SAMPLE PROBLEMS FOR FINAL EXAM 2601 Calculus III for CS Fall 2001

Dec 1, 2001
2601
Calculus III for CS
Fall 2001
SAMPLE PROBLEMS FOR FINAL EXAM
TIME: Dec 14, Friday, 11:30-2:20 in the usual classroom Skiles 249.
MATERIAL: Everything we covered:
• Sections 1, 2, 3, 4, 5, 6, 7 from the [Notes:Th]
• Sections 1, 2, 3, 4, 5, 6 from [Notes:Num] (see on the web if you have lost your copy).
Section 9 (Appendix on determinants) is not needed, but you should be able to compute 2
by 2 and 3 by 3 determinants.
• Salas-Hille-Etgen: Section 12.1–12.7, 13.1–13.4, 14.1–14.6, 15.1–15.6, 16.1–16.3
• Handout on the Chain rule and Jacobian matrix, Newton’s method and Handout on
Hessian and second partial derivative test.
(http://www.math.gatech.edu/~bourbaki/math2601/pdf/jacobian.pdf)
(http://www.math.gatech.edu/~bourbaki/math2601/pdf/hessian.pdf)
See also the “List of important problems” below.
The exam will be slightly biased towards the material we learned since the second
midterm. However, the exam is cumulative. The following sample problem set contains
problems ONLY from the last third of our syllabus. However, the exam covers EVERYTHING. Use the previous Sample Problems for Midterm I and II to review the earlier
material. If you lost your copy, look at:
http://www.math.gatech.edu/~bourbaki/math2601/pdf/sample1.pdf
1
http://www.math.gatech.edu/~bourbaki/math2601/pdf/sample2.pdf
All definitions and theorems should be learned and can be asked.
Review the basic differentiation rules. You MUST be able to differentiate basic functions
without errors (polynomials, trigonometric functions and inverses, exp, log, product rule,
quotient rule, chain rule, derivative of inverse function).
Review the basic integration formulas (including integrals of elementary functions, integration by parts, integration by substitution).
The exam is closed-book, only pen, pencil, eraser and a small (non-programmable)
calculator are allowed. You can use a “cheat-sheet” of size A4, one-sided, where you can
put any formula you wish in advance (like for Midterm II. just twice bigger sheet).
When solving the problems, include all side calculations, and clearly cross out those
parts which you consider irrelevant or wrong. Unjustified results are not accepted, you
have to show your work. The solution should contain sufficient details (side calculations,
explanatory words or signs indicating which equation follows from the other one etc.) so
that anyone could reproduce your argument based upon your writing. Showing the final
result only is not enough to get credit.
The final result should be reasonably simplified if possible. Results like 12 − 14 or 2x+3−
4x−6 will not be considered complete, you have to write 12 − 14 = 14 or 2x+3−4x−6 = −2x−3.
Usual OFFICE HOURS are on MWF 11:00-12:30 next week. Before the test, I will
have an office hour on Dec 13, Thursday 4:00– 7:00 PM and on Dec 14, Friday 10:00 – 11:30
PM. If these times are in conflict with your other schedules, let me know; other times are
also possible.
The graded finals will be available on Saturday morning from 10:00-11:00 AM. I can
send your final grade by email if you ask for it by email from your own Gatech account
before Saturday 10:00 AM. I will leave for Europe at 11:00 AM Saturday so every question
should be settled before that.
Do not jump onto the solutions immediately, try to solve the problems first by yourself.
If you find any mistake, typo etc. in this problem set, please let me know.
2
List of important basic problems
The following is a list of basic problems, which I expect everyone to be able to solve.
The exam is not restricted to only these problems, but these will play a major role.
• Basic geometry of lines and planes in space. Problems of type Problem 1 and 2 on
Sample Midterm I.
• Full solution of Ax = b with Gaussian elimination and its application (finding basis
in the null-space, column-space; dimension, deciding linear dependence, indepedence).
Problem 3 (i)-(v) and Problem 4 from Sample Midterm I.
• Change of basis matrix and matrix of linear transformations in general vectorspaces
and in Rn . Problem 5 and 6 from Sample Midterm I.
• QR-decomposition with Gram-Schmidt. Projections onto nullspace and columnspace.
Least square solutions. Problem 3 (vi)-(xii) from Sample Midterm I. (this problem
did not work out well at the first midterm and is very important...)
• Finding eigenvalues and eigenvectors of a square matrix. (e.g. Problem 1 from Sample
Midterm II.)
• Finding the spectral decomposition of a symmetric matrix. Solving systems of linear
ordinary differential equations. (Problem 2 from Sample Midterm II. This question
again did not work out well...)
• Finding the singular value decomposition of a matrix (Problem 3 from Sample Midterm
II.)
• Computing condition number and its application in error magnification. Importance
of orthogonal matrices. (e.g. Problem 4 from Sample Midterm II.)
• Finding LU-decomposition by “bookkeeping” Gaussian elimination. (Problem 4 from
Sample Midterm II.)
• Finding QR-decomposition with Householder and Givens. (Problems 5, 6, 7 from
Sample Midterm II.)
• Jacobi and Gauss-Seidel iterations. Including: prediction of the number of steps
needed (this latter part did not work out well...) e.g. Problem 9 from Sample Midterm
II.
• Power method for eigenvalues (Problem 10 from Sample Midterm II.)
• QR-iteration for diagonalizing symmetric matrices (Problem 11 from Sample Midterm
II.)
3
• Calculus of curves. Tangent and principal unit normal vectors. Arclength. (Problems
1–3 below)
• Computing partial derivatives and gradients with its application (directional derivatives, tangent lines to level curves, tangent planes to graphs etc.). Problems 4–9 and
14–16 below.
• Finding the Jacobian and applying chain rule for vector valued functions in several
variables. (Problems 10–12 below)
• Newton’s method in several variables. Finding all roots of complex equations. (Problem 13)
• Finding local and global extrema. Lagrange multiplier. (Problems 17–22)
• Evaluate multiple integrals. (Problem 23).
4
PROBLEMS
PROBLEM 1. The curves
r1 (t) = (et − 1)i + 2 sin tj + ln(t + 1)k
and
r2 (u) = (u + 1)i + (u2 − 1)j + (u3 + 1)k
intersect at P (0, 0, 0). Find the angle of intersection.
PROBLEM 2. Find the point(s) at which the curve
r(t) = ti + t2 j + t3 k
intersects the plane 4x + 2y + z = 24 and find the angle of intersection.
PROBLEM 3. Find the unit tangent vector, the principal normal vector and an
equation in x, y, z of the osculating plane at the point on the curve
r(t) = i + 2tj + t2 k
corresponding to t = 1.
