Name: 3 (9)0 1,497)
Exam 1
No calculators or notes are allowed on this test. Make sure to show all of your work, and leave no
answers blank.
Question 1
Simplify the following expressions.
a) szyxy3
: szxggf 2 2x33 ,/
(xZ) ; (xZ)
(2x+1) ' 5
()0?) _; Qt?) _
FINAL EXAM STUDY GUIDE
Sections: 1.1-1.7, 2.1-2.8, 3.1-3.6, 4.1-4.4, 5.1, 5.5
CHAPTER 1
1.1: Write the equation: the y-value is two more than the square of the x-value.
1.2: Describe the solution set for the equation 3 5 = 3( 2) + 1.
1.3: After a 20% pric
A is a subset of X
W is a subset of B
-X is a subset of W D
Show that A U B I sa subset of X
We need to know that if x is in A or in B, then it must be in X
Well if x is in A, we know that it is in X, because we know that A is a subset of X.
If something
The relation x = y^2 on all the integers:
Is not reflexive because (2, 2) is not in the relation
Not symmetric, because (4, 2) is in the relation, but (2, 4) is not
X = y^2
Y = x^2
y = (y^2)^2
y = y^4
y^4 y = 0
y(y^3 1) = 0
y = 0, y = 1
if y = 0, then x =
Is the reatlion on the reals, aRb iff
ab 0 an equivalence relation? I think so
No absolutely not, because if a is zero, then the relation doesnt hold
What if we define the relation as being on the positive reals?
aRb not equal to 0
a>0
a^2 > 0, which is n
We need to show p->q and p -> -q
P is n is odd; Q is 3n^2+6 is odd
If n is odd, n = 2k+1, so
3n^2+6=3(2k+1)^2+6=3(4k^2+4k+1)+6=12k^2+12k+3+6=12k^2+12k+8+1=2(6k^2+6k+4)+1
Now show that if n is even, so is 3n^2 +6
N = 2k
3(2k)^2+6=12k^2+6=2(6k^2+3)
If n is
Show that if a -> q
And b -> q
Then (a V b) -> q
(-a V q) A (-b V q)
(q V a) A (q V b)
Q V (-a A b)
-(a V b) V q
(a V b) -> q
a
T
T
T
T
F
F
F
F
b
q
T
T
F
F
T
T
F
F
Show that if
T
F
T
F
T
F
T
F
aq
bq
T
F
T
F
T
T
T
T
T
F
T
T
T
F
T
T
(a ( b c ) ( b) a
a ( b
Sqrt(3) is irrational
Assume sqrt(3) is rational
Sqrt(3) = a/b
3 = a^2/b^2
A^2 = 3b^2
if a^2 is divisible by 3, then so is a
suppose a is not divisible by 3; then are two cases: a = 3k+1 or a = 3k+2
verify (A U (A and B) = A
A U (A and B)
(A U A) and (A U
Prove that if a > 0 and b > 0, then ab > 0
a>0
ab > 0(b)
ab > 0
Prove if ac = bc, then a = b
a < b, c > 0
a < b, c < 0
a > b, c
(p A q ) -> r
-r -> -(ac = bc A c not equal 0)
If (a < b or a > b) then (ac not equal to bc or c = 0)
Let q be the proposition
(A and B) is a subset of A, and A is a subset of (A and B)
If x is an element of A, then it is also an element of A and B, which means that it is also an element of B.
If x is an element of A
pis x A ; q is x B ; r is x ( A B )= p q
we want show that if A
I want to show that if
B A , then B ( A Bc )=
First , show that if B A , then B A=B
So we must show that if B isaof A , then Bis aof A B , A Bis aof B
let x B be p ,let x A be q
we can write B is aof A as ( x B x A ) =p q
[ x B x ( A B ) ] [ x ( A B ) x B
Show that if a and b are both positive integers, then a = b + 9 means a > b
We know that a b = 9 > 0
If a b > 0, then a > b
Show that p -> (p V q) is a tautology
-p V (p V q), disjunctive form
(-p V p) V q, associative law
T V q, law of excluded middle
T,
MATH 265: MULTIVARIABLE CALCULUS: LESSON 9 PROBLEMS
Problems are worth 20 points each.
(1) Let f (x, y) = sin xy + x cos y. Compute fx (x, y), fy (x, y), and fxy (x, y).
(2) Let f (x, y, z) = x + ln(2y + 3z 2 ). Compute
@2f
(2, 1, 2).
@y@x
@ 2 + 2
(3) Com
MATH 265: MULTIVARIABLE CALCULUS: LESSON 7 PROBLEMS
Problems are worth 20 points each.
(1) Sketch the graph of f (x, y) = x + y 2 .
!
(2) Sketch the graph of f (x, y) = 4 x2 y 2 .
(3) Thinking about the shape of the graph (no calulus needed), what is the
MATH 265: MULTIVARIABLE CALCULUS: LESSON 11 PROBLEMS
Problems are worth 20 points each.
(1) Let f (u, v) = u + v + uv 2 and u = u(x, y) = ex + y, v = v(x, y) = xy. Use the
@f
@f
Chain Rule to compute
and
.
@x
@y
(2) Repeat problem 1, but this time first f
MATH 265: MULTIVARIABLE CALCULUS: LESSON 10 PROBLEMS
Problems are worth 20 points each.
(1) Let f (x, y) = x cos y.
(a) Calculate rf (x, y).
4!
!
(b) If !
u = 35 !
5 | , find D u f (2, 3 ).
(2) Let g(x, y, z) = ex y + z ln y. Calculate the gradient of g.
MATH 265: MULTIVARIABLE CALCULUS: LESSON 2 PROBLEMS
Remember: the dot and the cross used to denote the dot and cross products
are not optional.
Problems are worth 20 points each.
(1) Compute h3, 1, 0i (2h 1, 1, 2i).
(2) Calculate the angle between !
v = h
MATH 265: MULTIVARIABLE CALCULUS: LESSON 6 PROBLEMS
Problems are worth 20 points each.
!
(1) Determine the length of the curve !
r (t) = et cos t!
+ et sin t!
| + et k for 0 t 1.
(2) Determine the speed of a moving particle at time t = 2 whose position a