University of California, Santa Barbara
Campus Learning Assistance Services
Midterm Review Package for Mathematics 5B
Tutor: Alan Mak
This review package, which contains some key points, worked examples, and practice problems
on some selected topics, serves to help students prepare for the midterm exam. Please be reminded
that this review package is a supplement, but not a substitute, of your textbook and lecture notes.
To suceed in the midterm exam, it would be a good idea to also work on the problems in the
textbook, in addition to those in this review package.
This review package is prepared by Alan Mak for the students enrolled in the tutorial group
of the Campus Learning Assistance Services. No part of this review package may be reproduced
without the permission of the Campus Learning Assistance Services.
1 Basic vector operations
Magnitude and unit vector
A vector ~v = aî + bĵ + ck̂ (a,b and c are scalars) can also be written as ~v = (a, b, c). The arrow in
the symbol ~v indicates that it is a vector, which is a quantity that has a magnitude and a direction.
The magnitude of ~v = aî + bĵ + ck̂, represented by ~v can be found by the following formula:
√
|~v | = a2 + b2 + c2
A unit vector is a vector with magnitude equal to 1. The unit vector in the direction of ~v , represented
by v̂, can be found by the following formula:
v̂ =
~v
|~v |
Dot product
The dot product is also known as scalar product. The result of a scalar product is a scalar. One
can comupte the scalar product of two 3-component vectors by summing up the products of the
corresponding components in the two vectors. Here is an example. Given two vectors v~1 =
î + 2ĵ + 3k̂ and v~2 = 4î + 5ĵ + 6k̂. Their dot product would be:
v~1 · v~2 = 1 · 4 + 2 · 5 + 3 · 6 = 32
Here are some more facts about dot product that you should remember.
1. v~1 · v~2 = v~2 · v~1
2. The dot product can also be computed by:
v~1 · v~2 = |v~1 ||v~2 | cos θ
where θ is the angle between the vectors v~1 and v~2 . Conversely, if you know the magnitudes
and dot product of the two vectors v~1 and v~2 , you can use this formula to find out the angle
θ between the two vectors.
1
3. The magnitude of a vector ~v can be computed from its dot product with itself, according to
the following formula:
√
|~v | = ~v · ~v
4. Any two vectors v~1 and v~2 that are perpendicular to one another have:
v~1 · v~2 = 0
Cross product
The cross product is also known as vector product. The result of a vector product is a vector. One
can compute the vector product of two 3-component vectors by setting up a 3-by-3 determinant.
First, put the unit vectors î, ĵ and k̂ in the first row. Then, put each of the 3 components of the first
vector in the first row, and each components of the second vector in the second row. Expanding
the determinant gives the cross product. Here is an example. Given two vectors v~1 = î + 2ĵ + 3k̂
and v~2 = 4î + 5ĵ + 6k̂. Their cross product would be:
¯
¯
¯
¯
¯
¯
¯ î ĵ k̂ ¯
¯
¯
¯ 1 2 ¯
¯
¯ 2 3 ¯
¯ 1 3 ¯
¯
¯
¯
¯
¯
¯
¯
¯
¯
v~1 × v~2 = ¯ 1 2 3 ¯ = î ¯
¯ − ĵ ¯ 4 6 ¯ + k̂ ¯ 4 5 ¯ = −3î + 6ĵ − 3k̂
5
6
¯ 4 5 6 ¯
Here are some more facts about dot product that you should remember.
1. v~1 × v~2 = −v~2 × v~1 (not equal to + v~2 × v~1 in general)
2. The magnitude of the cross product can be computed by:
|v~1 × v~2 | = |v~1 ||v~2 | sin θ
where θ is the angle between the vectors v~1 and v~2 .
3. The vector product v~1 × v~2 of two vectors v~1 and v~2 is always perpendicular to both v~1 and
v~2 . In other words,
(v~1 × v~2 ) · v~1 = (v~1 × v~2 ) · v~2 = 0
Practice problems
1. Given ~v = î + 9ĵ + 2k̂. Find |~v |. (Answer:
√
86)
√
2. Given
w
~
=
3
î
+
6
ĵ
+
9
k̂.