PROBLEM 4. Compute the arclength of the following curves
a,
b,
r(t) = t3 i + t2 j
between
r(t) = (ln t)i + 2tj + t2 k
0≤t≤1
between
1≤t≤e
PROBLEM 5. Find the gradient of the following functions
f (x, y) = 3x2 − xy + y
g(x, y) = (x + y)ex−y
h(x, y, z) = ex+2y cos(z 2 + 1)
PROBLEM 6. Find the directional derivative of ln(x2 + y 2 ) at the point (0, 1) in the
direction of 8i + j.
5
PROBLEM 7. Find the directional derivative of f (x, y, z) = x2 + 2xyz − yz 2 at
(1, 1, 2) in a direction parallel to the straight line
x−1
z−2
=y−1=
2
−3
p
PROBLEM 8. a, Find the unit vector in the direction in which f (x, y, z) = x2 + y 2 + 2z 2
increases most rapidly at the point (1, −2, 1) and give the rate of change of f in this direction.
p
b, Find the unit vector in the direction in which f (x, y, z) = x2 + y 2 + 2z 2 decreases
most rapidly at the point (1, −2, 1) and give the rate of change of f in this direction.
x2
PROBLEM 9. a, Determine the path of the steepest descent along the surface z =
+ 3y 2 starting from the point (1, 2, 13).
b, Where does this path end?
c, What is the projection of this path on the (x, y) plane?
PROBLEM 10. Find the rate of change of f with respect to t along the given curve
r(t)
a, f (x, y) = x2 y, r(t) = et i + e−t j
b, f (x, y, z) = y 2 sin(x + z), r(t) = 2ti + (cos t)j + t3 k
PROBLEM 11. Find the rate of change of u with respect to t in the following two
problems:
√
a, u = x + 4 xy − 3y where x = t3 , y = t−1
b, u = x2 tan y, where x = s2 t, y = s + t2 .
PROBLEM 12. Let
f1 (u1 , u2 ) = u1 u2
f2 (u1 , u2 ) = u21 + u2
and
u1 (x1 , x2 , x3 ) = x1 − x2 x3
6
u2 (x1 , x2 , x3 ) = x22 + x3
a, Find the Jacobian of f =
f1
f2
and the Jacobian of u =
u1
u2
b, Find the Jacobian of f ◦ u
c, What is
∂f2
∂x3 ?
PROBLEM 13. Find a solution to the system of nonlinear equations
x + y3 = 2
x2 + y = 3
with Newton’s method starting from (x0 , y0 ) = (1, 1). Run two steps of the iteration. Plug
these approximate solutions to the original equations to see the error.
PROBLEM 14. Find the normal vector and the tangent vector to the curve x5 + y 5 =
at the point (1, 1). Write up the equation of the normal line and the tangent line.
(Recall that the equation of a line has two forms: normal form and parametric form. Write
up both.)
2x3
PROBLEM 15. a, Find an equation for the tangent plane of xy 2 + 2z 2 = 12 at the
point (1, 2, 2).
b, Find the parametric equation of the normal line.
c, Find the equation of the line in the tangent plane obtained in part a, which is parallel
to the plane x + y + z = 3 and goes through (1, 2, 2).
PROBLEM 16. The curve r(t) = 2ti+3t−1 j−2t2 k and the ellipsoid x2 +y 2 +3z 2 = 25
intersects at (2, 3, −2). What is the angle of intersection?
PROBLEM 17. Find the critical points and local extreme values of the following
functions. Decide whether the extreme point is local maximum or minimum.
a, f (x, y) = x2 + 2xy + 3y 2 + 2x + 10y + 1
b, g(x, y) = xy + x−1 + 8y −1
c, h(x, y) = (x − y)(xy − 1)
d, k(x, y) = xy −1 − yx−1
7
PROBLEM 18. Find the absolute maximum and minimum of the function f (x, y) =
y(x − 3) on the closed disk of radius 3 about the origin.
PROBLEM 19. Find the absolute maximum and minimum of the function f (x, y) =
(x + y − 2)2 on the closed square D = {(x, y) : 0 ≤ x ≤ 3, 0 ≤ y ≤ 3}.
PROBLEM 20. Minimize x + 2y + 4z on the sphere x2 + y 2 + z 2 = 7.
PROBLEM 21. Find the distance from the point (0, 1) to the parabola x2 = 4y.
PROBLEM 22. Show that all triangles inscribed in a circle of radius 1 the equilateral
one has the largest product of the lengths of the sides.
PROBLEM 23. Evaluate the following integrals on the given domain Ω
Z Z
(a)
Z Z
(b)
Ω
n
Z Z
(c)
Ω
(d)
Ω
Ω
Ω
yex dxdy
(x4 + y 2 ) dxdy
Z Z
(f )
2
ex dxdy
Ω = (x, y) : 0 ≤ x ≤
n
(x + 3y 3 ) dxdy
Z Z
(e)
Ω = (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x
sin(x + y) dxdy
Ω
Z Z
n
xy 3 dxdy
o
π
πo
,0 ≤ y ≤
2
2
o
Ω = (x, y) : x2 + y 2 ≤ 1
n
Ω = (x, y) : 0 ≤ y ≤ 1, 0 ≤ x ≤ y 2
o
Ω = bounded region between y = x2 and y = x3
Ω = triangle between the x-axis, x = 2 and x = 2y
PROBLEM 24. Decide whether the following statements true (T) or false (F). If
false, give a short argument. WARNING: Some of them are “If... then” statements. Such a
statement is true only when the “if” part surely implies the “then” part. Such a statement
is false when the “then” part is not satisfied even in one single case for which the “if” part
is valid.
(a) If a function f : R2 → R has first partial derivatives at the origin, then ∇f (0, 0)
exists.
8
(b) If a function f : R2 → R has continuous first partial derivatives in a neighborhood
of the origin, then ∇f (0, 0) exists.
(c) If a function f : R2 → R is continuous at the origin, then ∇f (0, 0) exists.
(d) If ∇f (0, 0) exists for a function f : R2 → R, then it is continuous at the origin.
(e) Let U be an open set of R2 . If f : U → R satisfies ∇f (x) = 0 for all x ∈ U , then f
is constant on U .
(f) Let U be an open connected set of R2 . If f : U → R satisfies ∇f (x) = 0 for all
x ∈ U , then f = 0 on U .
(g) Let U be an open connected set of R2 . If f : U → R and g : U → R are two
functions whose gradients are equal at every point of U , then the two functions are equal.
(h) If f : R2 → R has a local extreme value at x0 then ∇f (x0 ) = 0
(i) Let f : R2 → R. If ∇f (x0 ) = 0, then f has a local extremum at x0 .
