Problem sets and quizzes
Almost every Monday, there will be a quiz, based on an associated problem set.
(However, the first quiz will be on January 11 Thursday.)
Unless otherwise specified,
all exercises in the problem sets
are from the 3rd Edition
of University Calculus: Early Transcendentals
by Hass et al published by Addison Wesley (Pearson).
Here are the quizzes and their associated problem sets
(Quiz 1, Quiz 2, Quiz 3,
Quiz 4, Quiz 5, Quiz 6,
Quiz 7, Quiz 8, Quiz 9):
- Vectors and curves:
- Date taken: January 11 Thursday.
- Exercises from the Chapter 11 Practice Exercises (pages 638&639):
17, 19, 25, 29, 31, 35, 37, 43, 50.
- Exercises from §15.1 (page 826): 1–8.
- Exercises from §12.1 (pages 648–650):
1, 4, 6, 7, 11, 14, 15, 17, 19, 20, 23.
- Exercises from §12.2 (pages 654–656):
1, 4, 6, 12, 15, 17, 21, 22.
- Exercises from §12.3 (page 660): 1, 5, 8, 9, 11, 15, 18.
- Extra-credit essay question:
Explain your background in mathematics
and what you are going to use this course for.
- Functions of several variables:
- Date taken: January 22 Monday.
- Exercises from §13.1 (pages 682–684):
3, 4, 7, 9, 10, 15, 17, 18, 19, 23, 24, 30,
31–36, 39, 40, 42, 52, 54, 59, 62.
- Exercises from §13.2 (pages 690–693):
2, 6, 11, 18, 23, 28, 31, 32, 36, 39, 43, 46, 55.
- Exercises from §15.2 (pages 838–840): 5, 6, 39, 41, 43.
- Additional exercises
(you must attempt these to get full credit for the problem set):
- Given α =
3x dx +
4x2y dy,
evaluate α
at (x, y) = (2, 6)
along
〈dx, dy〉 = 〈0.003, 0.005〉.
(Answer.)
- Given α =
2xy dx +
2yz dy + 2xz dz,
evaluate α
at (x, y, z) =
(−1, 3, 2)
along
〈dx, dy, dz〉 =
〈0.01, 0.02, −0.01〉.
- Given β =
x2 dx +
xy dy + xz dz,
evaluate β
at (x, y, z) =
(4, 3, −2).
(Answer.)
- Given β =
5x2 dx −
3xy dy,
evaluate β
at (x, y) = (1, 2).
- Additional extra-credit exercise:
Prove that the two definitions of continuity
spanning pages 23 and 24 of my notes
are equivalent.
Actually, just do the direction that is not incredibly difficult:
Given a function f of several variables
and a point P0,
show that if the conditions
in the second definition of the continuity of f at P0
(the one in terms of ε and δ
in the first paragraph that's entirely on page 24)
are met,
then the conditions in the first definition
(the one in terms of the continuity of composite functions
in the paragraph that's split between pages 23&24)
must also be met
(but don't try to prove it the other way around).
To provide the link between these definitions,
you will need to refer to the ε-δ definition
of continuity for an ordinary function of one variable;
see the middle of page 7 in
my notes
from my Calculus 1 class last term
if you don't know precisely how that goes.)
- Differentiation:
- Date taken: January 29 Monday.
- Exercises from §13.3 (pages 702–704):
3, 4, 10, 12, 24, 26, 30, 39, 43, 46, 55, 57, 75, 82, 91.
- Exercises from §13.5 (pages 720&721):
2, 3, 7, 8, 14, 15, 16, 20, 23, 28.
- Exercises from §15.2 (pages 838): 1, 4.
- Exercises from §13.4 (pages 711&712):
- Use any method (including differentials or gradients):
2, 4, 7, 10;
- 19, 20, 27, 28, 33, 41.
- Exercises from §13.6 (pages 727–730):
3, 6, 10, 13, 14.
- Additional extra-credit exercise:
If you don't know about matrices (at least how to multiply them),
read
http://www.mathsisfun.com/algebra/matrix-multiplying.html
first.
Now, if you have m functions of n variables each,
then you can put their partial derivatives into an m-by-n matrix;
for example, if you have 2 functions of 3 variables each,
say u = f(x, y, z)
and v = g(x, y, z),
then the partial derivatives fit into a 2-by-3 matrix:
⎡ | ∂u/∂x |
∂u/∂y |
∂u/∂z | ⎤ |
⎣ | ∂v/∂x |
∂v/∂y |
∂v/∂z | ⎦ |
We may call this matrix
d(u, v)/d(x, y, z).