Find
ŵ,
the
unit
vector
in
the
direction
of
w.
~
(Answer:
(1/
14)î +
√
√
(2/ 14)ĵ + (3/ 14)k̂)
3. Given ~u = 2î + 10ĵ + k̂ , ~v = 15î + 20ĵ. Find ~u · ~v . (Answer: 230)
4. Given ~x = (8, 9, 12) , ~y = (2, 3, 1). Find ~x · ~y and ~y · ~x. (Answer: 55, 55)
5. Given v~1 = 5î + 5ĵ + 6k̂ , v~2 = 10î + 20ĵ + 15k̂. Find v~1 × v~2 . (Answer: −45î − 15ĵ + 50k̂)
6. Given u~1 = (1, 3, 2) , v~2 = (7, 8, 5). Find u~1 × u~2 and u~2 × u~1 . (Answer: (16,9,-13),
(-16,-9,13))
2
2
Linear Dependence
Criterion for linear independence
The criterion for n vectors v~1 , v~2 , · · · , v~n to be linearly indepedent is that for scalars c1 , c2 , · · · , cn ,
c1 v~1 + c2 v~2 + · · · + cn v~n = 0 ⇒ c1 = c2 = · · · = cn = 0
Here is an example. Consider the vectors v~1 = 3î, v~2 = î + 4ĵ + 8k̂ and v~3 = 2î + 9ĵ. Let c1 , c2 , c3
be three scalars. Now, we check the criterion for linear independence:
c1 v~1 + c2 v~2 + c3 v~3 = 0
⇒ c1 (3î) + c2 (î + 4ĵ + 8k̂) + c3 (2î + 9ĵ) = 0
⇒ (3c1 + c2 + 2c3 )î + (4c2 + 9c3 )ĵ + 8c2 k̂ = 0
⇒ 3c1 + c2 + 2c3 = 0 and 4c2 + 9c3 = 0 and 8c2 = 0
⇒ c1 = c2 = c3 = 0
Since the criterion is satisfied, we can now conclude that v~1 = 3î, v~2 = î+4ĵ +8k̂ and v~3 = 2î+9ĵ
are linearly independent.
Test for linear indenpence using determinant
We can also check whether vectors are linearly independent by computing a determinant whose
rows are the vectors to be tested. If the determinant turns out to be non-zero, then the vectors are
linearly independent. As an example, for the vectors v~1 = 3î, v~2 = î + 4ĵ + 8k̂ and v~3 = 2î + 9ĵ,
we can check if they are linearly independent by evaluating the following determinant:
¯
¯
¯ 3 0 0 ¯
¯
¯
¯ 1 4 8 ¯
¯
¯
¯ 2 9 0 ¯
The value of this determinant turns out to be -216, which is not equal to zero. Therefore, we can
conclude that the vectors v~1 = 3î, v~2 = î + 4ĵ + 8k̂ and v~3 = 2î + 9ĵ are linearly independent.
However, it should be noted that this method only works when the number of vectors and the number of components in each vectors are equal. Otherwise, say, if you include an addition vector
v~4 = î + 6ĵ + 5k̂, the determinant above will become a 4-by-3 one, which is undefined.
Practice problems
Check whether the following sets of vectors are linearly independent:
1. v~1 = î, v~2 = 2î + 8ĵ, v~3 = 30k̂ (Answer: linearly independent)
2. v~1 = 2î, v~2 = 3ĵ, v~3 = 5î + k̂ (Answer: linearly independent)
3. ~x = (1, 2, 5), ~y = (5, 0, 6), ~z = (3, 4, 2) (Answer: linearly dependent)
4. ~a = (8, 10, 90), ~b = (120, 56, 98), ~c = (0, 0, 0), d~ = (−8, 0, 300) (Answer: linearly dependent)
5. ~u = (5, 6, 9, 0), ~v = (2, 0, 3, 4), w
~ = (1, 0, 0, 0) (Answer: linearly independent)
3
3
Limits of Multivariable Functions
Examples
The idea is best illustrated with examples. First, let us take a look at a relatively straight forward
2
one. Here we want to evaluate: limx→0,y→0 ex sin(xy 2 ). The way to do this is by replacing the x’s
2
and y’s in the function ex sin(xy 2 ) by zero, and then simplify.