(j) If the function f : R2 → R has a local maximum at x0 then x0 is a critical point.
(k) Let Ω be a closed subset of R2 and f : Ω → R a continuous function. Then f
attains its maximum on Ω
(l) Let Ω be an open subset of R2 and f : Ω → R a continuous function. Then f attains
its maximum on Ω.
(m) Let Ω be a closed subset of R2 and f : Ω → R be a continuous function and let
∇f (x0 ) = 0 be the only stationary point in Ω. Then f attains its maximum or minimum
at x0 .
(n) Let f : R2 → R have continuous second partial derivatives and let x0 be a critical
2
point. If ∂∂ 2fx (x0 ) > 0 then f has a local minimum at x0 .
(o) Let f : R2 → R have continuous second partial derivatives and let x0 be a critical
point. If the discriminant D = (fxy )2 − (fxx )(fyy ) is positive at x0 , then x0 is a saddle.
(p) Let f : R2 → R have continuous second partial derivatives and let x0 be a critical
point. If the discriminant D = (fxy )2 − (fxx )(fyy ) is negative at x0 , then x0 is a local
maximum.
(q) Let f : R2 → R have continuous second partial derivatives and let x0 be a critical
point. If all second partial derivatives vanish at x0 , then the function has no local extremum
in x0 .
9
SOLUTIONS
PROBLEM 1. The curves
r1 (t) = (et − 1)i + 2 sin tj + ln(t + 1)k
and
r2 (u) = (u + 1)i + (u2 − 1)j + (u3 + 1)k
intersect at P (0, 0, 0). Find the angle of intersection.
SOLUTION TO PROBLEM 1. We have to compute the angle between the tangent vectors to these curves at the given point. First we determine the parameter values
corresponding to P (0, 0, 0): it is clear that r1 (0) = r2 (−1) = P (0, 0, 0) (for example you
solve et − 1 = 0 and u + 1 = 0).
Then we compute the derivatives:
r01 (t) = et i + 2 cos tj +
and
1
k
t+1
r02 (u) = i + 2uj + 3u2 k
i.e
u := r01 (0) = i + 2j + k
and
v := r02 (−1) = i − 2j + 3k
The cosine of the angle between these two vectors is
cos θ =
u·v
0
= √ √ =0
kuk kvk
6 14
hence θ = π/2.
PROBLEM 2. Find the point(s) at which the curve
r(t) = ti + t2 j + t3 k
intersects the plane 4x + 2y + z = 24 and find the angle of intersection.
SOLUTION TO PROBLEM 2. The t-parameter value of the points of intersection
is given by
4t + 2t2 + t3 = 24
10
This is a cubic equation, so no simple solution formula is available. One can apply Newton’s
method, but before that, it is advisable to check for simple (=rational) roots. The candidates
for rational roots of t3 + 2t2 + 4t − 24 = 0 are exactly the divisors of 24 (since the main
coefficient is 1). i.e. ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24. After a short trial you get that t = 2
is a root (you might have seen this immediately). Dividing by the factor t − 2 we have
t3 + 2t2 + 4t − 24 = (t − 2)(t2 + 4t + 12)
The quadratic equation t2 + 4t + 12 has no real roots, so the only intersection is at t = 2,
at the point P (2, 4, 8).
The tangent vector to r(t) at this point is


1
u := r0 (2) = i + 4j + 12k =  4 
12
from r0 (t) = i + 2tj + 3t2 k.
 
4
The normal vector to the given plane is n =  2  (coefficients of the equation). The
1
angle θ between u and n is obtained from
cos θ =
u·n
24
√
=√
kuk knk
161 21
i.e. θ ≈ 1.15(rad). However, this is the angle between the tangent to the curve and the
normal to the plane. The angle of intersection is π2 − 1.15 = 0.42. (This depends a bit on
the interpretation, but if the curve happens to lie within the plane, then we would rather
say that the “angle” of intersection is 0 and not π/2).
PROBLEM 3. Find the unit tangent vector, the principal normal vector and an
equation in x, y, z of the osculating plane at the point on the curve
r(t) = i + 2tj + t2 k
corresponding to t = 1.
SOLUTION TO PROBLEM 3. We have
r0 (t) = 2j + 2tk
hence
T(t) :=
r0 (t)
2j + 2tk
j + tk
=√
=√
0
2
kr (t)k
4t + 4
t2 + 1
11
(always simplify as much as possible!!!)
Thus at t = 1,
j+k
T(1) = √
2
For the normal vector we need
k
−tj + 1k
t
T0 (t) = √
(j + tk) = 2
− 2
3/2
2
(t + 1)3/2
t + 1 (t + 1)
(Chain rule!! Imagine T(t) =
√ 1 (j
t2 +1
+ tk)). Evaluate it at the point
T0 (1) =
−j + k
23/2
We have to normalize this vector. Before we divide by its norm, recall that we are
allowed to multiply it by any positive number, it will not change the normalized vector.
Hence we will rather normalize −j + k, i.e.
−j + k
√
2
N(1) =
Finally, the osculating plane goes through the point r(1) = (1, 2, 1) and its normal
vector is n := T(1) × N(1) = (1, 0, 0). Hence the equation is
 
n·
 
x
1 !
 y  −  2  = 0,
z
1
x−1=0
i.e.
PROBLEM 4. Compute the arclength of the following curves
a,
b,
r(t) = t3 i + t2 j
between
r(t) = (ln t)i + 2tj + t2 k
between
SOLUTION TO PROBLEM 4.
a, Compute
r0 (t) = 3t2 i + 2tj
12
0≤t≤1
1≤t≤e
Hence
Z
1
0
9t2
Use substitution: u =
Z
0
1
kr0 (t)k dt =
Z
1p
9t4 + 4t2 dt =
0
Z
1
0
p
t 9t2 + 4dt
+ 4, then du = 18tdt and don’t forget to change the limits:
p
t 9t2 + 4dt =
Z
1
18
13 √
4
u du =
1 3/2
1 √
13 − 43/2 =
13 13 − 8
27
27
b, We compute
1
r0 (t) = i + 2j + 2tk
t
Hence
Z
1
e
Z
0
kr (t)k dt =
t−2
1
Z
=
ep
e
1
+4
+ 4t2
Z
dt =
h
(t−1 + 2t) dt = ln t + t2
eq
(t−1 + 2t)2 dt
1
ie
1
= e2
PROBLEM 5. Find the gradient of the following functions
f (x, y) = 3x2 − xy + y
g(x, y) = (x + y)ex−y
h(x, y, z) = ex+2y cos(z 2 + 1)
SOLUTION TO PROBLEM 5.