- If you have an ordinary function y = f(x),
think of this as a group of only 1 function of only 1 variable each,
so that d(y)/d(x) in the notation above
is a 1-by-1 matrix, consisting of a single entry.
Check that the usual derivative dy/dx is this entry.
(You don't have to even write anything down for this if you don't want to;
all that you'd write down anyway
is d(y)/d(x) = [dy/dx].
But make sure that it makes sense before you go on.)
- If you have a parametrized curve in 3 dimensions,
say P = (x, y, z) =
(f(t), g(t), h(t)),
think of this as a group of 3 functions of only 1 variable each,
so that d(x, y, z)/d(t)
is a 3-by-1 matrix, consisting of a single column with 3 entries.
Check that the components of the velocity vector dP/dt
are the same as the entries of this matrix.
(For this reason, ordinary vectors that represent change of a point
are sometimes called column vectors.)
- If you have a function of 3 variables,
say u = F(x, y, z),
think of this as a group of only 1 function of 3 variables each,
so that d(u)/d(x, y, z)
is a 1-by-3 matrix, consisting of a single row with 3 entries.
Check that the components of the gradient vector
∇F(x, y, z)
are the same as the entries of this matrix.
(For this reason,
vectors such as gradients
that represent change with respect to a point
are sometimes called row vectors.)
- If you have both
(x, y, z) =
(f(t), g(t), h(t))
and u = F(x, y, z),
then composition makes u an ordinary function of t.
Show that d(u)/d(t) =
d(u)/d(x, y, z)
d(x, y, z)/d(t),
using matrix multiplication,
and check that this matches the defining property of the gradient
from page 26 of my notes.
- If you have both V =
〈f(u), g(u), h(u)〉
and u = F(x, y, z),
then composition in the other order makes a vector field;
that is, V is
a vector-valued function of
P = (x, y, z).
Use matrix multiplication to get a 3-by-3 matrix;
the fact that this equals the matrix dV/dP
is a very general version of the multivariable Chain Rule.
(This matrix-valued function
is called
the total derivative of this vector field.)
- Applications of differentiation:
- Date taken: February 5 Monday.
- Exercises from §13.6 (pages 727–730):
19, 21, 29, 30, 33, 35, 39, 50, 54.
- Exercises from §13.7 (pages 737–739):
2, 7, 9, 15, 27, 32, 34, 37, 43, 52, 57.
- Exercises from §13.8 (pages 746–748):
1, 5, 10, 11, 16, 23, 29.
- Additional extra-credit exercise:
Let f be the function of two variables
given by f(x, y) =
3 sin(x + y) +
4 cos(x − y).
Evaluate f, both of its partial derivatives,
and all four of its second partial derivatives
at (0, 0).
Then use these results to approximate f near (0, 0)
with a quadratic polynomial
(that is one whose degree is at most 2).
- Integration on curves:
- Date taken: February 12 Monday.
- Exercises from §15.2 (pages 838–840):
10, 11, 14, 16, 17.A&B,
19, 22, 23, 24, 29.
- Exercises from §15.1 (pages 826–828):
10, 13, 16, 22, 30, 35.
- Exercises from §15.3 (pages 849–851):
1, 3, 6, 7, 8, 11, 14, 17, 21, 25.
- Additional extra-credit exercise:
Suppose that F is a conservative vector field
defined on all of 3-dimensional space;
then there exists a scalar field f
such that F = ∇f.
Let U = −f.
In physics, if F is a force field,
then we call U a potential energy field for F.
Recall that,
if an object travels along a curve C in the force field F,
then the work done on that object by that force field,
or in other words the energy transferred to that object by that force field,
is the integral
∫P∈C F(P) ⋅ dP
(or ∫C F ⋅ dr
for short).
If the curve C
begins at the point P1
and ends at the point P2,
then use that F = −∇U
to express the value of this work using
values of the scalar field U at those points.
If you imagine that U(P)
is the amount of ‘potential’ energy held by an object at P
by virtue of its position within this force field,
then check that
the amount of energy transferred to the object by the field (the work)
is the opposite of the change in the object's potential energy.
(In other words, we have conservation of energy:
the total change in energy is zero.
This conservation
is why
conservative vectors fields
are called ‘conservative’.)
- Multiple integrals:
- Date taken: February 19 Monday.
- Exercises from §14.1 (pages 759&760): 3, 7, 10, 17, 22, 27.
- Exercises from §14.2 (pages 767–769):
- 1, 2, 7, 9, 12, 14, 17, 19, 23, 35, 41, 47, 51, 57, 61;
- Extra credit: 80.
- Exercises from §14.5 (pages 785–788):
3, 6, 9, 15, 21, 25, 29, 34, 37.
- Exercises from §14.3 (page 772):
1, 4, 7, 12, 13, 14, 17, 20, 21.