2
limx→0,y→0 ex sin(xy 2 ) = e0 sin(0)
=1·0
=0
Now, let’s take a look at another example. This time we want to evaluate:
1 + 8x3 y 2
x→∞,y→∞ 2x5 y 6 + 3y
lim
The function looks very complicate at the first glance. However, the problem would become much
easier if you divide both the numerator and the denominator by the highest powers of x and y that
you see in the function.
1
+ x28y4
(1/x5 y 6 )(1 + 8x3 y 2 )
1 + 8x3 y 2
0+0
x5 y 6
= lim
= lim
=0
lim
=
x→∞,y→∞ (1/x5 y 6 )(2x5 y 6 + 3y)
x→∞,y→∞ 2 + 53 5
x→∞,y→∞ 2x5 y 6 + 3y
2+0
x y
It should be noted that limit does not always exist. Here is an example in which the limit does not
exist:
ln(1 + xy)
lim
x→0,y→0 sin(x3 y 3 )
In this case, both the numerator and the denominator tend to zero. Limit does not exist.
Practice problems
Evaluate the following limits, if any:
1. limx→0,y→0 ex+9xy (Answer: 1)
2. limx→0,y→0 x2 e1+2x+3y+2x
3y
(Answer: 0)
3. limx→∞,y→∞ xy 9 e−x (Answer: limit does not exist)
4. limx→∞,y→∞
10x3 y 2 +x+100y
5x3 y 2 +30x2 y 2 +1000x
5. limx→0,y→0
x
x3 +y 2
6. limx→0,y→0
(1+x) sin y
y
(Answer: 2)
(Answer: limit does not exist)
Solution: limx→0,y→0
(1+x) sin y
y
= limx→0,y→0 (1 + x) · limx→0,y→0
= limx→0 (1 + x) · limy→0 siny
y
= (1 + 0) · 0
=0
4
siny
y
4
Finding Partial Derivatives
Techniques
Let us consider a function f of two variables x and y, defined as f (x, y) = x2 y 3 + x sin y. The
partial derivative ∂f
can be found by differentiating f with respect to x and treating y as a constant.
∂x
Therefore,
∂f
∂ 2 3
=
(x y + sin y) = 2xy 3 + sin y
∂x
∂x
can be found by differentiating f with respect to y and treating
Similarly, the partial derivative ∂f
∂y
x as a constant.
∂f
∂ 2
=
(x y + sin y) = 3x2 y 2 + x cos y
∂y
∂y
∂f
∂x
Now we know the partial derivatives
the function f .
df =
∂f
.
∂y
and
We can then find out the total differential df of
∂f
∂f
dx +
dy = (2xy 3 + sin y)dx + (3x2 y 2 + x cos y)dy
∂x
∂x
The partial derivatives can also be used to find out the Jacobian Matrix , which is defined as in
equation (2.32) on page 91 of the textbook by Wilfred Kaplan.
Practice problems
1. Given f (x, y) = 2x + 3y − cos(y). Find
∂f
∂x
3
∂f
.
∂y
2. Given f (x, y) = x2 ex sin x + 2x2 y 6 . Find
3. Given f (x, y) = ex tan y. Find
∂f
∂x
and
∂f
.
∂y
and
∂f
.
∂y
(Answer: 2, 3 + sin(y))
(Answer: 12x2 y 5 )
(Answer: ex tan y, ex sec2 y)
4. Pratice on chain rule: Given f (x, y) = ln(xy 3 ). Find
Solution:
∂f
∂x
∂f
∂y
=
=
∂
∂x
∂
∂y
ln(xy 3 ) =
ln(xy 3 ) =
∂f
∂x
and
1
xy 3
1
xy 3
· y 3 = x1
· 3xy 2 =
5. Practice on product rule: Given f (x, y) = exy sin(xy). Find
the total differential df .
Solution:
∂f
∂x
∂f
∂y
df
∂f
.
∂y
∂f
∂x
3
y
and
∂f
.