∇f (x, y) = (6x − y, −x + 1)
∇g(x, y) = ex−y (1 + x + y), ex−y (1 − x − y)
∇h(x, y, z) = ex+2y cos(z 2 + 1), 2 cos(z 2 + 1), −2z sin(z 2 + 1)
PROBLEM 6. Find the directional derivative of ln(x2 + y 2 ) at the point (0, 1) in the
direction of 8i + j.
SOLUTION TO PROBLEM 6. The gradient of f (x, y) = ln(x2 + y 2 ) is
∇f (x, y) =
2x
2y ,
x2 + y 2 x2 + y 2
At the given point (x, y) = (0, 1):
∇f (0, 1) = (0, 2) = 2j.
13
The unit vector in the given direction is
u=
8i + j
8
8i + j
1
= √ i+ √ j
=√
2
2
k8i + jk
65
65
8 +1
The directional derivative
2
fu0 (0, 1) = ∇f (0, 1) · u = √ .
65
PROBLEM 7. Find the directional derivative of f (x, y, z) = x2 + 2xyz − yz 2 at
(1, 1, 2) in a direction parallel to the straight line
x−1
z−2
=y−1=
2
−3
SOLUTION TO PROBLEM 7.
∇f (x, y, z) = (2x + 2yz, 2xz − z 2 , 2xy − 2yz),
at the given point
∇f (1, 1, 2) = (6, 0, −2).
The direction is given in the parametric form of a straight line. Although the parameter
is not explicitly written, you should view the common value of the given three fractions as
the parameter, say t. Hence the line is
t=
x−1
2
t=y−1
t=
z−2
−3
where t ∈ R. Expressing x, y, z we obtain the more familiar parametric form:
x = 2t + 1
y =t+1
z = −3t + 2
We need only the direction of this line, which is obtained from the coefficients of t, i.e. it is
the vector (2, 1, −3). The unit vector in this direction is
u=
2i + j − 3k
2i + j − 3k
√
=
k2i + j − 3kk
14
or its opposite, −u (there are two unit vectors parallel with a given line).
Hence the directional derivative is
18
fu0 (1, 1, 2) = ∇f (1, 1, 2) · u = √
14
14
or its opposite, which is − √1814 .
p
PROBLEM 8. a, Find the unit vector in the direction in which f (x, y, z) = x2 + y 2 + 2z 2
increases most rapidly at the point (1, −2, 1) and give the rate of change of f in this direction.
p
b, Find the unit vector in the direction in which f (x, y, z) = x2 + y 2 + 2z 2 decreases
most rapidly at the point (1, −2, 1) and give the rate of change of f in this direction.
SOLUTION TO PROBLEM 8. a, Compute the gradient
xi + yj + 2zk
∇f (x, y, z) = p 2
x + y 2 + 2z 2
at the given point
∇f (1, −2, 1) =
i − 2j + 2k
√
7
This gives the direction of biggest increase, but it is not a unit vector. We have to find its
norm
3
k∇f (1, −2, 1)k = √
7
Hence the unit vector in the direction of biggest increase is
∇f
k∇f k
at the given point, i.e.
i − 2j + 2k
3
The rate of change in this direction is the norm of the gradient, i.e. it is
√3 .
7
b, The biggest decrease is in the opposite direction as the biggest increase and the rate
of change is also the opposite. Hence the answer to part b, is − i−2j+2k
for the direction
3
and − √37 for the rate.
x2
PROBLEM 9. a, Determine the path of the steepest descent along the surface z =
+ 3y 2 starting from the point (1, 2, 13).
b, Where does this path end?
c, What is the projection of this path on the (x, y) plane?
SOLUTION TO PROBLEM 9 a, The path of steepest descent r(t) = x(t)i + y(t)j
(with t being a real parameter along the path) has the property that its derivative is always
15
parallel with the gradient of the given function f (x, y) = x2 + 3y 2 , and it points in the
opposite direction. Since
∇f (x, y) = (2x, 6y)
the coordinate functions x(t), y(t) of the curve r(t) must satisfy the following relations
x0 (t) = −2x(t)
y 0 (t) = −6y(t)
In addition, the curve starts at the point x(0) = 1, y(0) = 2 (you can always choose the
parametrization in such a way that the given point correspond to t = 0). Hence we have
to find a function x(t) such that x0 (t) = −2x(t) and x(0) = 1. The relation x0 (t) = −2x(t)
is satisfied by all functions of the form x(t) = Ce−2t with arbitrary constant C, which is
fixed by the condition x(0) = 1. Hence C = 1 and x(t) = e−2t Similarly you find that
y(t) = 2e−6t . The path of steepest descent is
r(t) = (e−2t , 2e−6t )
with a parameter t ≥ 0. (If you go in the opposite direction, i.e. you run the path for t < 0,
or equivalently you consider the path (e2t , 2e6t ) for t > 0, then you get the path of steepest
ascent.)
b, After a very long time (t → ∞) the particle on this path gets arbitrarily close to the
origin. This is the other endpoint of the path.
c, To see the projection of this curve onto the (x, y) plane, then you have to express
y = 2e−6t in terms of x = e−2t , which is clearly y = 2x3 . Only that segment of this curve
counts which is covered by the regime t ≥ 0, i.e. x ∈ (0, 1]. Hence the projection of this
path is the piece of the graph of the function y = 2x3 in the (x, y) plane between (1, 2) and
(0, 0).
PROBLEM 10. Find the rate of change of f with respect to t along the given curve
r(t)
a, f (x, y) = x2 y, r(t) = et i + e−t j
b, f (x, y, z) = y 2 sin(x + z), r(t) = 2ti + (cos t)j + t3 k
SOLUTION TO PROBLEM 10. We need to compute
d
f (r(t)) = ∇f (r(t)) · r0 (t)
dt
a, Here
∇f (x, y) = (2xy, x2 ) = 2xyi + x2 j
16
r0 (t) = et i − e−t j
hence the rate of change is
2et e−t et − (et )2 e−t = et
(Don’t forget to express x, y in terms of t; the final result must contain only t, since we
asked for the rate of change with respect to this variable.)
b, Here
∇f (x, y, z) = y 2 cos(x + z)i + 2y sin(x + z)j + y 2 cos(x + z)k
r0 (t) = 2i − (sin t)j + 3t2 k
hence the rate of change is
2 cos2 t cos(2t + t3 ) − 2 cos t sin t sin(2t + t3 ) + 3t2 cos2 t cos(2t + t3 )
PROBLEM 11. Find the rate of change of u with respect to t in the following two
problems:
√
a, u = x + 4 xy − 3y where x = t3 , y = t−1
b, u = x2 tan y, where x = s2 t, y = s + t2 .