- Applications of multiple integrals:
- Date taken: February 26 Monday.
- Exercises from §14.6 (pages 793–795): 3, 14, 19, 25, 29.
- Extra-credit exercise from §14.8 (page 815):
To show work, show at least
the integral in u and v that you evaluate:
16.
- Exercises from §14.4 (pages 777–786):
1, 3, 5, 7, 9, 17, 20, 23, 24, 28, 29, 34, 37.
- Exercises from §14.7 (pages 803–806):
1, 2, 8, 12, 14, 23, 37, 43, 46, 57, 77.
- Integration on surfaces:
- Date taken: March 5 Monday.
- Exercises from §15.5 (pages 872–874):
2, 3, 6, 9, 13, 20, 23.
- Exercises from §15.6 (pages 883–884):
1, 5, 8, 11, 16, 17, 19, 23, 25, 34, 35, 37, 41, 45.
- Additional extra-credit exercise:
Consider the surface given by
r = f(z) in cylindrical coordinates,
where f is a differentiable function
defined on the interval [a, b].
Use the methods of §6.6 of my notes (or §15.5 of the textbook)
to show that the area of this surface is
2π ∫ab f(z) √(f′(z)2 + 1) dz.
- The Stokes theorems:
- Date taken: March 12 Monday.
- Exercises from §15.4 (pages 861–863):
1, 4, 7, 9, 12, 15, 21, 24, 26, 33.
- Exercises from §15.7 (pages 895–897):
1, 3, 5, 6, 9, 14, 17, 21, 28.
- Exercises from §15.8 (pages 906–908):
1, 2, 6, 7, 8, 13, 17.
- Extra credit exercise based on Exercise 15.8.31 from the 2nd Edition:
To describe a scalar quantity
that not only takes values throughout space but also changes with time,
we need a function of four variables,
one (say t) to represent time
and three (say x, y, and z) to represent space.
The overall 4-dimensional space
whose coordinates are t, x, y, and z
is called spacetime.
Let δ =
p(t, x, y, z)
be the mass density, at a given time and place, of some fluid substance,
and let v =
F(t, x, y, z)
be the velocity, at a given time and place, of the fluid.
The components of δ v
tell you the speed at which mass is flowing in particular directions,
and we can put all of this information together into
a single exterior differential pseudoform Φ
of rank 3 in 4 variables:
Φ =
δ dV −
δ v ⋅ dt ∧ dS.
(More explicitly, if v =
〈v1, v2, v3〉,
then Φ =
δ dx ∧
dy ∧ dz −
δ v1 dt ∧
dy ∧ dz −
δ v2 dt ∧
dz ∧ dx −
δ v3 dt ∧
dx ∧ dy
using the right-hand rule.)
If the fluid is just flowing
and not undergoing any physical, chemical, or nuclear reactions,
then its mass should be conserved.
This means not only that its total mass should be the same at any two times,
but also that if its mass in any region of space changes between two times,
then that change should be accounted for by flow through the region's boundary.
That is, if D is any region
of (x, y, z)-space,
S is its boundary (pseudooriented outwards as usual),
and t1 and t2
are two times (two values of t),
then the integral of δ on D at t2
minus the integral of δ on D at t1
should equal the integral
over the time from t1 to t2
of the integral of δ v across S.
In symbols,
∫(x,y,z)∈D p(t2, x, y, z) dV −
∫(x,y,z)∈D p(t1, x, y, z) dV =
∫t=t1t2 (∫(x,y,z)∈S p(t, x, y, z) F(t, x, y, z) ⋅
dS) dt.
We can combine the region D at time t1,
the region D at time t2,
and the boundary surface S at the times in between
into a single closed 3-dimensional hypersurface H
in the 4-dimensional spacetime,
and then this equation states
that the integral of Φ out of H is 0.
By the generalized Stokes Theorem,
this integral is equal to the integral
of the exterior differential d ∧ Φ
on the 4-dimensional region of spacetime bounded by H;
by making D arbitrarily small around a given point in space
and making t1 and t2 arbitrarily close
on either side of a given time,
this means that d ∧ Φ itself must be 0.
Now here is your assignment:
Work out d ∧ Φ
and write the equation d ∧ Φ = 0
in terms of the partial derivatives with respect to time
and the gradient, curl, and/or divergence with respect to space
of δ =
p(t, x, y, z)
and v =
F(t, x, y, z).
(The equation that you should end up with
is called the continuity equation
of conserved fluid flow.)
That's it!
Go back to the the course homepage.
This web page was written between 2003 and 2018 by Toby Bartels,
last edited on 2018 January 19.
Toby reserves no legal rights to it.
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