∂y
Hence write down
∂ xy
= ∂x
e sin(xy)
= exy · cos(xy) · y + sin(xy) · exy · y
= yexy (sin(xy) + cos(xy))
∂ xy
= ∂y
e sin(xy)
= exy · cos(xy) · x + sin(xy) · exy · x
= xexy (sin(xy) + cos(xy))
= ∂f
dx + ∂f
dy
∂x
∂y
xy
= (ye (sin(xy) + cos(xy)))dx + (xexy (sin(xy) + cos(xy)))dy
= exy (sin(xy) + cos(xy))(ydx + xdy)
5
5
Partial Derivatives of Implicit Functions
Example 1
Consider the implicit function z(x, y) defined by x2 = z sin(yz). Find
∂z
∂x
and
∂z
.
∂y
We start from the given equation that relates x, y and z:
x2 = z sin(yz)
Now, we take the differential on both sides. Remember to apply product rule on the right hand
side.
⇒
⇒
⇒
⇒
⇒
2x dx = sin(yz) dz + z cos(yz)(y dz + z dy)
2x dx = sin(yz) dz + yz cos(yz) dz + z 2 cos(yz) dy
2x dx = (sin(yz) + yz cos(yz)) dz + z 2 cos(yz) dy
2x dx − z 2 cos(yz) dy = (sin(yz) + yz cos(yz)) dz
2x dx−z 2 cos(yz) dy
dz = (sin(yz)+yz cos(yz))
z 2 cos(yz)
2x
dz = (sin(yz)+yz cos(yz)) dx− (sin(yz)+yz cos(yz)) dy
Finally, read off the partial derivatives from the expression for the differential of z. This gives:
∂z
z 2 cos(yz)
=−
∂y
(sin(yz) + yz cos(yz))
∂z
2x
=
∂x
(sin(yz) + yz cos(yz))
Example 2
Consider the implicit functions y1 (x1 , x2 ) and y2 (x1 , x2 ) defined by:
2x21 + x22 + y12 − y1 y2 = 1
x21 + x22 + 2y12 + y1 y2 = 3
First, we move all the terms in each of the two equations to the left side, leaving the right side zero,
and then define the left sides of the two equations to be two functions, F1 and F2 , respectively.
F1 (x1 , x2 , y1 (x1 , x2 ), y2 (x1 , x2 )) = 2x21 + x22 + y12 − y1 y2 − 1
F2 (x1 , x2 , y1 (x1 , x2 ), y2 (x1 , x2 )) = x21 + x22 + 2y12 + y1 y2 − 3
Then, by the first formula in (2.61) on page 108 of the textbook by Wilfred Kaplan,
¯
¯
¯ 4x1 −y1 ¯
¯
¯
∂(F1 ,F2 )
¯ 2x1 y1 ¯
−6x1 y1
x1
∂y1
∂(x1 ,y2 )
¯=
= ∂(F1 ,F2 ) = ¯¯
=
−
2
¯
∂x1
6y1
y1
¯ 2y1 − y2 −y1 ¯
∂(y1 ,y2 )
¯ 4y1 + y2 y1 ¯
Practice Problems (on the next page)
6
1. An implicit function z(x, y) is defined by the equation x2 sin z = yz 2 . Find
(Answers:
2x sin z
−z 2
,
)
2yz−x2 cos z 2yz−x2 cos z
2. An implicit function z(x, y) is defined by the equation xz 3 = y tan z. Find
(Answers:
−z 3
tan z
,
)
3xz 2 −y sec2 z 3xz 2 −y sec2 z
∂z
∂x
and
∂z
.
∂y
∂z
∂x
and
∂z
.
∂y
3. Consider the same inplicit functions in the example 2. Use the other three formulae in (2.61)
on page 108 of the textbook by Wilfred Kaplan,
∂y1
=
∂x2
∂(F1 ,F2 )
∂(x2 ,y2 )
∂(F1 ,F2 )
∂(y1 ,y2 )
∂y2
=
∂x1
∂(F1 ,F2 )
∂(y1 ,x1 )
∂(F1 ,F2 )
∂(y1 ,y2 )
∂y2
=
∂x2
∂(F1 ,F2 )
∂(y1 ,x2 )
∂(F1 ,F2 )
∂(y1 ,y2 )
to show that
∂y1
2x2 y1
=
∂x2
3x21
6
∂y2
2y1 + y2
=
∂x1
6x1
∂y2
2x2 (y1 + y2 )
=
∂x2
3x21
Partial Derivatives of Inverse Functions
Example
Consider the mapping
x1 (u1 , u2 ) = u21 − u22 , x2 (u1 , u2 ) = 2u1 u2 .