SOLUTION TO PROBLEM 11. a, Here u depends only on t (via x and y). The
rate of change is
du
∂u dx ∂u dy
=
+
dt
∂x dt
∂y dt
r y
= 1+2
x
rx
3t2 + 2
y
− 3 (−t−2 )
= (1 + 2t−2 )3t2 + (2t2 − 3)(−t−2 ) = 3t2 + 4 + 3t−2
b, Here u depends on both variables s, t. The rate of change with respect to t is given
by
∂u
∂u ∂x ∂u ∂y
=
+
∂t
∂x ∂t
∂y ∂t
= (2x tan y)s2 + (x2 sec2 y)2t = 2s4 t tan(s + t2 ) + 2s4 t3 sec2 (s + t2 )
∂u
(Notice the difference between the notation du
dt and ∂t . The first notation is used for simple
derivative, when u depends only on the variable t. If u depends on several variables, we use
the ∂u
∂t partial derivative notation).
17
PROBLEM 12. Let
f1 (u1 , u2 ) = u1 u2
f2 (u1 , u2 ) = u21 + u2
and
u1 (x1 , x2 , x3 ) = x1 − x2 x3
u2 (x1 , x2 , x3 ) = x22 + x3
a, Find the Jacobian of f =
f1
f2
and the Jacobian of u =
u1
u2
b, Find the Jacobian of f ◦ u
c, What is
∂f2
∂x3 ?
SOLUTION TO PROBLEM 12.
a,
Jf (u1 , u2 ) =
u2
2u1
u1
1
Ju (x1 , x2 , x3 ) =
1 −x3
0 2x2
ih
i
−x2
1
b,
h
Jf ◦u (x1 , x2 , x3 ) = Jf (u1 , u2 ) Ju (x1 , x2 , x3 ) =
=
=
c,
∂f2
∂x3
x22 + x3
x1 − x 2 x3
2(x1 − x2 x3 )
1
u2
2u1
1 −x3
0 2x2
u1
1
−x2
1
1 −x3
0 2x2
−x2
1
x22 + x3
−x3 (x22 + x3 ) + 2x2 (x1 − x2 x3 )
−x32 + x1 − 2x2 x3
2(x1 − x2 x3 )
−2x3 (x1 − x2 x3 ) + 2x2
−2x2 (x1 − x2 x3 ) + 1
is just the third entry of the second row of Jf ◦u (x1 , x2 , x3 ), i.e.
∂f2
= −2x2 (x1 − x2 x3 ) + 1
∂x3
You could have obtained it of course without computing the full Jacobian:
∂f2
∂f2 ∂u1
∂f2 ∂u2
=
+
= 2u1 (−x2 ) + 1 · 1 = −2x2 (x1 − x2 x3 ) + 1
∂x3
∂u1 ∂x3 ∂u2 ∂x3
18
PROBLEM 13. Find a solution to the system of nonlinear equations
x + y3 = 2
x2 + y = 3
with Newton’s method starting from (x0 , y0 ) = (1, 1). Run two steps of the iteration. Plug
these approximate solutions to the original equations to see the error.
SOLUTION TO PROBLEM 13. We have to find the root of
f (x, y) =
The Jacobian is
Jf (x, y) =
Let x0 =
1
1
x + y3 − 2
x2 + y − 3
3y 2
1
1
2x
be the initial guess. Then we have to solve
Jf (x0 ) x − x0 = −f (x0 )
i.e.
1 3
2 1
since
Jf (1, 1) =
x−
1 3
2 1
1
1
0
1
=
and
f (1, 1) =
0
−1
The solution will be the first approximation, x1 . Temporarily let u = x −
solution of
1 3
2 1
1
, and the
1
0
1
u=
requires solving the system
u1 + 3u2 = 0
2u1 + u2 = 1
You can use your favorite method to do that, the result is u1 = 3/5, u2 = −1/5, hence
x1 =
3/5
−1/5
+
1
1
=
8/5
4/5
=
1.6
0.8
1
3.2
1.92
1
This is the result of the first Newton iteration.
In the second step we have
Jf (x1 ) =
1
2(1.6)
3(0.8)2
1
19
=
and
f (x1 ) =
1.6 + 0.83 − 2
1.62 + 0.8 − 3
=
0.112
0.36
and we have to solve
1
3.2
1.92
1
The solution of
x−
1.6
0.8
=−
0.112
0.36
u1
+ 1.92u2 = −0.112
3.2u1 +
u2
= −0.36
is u1 = −0.1125, u2 = 0.00031.
Hence
x2 =
−0.1125
0.00031
+
1.6
0.8
=
1.4875
0.80031
This is second approximation.
Let’s plug back x0 , x1 and x2 to the original equation. This is the same as computing
the value of f on these points; the true solution would give zero. Here we have
f (x0 ) =
0
−1
f (x1 ) =
0.112
0.36
f (x2 ) =
0.0000762
0.00126
so after two steps we are already pretty close.
PROBLEM 14. Find the normal vector and the tangent vector to the curve x5 + y 5 =
at the point (1, 1). Write up the equation of the normal line and the tangent line.
(Recall that the equation of a line has two forms: normal form and parametric form. Write
up both.)
2x3
SOLUTION TO PROBLEM 14. Represent the curve as a level curve of a function,
hence choose f (x, y) = x5 + y 5 − 2x3 . Compute the gradient
∇f (x, y) = (5x4 − 6x2 )i + 5y 4 j
at the given point
∇f (1, 1) = −i + 5j
Hence the normal vector is n = −i + 5j (not normalized), the tangent vector is t = 5i + j.
Notice that the orientation of these two vectors are not determined, i.e. i − 5j would also
be a normal vector and −5i − j is also a tangent vector.
The equation of the tangent line is n · (x − x0 ) = 0 where x0 = (1, 1), i.e. it is
−x + 5y − 4 = 0.
20
The equation of the normal line is t · (x − x0 ) = 0 i.e. it is 5x + y − 6 = 0.
Be careful: if you use the normal equation v · (x − x0 ) = 0 for these lines (where v
is the vector orthogonal to the line), then you have to use v = n for the tangent line and
v = t for the normal line. It sounds a bit confusing but think about it: the normal line is
orthogonal to the tangent vector and the tangent line is orthogonal to the normal vector.
You can equivalently use the parametric equation of a line. If e = (e1 , e2 ) is a vector
along the line, then the equation has the form x = et + x0 , or in coordinates
x = e1 t + x0
y = e2 t + y0
where t is a parameter along the line. If you use this form, then you have to use the normal
vector for the normal line, i.e.
x = −t + 1
y = 5t + 1
and the tangent vector for the tangent line
x = 5t + 1
y =t+1
You can check that the normal equation 5x + y − 6 = 0 and the parametric equation
(x, y) = (−t + 1, 5t + 1) of the normal line really determine the same line:
5(−t + 1) + (5t + 1) − 6 = 0
Similarly you can check that the two given forms of the tangent lines are equivalent.
