1
→
−
(3 cos 3, −3 sin 3, 5). Thus we obtain the tangent line equation ` (u) =
(sin 3, cos 3, 2) + (3 cos 3, −3 sin 3, 5)u.
→
−
→
−
0
2
(b) We have c (0) = (1, 0, 0), c (0) = (−2 cos t sin t, 3 − 3t , 1) =
t=0
→
−
(0, 3, 1). Thus we obtain the tangent line equation ` (u) = (1, 0, 0) +
→
−
(0, 3, 1)u, or ` (u) = (1, 3u, u).
→
−
−
Problem 4. Let h : R3 → R5 and →
g : R2 → R3 be given by
√
→
−
−
h (x, y, z) = (xyz, exz , x sin y, − x9 , 17) and →
g (u, v) = (v 2 + 2u, π, 2 u).
Find D(h ◦ g)(1, 1) using the chain rule.
→
− −
→
− −
−
Solution. By the chain rule, D( h ◦→
g )(1, 1) = D h (→
g (1, 1))D→
g (1, 1).
→
−
We compute g (1, 1) = (3, π, 2),




2 2v
2 2
−
−
D→
g (u, v) =  0 0  , D→
g (1, 1) = 0 0 ,
1
√
0
1 0
u




yz
xz
xy
2π 6 3π
2e6 0 3e6 
 zexz
0
xexz 




→
−
→
−
 , D h (3, π, 2) =  0 −3 0  ,
sin
y
x
cos
y
0
D h (x, y, z) = 



 9
 1
 2
0
0 
0
0 
x
0
0
0
0
0
0




2π 6 3π 
7π 4π

2e6 0 3e6  2 2
7e6 4e6 




→
− −
 0 0 =  0
.
0
−3
0
0
D( h ◦ →
g )(1, 1) = 




 1
 2
2 
0
0  1 0
0
0
0
0
0
Problem 5. Compute the directional derivative of f in the given
−
direction →
v at the given point P :
→
−
→
−
−
(a) f (x, y, z) = xy 2 + y 2 z 3 + z 3 x, P (4, −2, −1), →
v = √114 ( i + 3 j +
→
−
2 k ).
→
−
→
−
→
−
−
(b) f (x, y, z) = xyz , P (e, e, 0), →
v = 12 i + 3 j + 4 k .
13
13
13
3
−
Solution. (a) Df (P )→
v
−
= ∇f (P )→
v = y 2 + z 3 2xy + 2yz 3
 √ 
1/√14
= 3 −12 24 3/√14 =
2/ 14
→
−
(b) Df (P ) v
−
= ∇f (P )→
v = yzxyz−1


12/13
= 0 0 e  3/13  =
4/13
=
 √ 
1/√14
3/ 14
√
(x,y,z)=(4,−2,−1)
2/ 14
√15 .
14
yz
yz
zx ln x yx ln x 

12/13
 3/13 
(x,y,z)=(e,e,0)
4/13
4e
.
13
Problem 6. Verify that
yz 3 x2 .
3
3−36+48
√
14
3y 2 z 2 + 3xz 2 ∂3f
∂x ∂y ∂z
=
∂3f
∂z ∂y ∂x
for f (x, y, z) = zexy +
2
∂
Solution. ∂x∂∂yf∂z = ∂x∂ ∂y (exy + 3yz 2 x2 ) = ∂x
(xexy + 3z 2 x2 ) = exy +
xyexy + 6xz 2 .
2
∂3f
∂
= ∂z∂ ∂y (yzexy + 2xyz 3 ) = ∂z
(zexy + xyzexy + 2xz 3 ) = exy +
∂z ∂y ∂x
xyexy + 6xz 2 .
3
3
Therefore ∂x∂∂yf∂z = ∂z∂∂yf∂x .
Problem 7. (a) Suppose f, g, h : R → R are differentiable. Show
→
−
that the vector field F (x, y, z) = (f (x), g(y), h(z)) is irrotational.
→
−
→
−
→
−
(b) Show that F (x, y) = (x2 + y 2 ) i − 2xy j is not a gradient field.
→
→
− →
−
−i
j
k →
−
→
−
∂
∂ = ( ∂ h(z) −
∂
Solution. (a) curl F = ∇ × F = ∂x
∂y
∂y
∂z f (x) g(y) h(z)
→
−
→
−
→
−
→
−
∂
∂
∂
∂
∂
g(y)) i −( ∂x
h(z)− ∂z
f (x)) j +( ∂x
g(y)− ∂y
f (x)) k = 0 . Therefore
∂z
the field is irrotational.
→
−
∂
∂
(b) The scalar curl of F is equal to ∂x
(−2xy) − ∂y
(x2 + y 2 ) = −4y
→
−
which is not identically zero. Therefore F is not a gradient field.