Its Jacobian matrix is:
·
[xu ] =
2u1 −2u2
2u1 2u2
¸
According to page 120 of the textbook by Wilfred Kaplan, the Jacobian matrix of the inverse
mapping is simply the inverse of the Jacobian matrix of the mapping . In other words,
[ux ] = [xu ]−1
Now, recall the rule for inverting a 2 × 2 matrix, which you learned in previous Mathematics
courses:
·
¸−1
·
¸
1
a b
d −b
=
c d
ad − bc −c a
Using this rule, we have:
1
[ux ] = 2
4u1 + 4u22
·
2u1 2u2
−2u2 2u1
"
¸
=
u1
2u21 +2u22
2
− 2u2u+2u
2
1
2
u2
2u21 +2u22
u1
2u21 +2u22
#
This is is Jacobian matrix of the inverse mapping u1 (x1 , x2 ), u2 (x1 , x2 ). We can read off from the
Jacobian matrix that:
u1
∂u1
u2
∂u1
= 2
,
= 2
2
∂x1
2u1 + 2u2
∂x2
2u1 + 2u22
∂u2
u2
∂u2
u1
=− 2
,
= 2
2
∂x1
2u1 + 2u2
∂x2
2u1 + 2u22
7
Practice Problem
Consider the mapping x1 (u1 , u2 ) = u1 u2 , x2 (u1 , u2 ) = u31 .
(a) Find the Jacobian matrix of this mapping, [xu ].
(b) Find the Jacobian matrix of the inverse mapping, [ux ].
∂ui
(c) Find the partial derivatives of the inverse mapping, ∂x
.
j
∂u1
∂u1
∂u2
−u2
1
1 ∂u2
(Answers: ∂x1 = 0, ∂x2 = 3u2 , ∂x1 = u1 , ∂x2 = 3u3 )
1
7
1
Curvilinear Coordinates
Facts
Transforming from spherical coordinates to rectangular coordinates:
x = ρ sin φ cos θ
y = ρ sin φ sin θ
z = ρ cos φ
Transforming from rectangular coordinates to spherical coordinates:
ρ
θ
p
= x2 + y 2 + z 2
= arccos √ 2 x 2
φ
= arccos √
x +y +z 2
z
x2 +y 2 +z 2
= arcsin √ 2 y 2 2 = arctan xy
√x +y +z
x2 +y 2
= arctan
z
Transforming from cylindrical coordinates to rectangular coordinates:
x = r cos θ
y = r sin θ
z=z
Transforming from rectangular coordinates to cylindrical coordinates:
r=
p
x2 + y 2
θ = arctan
y
x
z=z
Practice Problems
1. Using the transfomration from spherical coordinates to rectangular coordinates, show that
the Jacobian matrix is:


sin φ cos θ −ρ sin φ sin θ ρ cos φ cos θ
 sin φ sin θ ρ sin φ cos θ ρ cos φ sin θ 
cos φ
0
−ρ sin φ
Hence, show that the Jacobian determinant is:
∂(x, y, z)
= ρ2 sin φ
∂(ρ, θ, φ)
8
2. Using the transfomration from cylindrical coordinates to rectangular coordinates, show that
the Jacobian matrix is:


cos θ −r sin θ 0
 sin θ r cos θ 0
0
0
1
Hence, show that the Jacobian determinant is:
∂(x, y, z)
=r
∂(r, θ, z)