PROBLEM 15. a, Find an equation for the tangent plane of xy 2 + 2z 2 = 12 at the
point (1, 2, 2).
b, Find the parametric equation of the normal line.
c, Find the equation of the line in the tangent plane obtained in part a, which is parallel
to the plane x + y + z = 3 and goes through (1, 2, 2).
SOLUTION TO PROBLEM 15. a, View the given surface as the level surface of
f (x, y, z) = xy 2 + 2z 2 (at level c = 12).
∇f (x, y, z) = y 2 i + 2xyj + 4zk
∇f (x0 ) = ∇f (1, 2, 2) = 4i + 4j + 8k
The tangent plane is given by ∇f (1, 2, 2) · (x − x0 ) where x0 = (1, 2, 2), and it is
4(x − 1) + 4(y − 2) + 8(z − 2) = 0
21
i.e. x + y + 2z = 7.
b, The parametric equation of the normal line is given as x = t∇f (x0 ) + x0 , hence it is
x = 4t + 1,
y = 4t + 2,
z = 8t + 2
where t ∈ R (you can reparametrize it, and use s = 4t as a new parameter to get slightly
simpler formulas: x = s + 1, y = s + 2, z = 2s + 2).
c, The vector in the direction of the line in question must be orthogonal both to the
normal vector of the plane x + y + z = 3, hence to (1, 1, 1) and must also be orthogonal to
the gradient, i.e. to (4, 4, 8). Hence this vector is given by the vector product of these two
vectors, i.e. it is e = (4, −4, 0). The parametric equation of this line is
x = 4t + 1,
y = −4t + 2,
z=2
PROBLEM 16. The curve r(t) = 2ti+3t−1 j−2t2 k and the ellipsoid x2 +y 2 +3z 2 = 25
intersects at (2, 3, −2). What is the sine of the angle of intersection?
SOLUTION TO PROBLEM 16. The given point correspond to the t = 1 parameter
value. The tangent vector to this curve is the value of r0 (t) at this point, i.e. from
r0 (t) = 2i − 3t−2 j − 4tk
r0 (1) = 2i − 3j − 4k
The normal vector to the surface at this point is given by the gradient of f (x, y, z) =
x2 + y 2 + 3z 2 (whose level curve is the surface).
∇f (x, y, z) = (2x, 2y, 6z)
∇f (2, 3, −2) = (4, 6, −12) = 4i + 6j − 12k
The angle θ between the normal vector to the surface and the tangent vector to the curve
satisfies
2i − 3j − 4k
4i + 6j − 12k
38
19
cos θ =
·
= √ = √
k2i − 3j − 4kk k4i + 6j − 12kk
14 29
7 29
(in order to simplify the calculation, you could have considered half of the gradient vector
2i + 3j − 6k because only its direction matters).
The angle of intersection is the angle between the tangent plane to the ellipsoid and
19
the tangent vector to the curve, hence it is π2 − θ. Its sine is the same as cos θ = 7√
.
29
22
(You can easily compute the angles with a calculator by taking inverse cosine, but
calculator is not allowed on the exam, hence I asked for the sine instead of the angle itself.)
PROBLEM 17. Find the critical points and local extreme values of the following
functions. Decide whether the extreme point is local maximum or minimum.
a, f (x, y) = x2 + 2xy + 3y 2 + 2x + 10y + 1
b, g(x, y) = xy + x−1 + 8y −1
c, h(x, y) = (x − y)(xy − 1)
d, k(x, y) = xy −1 − yx−1
SOLUTION TO PROBLEM 17. a, The function is differentiable everywhere, the
critical points are the stationary points (where the gradient vanishes). Since
∇f (x, y) = (2x + 2y + 2, 2x + 6y + 10)
we have to solve the system of equations
2x + 2y = −2
2x + 6y = −10
We easily find that (x, y) = (1, −2) is the only critical point. To check its type, we use the
second partial derivative test. We compute
A=
∂2f
=2
∂x2
B=
∂2f
=2
∂x∂y
C=
∂2f
=6
∂y 2
hence D = 22 − 2 · 6 < 0 and A > 0, so it is a local minimum. The value of the function at
this point is −8.
b, The function is differentiable everywhere on its domain, hence the only critical points
are the stationary points. Since ∇g(x, y) = (y − x−2 )i + (x − 8y −2 )j, this is zero only at
(x, y) = ( 12 , 4). The second partials at this point are
fxx = 2x−3 = 16
fxy = 1
fyy = 16y −3 =
1
4
2 −f f
hence D = fxy
xx yy = −3. Since A > 0, we have a local minimum, and its value is 6.
c, All critical points are stationary points. Multiply out the given expression: h(x, y) =
x2 y − xy 2 − x + y
∇h(x, y) = (2xy − y 2 − 1)i + (x2 − 2xy + 1)j
23
We have to solve
2xy − y 2 = 1
x2 − 2xy = −1
This is not a linear equation, but it is fairly simple, so one can try to solve it with some
pedestrian way. The easiest if you add the two equations, then you get x2 − y 2 = 0, i.e.
(x − y)(x + y) = 0. Hence either x = y or x = −y. In the first case from the first equation
we have 2y 2 − y 2 = 1, i.e. x = y = ±1, in the second case the first equation becomes
−3y 2 = 1 which has no solution. Hence there are two critical points: (1, 1) and (−1, −1).
The second partials are
hxx = 2y
hxy = 2x − 2y
hyy = −2x
and D = h2xy − hxx hyy is positive at both points. Hence both stationary points are saddle.
d, Again, the function is differentiable at every point of the domain.
∇k(x, y) =
x2 + y 2
x2 + y 2
j
i
−
x2 y
xy 2
This gradient is never zero. The only possibility would be if the numerator x2 + y 2 were
zero, but that would mean x = y = 0 which is not in the domain of k. Hence no critical
points, no stationary points, no local extrema.
PROBLEM 18. Find the absolute maximum and minimum of the function f (x, y) =
y(x − 3) on the closed disk of radius 3 about the origin.
SOLUTION TO PROBLEM 18. The function is differentiable, hence all critical
points are stationary points. First compute the gradient to see if the stationary points are
possible candidates for absolute minimum or maximum. Since ∇f (x, y) = (y, x − 3), the
only stationary point is at (3, 0) which is not in the interior of the domain. Hence it does
not count (recall local max. or min. must be in the interior of the domain).
