Homework 6
Problem 1.
Problem 2.
4x 3y + 12 = 0 y = 0, 4x = 12, x = 3 x = 0, 3y = 12, y = 4
1
Problem 3.
By examining the graph in problem 2, you can eyeball points that are 5 apart,
for example:
(3, 0) (0, 4)
The way we guess that these are distance 5 ap
Homework 1
Problem 1a.
am
1
am
= n = amn
an
a
Problem 1b.
5 (x4 y 3 z 2 )2 (3 z 1 x5 y 3 )1
=
5 (x4 y 3 z 2 )2
5 x8 y 6 z 4
5
5
=
= (x85 y 6(3) z 4(1) ) = x3 y 9 z 5
3 z 1 x5 y 3
3 x5 y 3 z 1
3
3
Problem 1c.
(
a2n
2a3 n
a2n
bn
a2n
a2n3n
1
)(
) =( n )( n
Homework 5
Problem 1a.
We know that
8x6 = (2x2 )3
and
27
3
= ( )3
8
2
We apply the formula for dierence of two cubes:
27
3
9
= (2x2 )(4x4 + 3x2 + )
8
2
4
We can still factor the rst factor in the above, so we have:
3
9
3
3
9
2
4
2
f1 (x) = 2(x )(4x + 3x +
Homework 8
Problem 1.
One should observe that (0.25,0.5) lies on the given line 2x + y = 1. Thus,
the reection is itself, i.e. (0.25,0.5).
Problem 2. We should start by drawing the line L through our point (0, 3)
that is perpendicular to our line 2x + y =
Math W1003, Fall 2013
Lecture 21
Last time, we talked about exponential functions, and dened what we
mean by expressions like:
exp2 (x) = 2x
But quite often, we may be faced with equations like: what is the exponent x
such that 2x = 5? The answer, which w
Math W1003, Fall 2013
Lecture 23
Our last topic will be sequences, which will probably also be the rst
topic youll study in calculus. A sequence is a bunch of numbers placed in
a certain order:
2, , e2 3, 7
The above sequence has four elements, which can
Homework 14
Problem 1a.
If we shift the graph of sin a by units to the left, we get back the graph of
2
cos a:
sin a +
= cos a
2
If instead we shift the graph of cos a by units to the right, we get sin a
2
instead of sin a:
cos a +
= sin a
2
Problem 1b.
I
Homework 18
Problem 1a.
arcsin(1) =
2
2
= 1
5
6
=
4
= 1
sin
because
Problem 1b.
arccos
3
2
=
5
6
because
cos
3
2
Problem 1c.
4
because
tan
2
3
because
cot
arctan(1) =
Problem 1d.
1
arccot
3
=
2
3
1
=
3
In all the above problems, one needs to also e
Math 4 Unit 4: Trig Identities
_
Wksht. 4.02-Verifying Identities
Verify the identity algebraically.
1. cscqtan q = secq
3.
csc cos
+
= 2cot
sec sin
5. sin t csct = 1
7.
9.
11.
csc2 x
= csc x sec x
cot x
cos2 b - sin2 b = 2cos2 b - 1
2- csc2 z = 1- cot2 z
Homework 11
Problem 1.
1) at 11:24 pm - thats about the time when the population stopped increasing
2) around 20 million - thats by eyeballing the yvalue when x is about 11:24 pm
3) at 3:36 am - thats when the population starts increasing drastically agai
Homework 19
Problem 1.
This is an even function because it is symmetric across the yaxis: f (x) =
f (x).
This is an odd function because it is symmetric across the origin: f (x) =
f (x).
Problem 2.
We have:
f (x) = x3 + x2
Since this is neither equal to f
Homework 12
Problem 1
= 30 :
= 45 :
= 60 :
x = x cos(30 ) y sin(30 ) = x 3 y
2 2
y = x sin(30 ) + y cos(30 ) = x + y 3
2
2
x = x cos(45 ) y sin(45 ) = x 2 y 2
2
2
y = x sin(45 ) + y cos(45 ) = x 2 + y 2
2
2
x = x cos(60 ) y sin(60 ) = x y 3
2 2
y = x s
Homework 15
Problem 1a.
5
)
4
x = r cos , y = r sin
(2,
hence:
x = 2 cos
5
4
=2
y = 2 sin
5
4
=2
2
2
2
2
= 2
= 2
So the point has Cartesian coordinates ( 2, 2)
1
Problem 1b.
5
)
3
x = r cos , y = r sin
(6,
hence:
x = 6 cos
y = 6 sin
5
3
5
3
1
2
=6
=6
Math W1003, Fall 2013
Practice Midterm Exam
2 hours, 4 problems, 100 points. No calculators or notes/books, but
you are welcome (though not required) to use graph paper/ruler.
Explain all steps in your reasoning, or you may lose up to 75% of
points. Dont
Math W1003, Fall 2013
Practice Final 1
3 hours, 6 problems, 150 points. No calculators or notes/books, but
you are welcome (though not required) to use graph paper/ruler.
Explain all steps in your reasoning, or you may lose up to 75% of
points. Dont forge
Math W1003, Fall 2013
Lecture 13
We will now start talking about the modern way to think about trigonometric functions, and the rst step in doing so is to stop thinking of angles
as measured in degrees. The reason why this is not such a mathematical idea
Math W1003, Fall 2013
Lecture 7
In our last class we discussed lines in the coordinate plane. The equation of
a line is:
ax + by + c = 0
(1)
for some real constants a, b, c. This should be read as:
the line cut out by the above equation is the set of poin
Math W1003, Fall 2013
Lecture 8
We will start this class by doing a bonus topic to our previous lecture.
Namely, we gave formulas for the reection in certain special lines (the axes
x = 0 and y = 0, and the diagonal x = y). These formulas were all in the