8
Gradient, Divergence, Curl and Laplacian
Definitions
∂f
∂f
∂f
+ ĵ
+ k̂
∂x
∂y
∂z
∂Fx ∂Fy ∂Fz
div F~ = ∇ · F~ =
+
+
∂x
∂y
∂z
¯
¯
¯ î ĵ k̂ ¯
¯ ∂
¯
∂
∂ ¯
curl F~ = ∇ × F~ = ¯¯ ∂x ∂y
∂z ¯
¯ F F F ¯
x
y
z
2
2
∂ f
∂f 2
∂ f
+
+
∇2 f = ∇ · (∇f ) =
∂x2
∂y 2
∂z 2
grad f = ∇f = î
Practice Problems
1. Given f (x, y, z) = 3y 2 + x sin z + ln z, find the gradient of f .
(Answer: (sin z)î + 6y ĵ + (x cos z + z1 )k̂)
2. Given f (x, y, z) = 3 exp(tan(xyz)), find the gradient of f .
(Answer: 3 exp(tan(xyz)) sec2 (xyz)(yz î + xz ĵ + xy k̂))
3. Given f (r) = 1r , find the gradient of f . (Answer: − r~r3 )
4. Given F~ (x, y, z) = exy î − exy ĵ + eyz k̂, find the divergence of F~ .
(Answer: (y − x)exy + yeyz )
5. Given F~ (x, y, z) = xî + (y + cos x)ĵ + (z + exy )k̂,
find the divergence of F~ . (Answer: 3)
6. Given F~ (x, y, z) = x3 î + (x sin(xy))ĵ, find the divergence of F~ .
(Answer: (3 − cos(xy))x2 )
7. Given F~ (x, y, z) = î sin(xy) − ĵ cos(x2 y), find the divergence of F~ .
(Answer: y cos(xy) + x2 sin(x2 y))
9
8. Given F~ (x, y, z) = (x2 + y 2 + z 2 )(3î + 4ĵ + 5k̂), find the curl of F~ .
(Answer: (10y − 8z)î + (6z − 10x)ĵ + (8x − 6y)k̂)
9. Given F~ (x, y, z) = (x + y)î + (y + z)ĵ + (x + z)k̂, find the curl of F~ .
(Answer: −~r)
10. Given F~ (x, y, z) = î sin x + ĵ cos y, find the scalar curl of F~ .
(Answer: − sin x)
11. Given F~ (x, y, z) = xy î + (x2 − y 2 )ĵ, find the scalar curl of F~ .
(Answer: x)
12. Given f (x, y, z) =
x2 −y 2
,
x2 +y 2
verify that ∇ × ∇f = 0.
13. Given F~ (x, y, z) = îx2 z 3 + ĵx cos y + k̂zey , verify that ∇ · (∇ × F~ ) = 0.
14. Given F~ (x, y, z) = xî + y ĵ + z k̂, show that F~ is irrotational .
15. Given f (x, y, z) = √
9
1
,
x2 +y 2 +z 2
show that f is harmonic .
Geometrical Applications of the Gradient
Facts: Let S be a surface consisting of the points (x, y, z) such that f (x, y, z) = 0.
1. The tangent plane to S at the point (x0 , y0 , z0 ) of S is given by
∇f (x0 , y0 , z0 ) · (x − x0 , y − y0 , z − z0 ) = 0
2. The line normal to S at the point (x0 , y0 , z0 ) of S is given by
~r(t) = (x0 , y0 , z0 ) + t∇f (x0 , y0 , z0 )
Practice Problems
Find the equation for the tangent plane to each of the following surfaces at the indicated point.
1. x2 + y 2 + z 2 = 3, (1, 1, 1)
(Answer: x + y + z = 3)
2. x3 + z 3 = 2y 3 , (1, 1, 1)
3. ez cos x cos y = 0, ( π2 , 1, 0)
4. exyz = 1,
(1, 2, 0)
Find the parametric equations for the normal line to each of the following surfaces at the indicated
point.