Next, we have to check the boundary points. We give two solutions.
First solution: Parametrize the boundary, i.e. the circle of radius 3 about the origin.
The most convenient parametrization is
r(t) = (3 cos t, 3 sin t)
t ∈ [0, 2π)
Therefore the function on the boundary is
f (r(t)) = 3 sin t(3 cos t − 3)
24
Now we have to minimize and maximize the function g(t) = 9 sin t(cos t − 1) of one variable
on the closed interval [0, 2π] (you can add the other endpoint, since g(0) = g(2π) anyway).
Compute g 0 (t) = 9 cos t(cos t − 1) − 9 sin2 t = 9(2 cos 2 t − cos t − 1) (notice the trick: using
sin2 t + cos2 = 1 we expressed everything in terms of cos t). The equation g0 (t) = 0 is
a quadratic equation for z = cos t, i.e. we have to solve 2z 2 − z − 1 = 0, which gives
cos t = z = 1, − 12 . There is one point corresponding to cos t = 1, t = 0, this is A = (3, 0)
on the circle.
There are two points
corresponding to cos t = − 12 , namely t = π3 , 2π
3 , i.e.
√
√
√
B = (− 32 , 3 2 3 )√and C = (− 32 , − 3 2 3 ). We can easily compute that f (A) = 0, f (B) = − 274 3
and f (C) = 274 3 . Hence B is the absolute minimum point and C is the absolut maximum
point with values given above.
Second solution: One can find the extremum points and values on the circle by
Lagrange’s method. Let g(x, y) = x2 + y 2 , then the circle can be viewed as a constraint
g(x, y) = 9 and we have to maximize and minimize f (x, y) under this condition. Since
∇g(x, y) = (2x, 2y), we have
y = λ2x
x − 3 = λ2y
in addition to the constraint x2 + y 2 = 9. From these two equations we get x − 3 = 4λ2 x,
i.e.
3
6λ
x=
y=
1 − 4λ2
1 − 4λ2
Putting this back into the constraint
i.e.
3 2 6λ 2
+
=9
1 − 4λ2
1 − 4λ2
1 + 4λ2 = (1 − 4λ2 )2
It is clearly an equation for c = 4λ2 , hence first we solve 1 + c = (1 − c)2 , or c2 − 3c = 0,
which gives c = 0, 3. If c = 0, then λ = 0 and (x, y) = (3, 0) giving
the point A in the first
√
3
3
2
solution. If c = 3, then λ = 4 , which gives two solutions λ = ± 2 . It is easy to see that
these give the points B and C above. Finally we have to compute the value of the function
at these three points, as above.
PROBLEM 19. Find the absolute maximum and minimum of the function f (x, y) =
(x + y − 2)2 on the closed square D = {(x, y) : 0 ≤ x ≤ 3, 0 ≤ y ≤ 3}.
SOLUTION TO PROBLEM 19. First we look for local extrema. The function is
differentiable everywhere, hence we look for stationary points. Since ∇f (x, y) = 2(x + y −
2)(i + j), all points on the line y = 2 − x are stationary points. (Yes, it could happen that
25
there are infinitely many stationary points). The value of f is zero at all of these points.
Since f ≥ 0 always (it’s a square of something), all these points which fall in the iterior
of D are local minima. Hence the local minima are (x, 2 − x) for all 0 < x < 2 and these
are clearly absolute minima as well. Since we exhausted all stationary points, there are no
local maxima.
To find the absolute extrema, one has to look at the boundary. It is clear that there
are two more absolute minima on the boundary, (0, 2) and (2, 0) (the endpoints of the line
y = 2 − x), which do not count as local minima since they are not interior points, but they
count as absolute minima.
For the absolute maxima in principle you should parametrize the boundary of the square
and maximize four functions. However, there is a shorter way if you think a bit: (x + y − 2)2
is clearly the biggest if either x + y is the biggest or it is the smallest (in principle it could
be negative). The biggest value of x + y is 6 at (x, y) = (3, 3) and the smallest value is 0
at (0, 0). The point (3, 3) gives the absolute maximum of f with value 16, while the point
(0, 0) does not give maximum (the value of the function is only 4).
PROBLEM 20. Minimize x + 2y + 4z on the sphere x2 + y 2 + z 2 = 7.
SOLUTION TO PROBLEM 20. Since the sphere is closed, bounded and f (x, y, z) =
x + 2y + 4z is continuous, the minimum exists. Let g(x, y, z) = x2 + y 2 + z 2 , then the sphere
is a level curve of g. Compute
∇f = (1, 2, 4)
∇g = (2x, 2y, 2z)
and from ∇f = λ∇g we have 1 = 2λx, 2 = 2λy and 4 = 2λz. Clearly λ 6= 0 and after
eliminating it we get express everything in terms of x as y = 2x, z = 4x. From g(x, y, z) = 7
we get 21x2 = 7, i.e. x = ± √13 , y = ± √23 , z = ± √43 and the minimum of f (x, y, z) is clearly
√
attained when you choose the minus signs. The value of the minimum is −7 3.
PROBLEM 21. Find the distance from the point (0, 1) to the parabola x2 = 4y.
SOLUTION TO PROBLEM 21. The distance is attained at the point closest to
the parabola from (0, 1). Such point clearly exists. Also, it is simpler the minimize the
distance square than the distance itself (to get rid of the square root). Hence we minimize
f (x, y) = x2 + (y − 1)2 under the constraint g(x, y) = x2 − 4y = 0. Compute
∇f = (2x, 2(y − 1))
∇g = (2x, −4)
hence from ∇f = λ∇g we have 2x = 2λ2x and 2y − 2 = −4λ. We can express 2λ = 1 − y
from the second equation and insert into the first: 2x = 2x(1 − y) or 2xy = 0. Hence either
26
x or y is zero, but then the other one is also zero from the constraint x2 = 4y. So the closest
point is (x, y) = (0, 0) and the minimum distance square is f (0, 0) = 1.
PROBLEM 22. Show that all triangles inscribed in a circle of radius 1 the equilateral
one has the largest product of the lengths of the sides.
SOLUTION TO PROBLEM 22. Let O be the center of the circle and A, B, C be
the vertices of the triangle. The angles between the radii from O to the vertices are x, y, z.