1. x2 + y 2 + 2z = 4, (1, 1, 1)
(Answer: x(t) = 1 + 2t, y(t) = 1 + 2t, z(t) = 1 + 2t)
2. tan(xy) = z,
( π2 , π4 , 1)
10
10
Directional Derivatives
Fact: The directional derivative of a function f along the direction ~v is found by
∇v f = ∇f · v̂ with v̂ =
~v
|~v |
Practice Problems
For each of the following functions f (x, y), find the directional derivatives at the given point
(x0 , y0 ) in the direction ~v .
p
1. f (x, y) = cos( √x2 + y 2 ), ~v = î + ĵ, (x0 , y0 ) = (1, 1)
(Answer: − sin 2)
2
2. f (x, y) = exp(−x
− y 2 ), ~v = î + ĵ,
√
(Answer: −2 2 exp(−2))
3. f (x, y) = ex + yz, ~v = î − ĵ + k̂,
(Answer: √e3 )
4. f (x, y) = xyz, ~v = î − k̂,
(Answer: 0)
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(x0 , y0 ) = (1, 1)
(x0 , y0 ) = (1, 1)
(x0 , y0 ) = (1, 0)
Critical Points of Multi-variable Functions
Techniques
If one wants to study about the critical points of a two-variable function f (x, y), one should first
locate the critical points. This can be achieved by setting
∂f
∂f
= 0 and
=0
∂x
∂y
and solve for thecoordinates, (x0 , y0 ), of the critical points. For each of the critcal points (x0 , y0 ),
check the discriminant
D(x0 , y0 ) =
∂2f
∂ 2f
∂ 2f
(x
,
y
)
(x
,
y
)
−
(
(x0 , y0 ))2
0
0
0
0
∂x2
∂y 2
∂x∂y
1. If D(x0 , y0 ) > 0 and
∂2f
(x0 , y0 )
∂x2
> 0, then (x0 , y0 ) is a local minimum.
2. If D(x0 , y0 ) > 0 and
∂2f
(x0 , y0 )
∂x2
< 0, then (x0 , y0 ) is a local maximum.
3. If D(x0 , y0 ) < 0, then (x0 , y0 ) is a saddle point.
Practice Problems
For each of the following functions, locate the critical point(s), and determine if each of them is a
maximum, a minimum or a saddle point.
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1. f (x, y) = ln(x2 + y 2 + 1) (Answer: Local minimum at (0, 0))
2. f (x, y) = x2 − y 2 + xy (Answer: Saddle point at (0, 0))
3. f (x, y) = x5 y + xy 5 + xy (Answer: Saddle point at (0, 0))
4. f (x, y) = exp(1 + x2 − y 2 ) (Answer: Saddle point at (0, 0))
5. f (x, y) = 3x2 + 2xy + 2x + y 2 + y + 4 (Answer: Local minimum at (− 41 , − 14 ))
6. f (x, y) = xy +
1
x
+
1
y
(Answer: Local minimum at (1, 1))
7. f (x, y) = x3 + y 2 − 6xy + 6x + 3y (Answer: Saddle point at (1, 23 ), Local minimum at
(5, 27
))
2
8. f (x, y) = (x2 + 3y 2 ) exp(1 − x2 − y 2 ) (Answer: Local minimum at (0, 0), Local maxima at
(0, 1) and (0, −1), Saddle points at (1, 0) and (−1, 0))
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Lagrange Multipliers
Techniques
If we want to find the maximum and the minimum of a function f (x, y, z), subjected to a contraint
g(x, y, z) = 0, we can use the Lagrange-multiplier method. This is done by setting up the following
four equations, and solve for the coordinate (x, y, z) of the extremum points.
∂f
∂g
=λ
∂x
∂x
∂f
∂g
=λ
∂y
∂y
∂f
∂g
=λ
∂z
∂z
g(x, y, z) = 0
Practice Problems
In each of the following, find the extrema of the function f subjected to the stated constraint.
q
q q
2
2
2
2
1. f (x, y, z) = x−y+z, subjected to x +y +z = 2 (Answer: Maximum at ( 3 , − 23 , 23 )
q q
q
2
2
and Minimum at (− 3 , 3 , − 23 ))
√
2
2
2. f (x,
y)
=
x,
subjected
to
x
+
2y
=
3
(Answer:
Maximum
at
(
3, 0) and Minimum at
√
(− 3, 0))
3. f (x, y) = 3x + 2y, subjected to 2x2 + 3y 2 = 3 (Answer: Maximum at ( √970 , √470 ) and
Minimum at (− √970 , − √470 ))
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