The constraint is g(x, y, z) = x + y + z = 2π. The lengths of the sides are 2 sin(x/2),
2 sin(y/2) and 2 sin(z/2) from basic trigonometry. We have to maximize
f (x, y, z) = 8 sin
x
y
z
· sin · sin
2
2
2
and clearly we can assume that none of x, y, z is zero. We compute
∇f = 4 sin
x
x
y
z
y
z
cot , cot , cot
· sin · sin
2
2
2
2
2
2
and ∇g = (1, 1, 1). Hence from ∇f = λ∇g we see that cot x2 = cot y2 = cot z2 . Since x, y, z
are between 0 and 2π, the half angles are between 0 and π and in this interval the cot
function is one-to-one. Hence it has well defined inverse, so from the cotangents being equal
it follows that x/2 = y/2 = z/2, i.e. the triangle is equilateral.
PROBLEM 23. Evaluate the following integrals on the given domain Ω
Z Z
(a)
Z Z
(b)
Ω
n
Z Z
(c)
Ω
(d)
Ω
Ω
Ω
yex dxdy
(x4 + y 2 ) dxdy
Z Z
(f )
Ω = (x, y) : 0 ≤ x ≤
2
ex dxdy
n
(x + 3y 3 ) dxdy
Z Z
(e)
Ω = (x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ x
sin(x + y) dxdy
Ω
Z Z
n
xy 3 dxdy
o
π
πo
,0 ≤ y ≤
2
2
o
Ω = (x, y) : x2 + y 2 ≤ 1
n
Ω = (x, y) : 0 ≤ y ≤ 1, 0 ≤ x ≤ y 2
o
Ω = bounded region between y = x2 and y = x3
Ω = triangle between the x-axis, x = 2 and x = 2y
SOLUTION TO PROBLEM 23.
27
(a)
Z Z
Ω
Z
xy 3 dxdy =
1Z x
0
0
xy 3 dy dx =
Z
1
x
0
h y 4 ix
4
0
1
4
dx =
Z
1
0
x5 dx =
1
24
(b)
Z Z
Z
sin(x + y) dxdy =
Ω
Z
=
(c)
Z Z
0
π/2 h
Z
3
1
Z
sin(x + y) dydx =
0
0
(x + 3y ) dxdy =
Ω
π/2 Z π/2
cos x − cos x +
Z √1−y2
−1 −
√
1−y 2
π/2
π/2
[− cos(x + y)]0 dx
0
π i
dx = 2
2
(x + 3y 3 ) dxdy =
Z
q
1
−1
6y 3 1 − y 2 dy = 0
since the integrand is odd on a symmetric interval.
(d)
Z Z
Ω
(e)
Z
yex dxdy =
Z Z
Ω
4
Z Z
Ω
0
0
Z
yex dxdy =
Z
2
(x + y ) dxdy =
Z
=
(f)
1 Z y2
x2
e
0
1 4x6
3
Z
dxdy =
0
1 Z x2
− x7 −
2 Z x/2
0
x3
0
x9 3
x2
e
1
0
2
y(ey − 1)dy =
4
Z
2
(x + y ) dydx =
dx =
h 4x7
21
Z
dydx =
0
2
−
x8
8
−
0
h ey2
2
1h
−
x4 y +
y 2 i1 e − 2
=
2 0
2
y 3 i x2
dx
3 x3
9
x10 i1
=
30 0 280
h 1 2 i2
1 x2
e4 − 1
=
xe dx = ex
0
2
4
4
PROBLEM 24. Decide whether the following statements true (T) or false (F). If
false, give a short argument. WARNING: Most of them are “If... then” statements. Such a
statement is true only when the “if” part surely implies the “then” part. Such a statement
is false when the “then” part is not satisfied even in one single case for which the “if” part
is valid.
(a) If a function f : R2 → R has first partial derivatives at the origin, then ∇f (0, 0)
exists.
(b) If a function f : R2 → R has continuous first partial derivatives in a neighborhood
of the origin, then ∇f (0, 0) exists.
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(c) If a function f : R2 → R is continuous at the origin, then ∇f (0, 0) exists.
(d) If ∇f (0, 0) exists for a function f : R2 → R, then it is continuous at the origin.
(e) Let U be an open set of R2 . If f : U → R satisfies ∇f (x) = 0 for all x ∈ U , then f
is constant on U .
(f) Let U be an open connected set of R2 . If f : U → R satisfies ∇f (x) = 0 for all
x ∈ U , then f = 0 on U .
(g) Let U be an open connected set of R2 . If f : U → R and g : U → R are two
functions whose gradients are equal at every point of U , then the two functions are equal.
(h) If f : R2 → R has a local extreme value at x0 then ∇f (x0 ) = 0
(i) Let f : R2 → R. If ∇f (x0 ) = 0, then f has a local extremum at x0 .
(j) If the function f : R2 → R has a local maximum at x0 then x0 is a critical point.
(k) Let Ω be a closed subset of R2 and f : Ω → R a continuous function. Then f
attains its maximum on Ω
(l) Let Ω be an open subset of R2 and f : Ω → R a continuous function. Then f attains
its maximum on Ω.
(m) Let Ω be a closed subset of R2 and f : Ω → R be a continuous function and let
∇f (x0 ) = 0 be the only stationary point in Ω. Then f attains its maximum or minimum
at x0 .
(n) Let f : R2 → R have continuous second partial derivatives and let x0 be a critical
2
point. If ∂∂ 2fx (x0 ) > 0 then f has a local minimum at x0 .
(o) Let f : R2 → R have continuous second partial derivatives and let x0 be a critical
point. If the discriminant D = (fxy )2 − (fxx )(fyy ) is positive at x0 , then x0 is a saddle.
(p) Let f : R2 → R have continuous second partial derivatives and let x0 be a critical
point. If the discriminant D = (fxy )2 − (fxx )(fyy ) is negative at x0 , then x0 is a local
maximum.
(q) Let f : R2 → R have continuous second partial derivatives and let x0 be a critical
point. If all second partial derivatives vanish at x0 , then the function has no local extremum
in x0 .
SOLUTION TO PROBLEM 24.
(a) F. The function even does not have to be continuous, for example let f (x, y) = 0 if
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x = 0 or y = 0 and f (x, y) = 1 otherwise.
(b) T.
(c) F. Continuity does not imply any differentiability.
(d) T.
(e) F. The set has to be connected. Otherwise you can choose different values of f on
different components of f .
(f) F. f can be constant, it does not have to be zero.
(g) F. The functions can differ by a additive constant.
(h) F. The gradient may not exist
(i) F. Saddle is a counterexample.
(j) T.
(k) T.
(l) F. Think of the paraboloid f (x, y) = x2 + y 2 on the open unit disk.
(m) F. The extrema can be on the boundary, For example think of the saddle x2 − y 2
on the closed unit disk.
(n) F. You also need the discriminant to be negative.
(o) T.
(p) F. It can be local minimum.
(q) F. Think of f (x, y) = x4 + y 4 which has local minimum at the origin but all second
partials vanish.
30