TEC 101: BASIC ENGINEERING MATHEMATICS I
Course Content
1.Algebra:unions,intersections,complements.Algebraic structures such as rational indices,multiplication,addition and partial fractions.
2.Polynomial functions:constant,linear,quadratic,division,remainder factor;functions and mappings;inverse,constant,step,even,odd,composite.
3.Trigonometric and hyperbolic functions,logarithms and exponential functions.
4.Vectors:scalars and vectors,components,addition,multiplication,vector spaces.Ration theorem,scalar and vector products, unit vectors,geometric interpretations,applications to mechanics.
5.Matrices:matrix algebra,determinants,rank of a matrix,transpose,inverse of an n x n matrix,eigenvalues,eigenvectors.
6.Solution of linear equations,crammer's rule, elementary row operations;Gauss' elimintion method;lower-upper decomposition.Solutions of homogenous equations.
References
1. Engineerin Mathematics, 5th edition by K.A Stroud
2. Mathematics for scientists n Engineers, vol 1 by Phiri PA Vuma D and Luboobi L.S.
3.Further Pure maths 1 vol 1 bz Sam Boardman,Tony Clouph,David Evans

wedges and screws

Wedges and screws

Wedge: A wedge is in general a triangular object which is placed between two objects to either hold them in place or is used to move one relative to the other. For example, the following shows a wedge under a block that is supported by the wall.

If the force P is large enough to push the wedge forward, then the block will rise and the following is an appropriate free-body diagram. Note that for the wedge to move one needs to have slip on all three surfaces. The direction of the friction force on each surface will oppose the slipping.

Since before the wedge can move each surface must overcome the resistance to slipping, one can assume that

These equations and the equations of equilibrium are combined to solve the problem. If the force P is not large enough to hold the top block from coming down, then the wedge will be pushed to the left and the appropriate free-body diagram is the following. Note that the only change is the direction of the frictional forces. A similar analysis to the above yield the solution to the problem.

Screws: One can consider a screw and a bolt as a combination of two wedges. One wedge is obtained from opening the helical treads of the screw and the other will come from opening the helical threads of the bolt. For example, if one opens one revolution of the thread of a screw having a lead of l and mean thread diameter 2r, one gets the following wedge where is the lead angle.

Now consider the situation where a screw is in a bolt or threaded hole. In the picture the screw is being pushed into the hole as the screwing moment is trying to unscrew it. Depending on the lead angle, the magnitude of the axial load W, and the magnitude of the applied moment M, either the axial load will dominate and the screw will move down or the moment will dominate and the screw will move up.

If the axial moment is sufficient to unscrew the screw, then the frictional force will oppose moving up of the screw threads and one will use the following free-body diagram for the wedge.

If the axial load is large enough to screw the screw into the bolt, then the frictional force will oppose the downward motion of the screw threads and one must use the following free-body diagram.

One can reverse the axial load W to be pulling the screw out or one can reverse the direction of the screwing moment. These cases can be studied in a similar way. If you need to know how much axial force W or screwing moment M is needed to make the screw turn in a given way, then you assume that the threads are slipping and set . Take the static coefficient of friction if you need the initial value to get the screw started turning, and take the kinetic coefficient of friction if you would like to know what is needed to keep the screw turning.

Self-locking screw: A screw is considered self-locking if the lead angle is selected such that in the absence of a screwing moment (i.e., M=0) the frictional force will remain less than so that the threads will not slip relative to each other. This can be studied using the following free-body diagram. The maximum thread angle for the screw to be self-locking is given by setting .

he general technique for solving bigger-than-quadratic polynomials is
pretty straightforward, but the process can be time-consuming.

The first step is to apply the Rational Roots Test to the polynomial
to get a list of values that might possibly be solutions to the
polynomial equation. You can follow this up with an application of
Descartes' Rule of Signs, if you like, to narrow down which possible
zeroes might be best to check. Of course, if you've got a graphing
calculator, it's a good idea to do a quick graph, since x-intercepts
of the graph are the same as zeroes of the equation. Seeing where the
graph looks like it crosses the axis can quickly narrow down your list
of possible zeroes.

Once you've found a value you want to test, you use synthetic division
to see if you can get a zero remainder. If you get a zero remainder,
you've not only found a zero, but you've also reduced your polynomial
by one degree. Remember that synthetic division is, among other
things, division, so checking if x = a is a solution is the same as
dividing out the linear factor x – a. This means that you should not
return to the original polynomial for your next computation (for
finding the other zeroes); you should instead work with the output of
the synthetic division. It's smaller, so it's easier to work with.

You should not be surprised to see some complicated solutions to your
polynomials (that is, solutions containing square roots or complex
numbers, or both); these zeroes will come from applying the Quadratic
Formula to the last (quadratic) factor of your polynomial. Here's how
the process plays out:

* Find all the zeroes of the following polynomial: 2x5 + 3x4 –
30x3 – 57x2 – 2x + 24

First, I'll apply the Rational Roots Test—

Wait. Actually, the first thing I'll do is check to see if x = 1
or x = –1 is a root, because these are the simplest roots to test for.
This isn't an "official" first step, but it can often be a timesaver,
because you can just look at the powers and the numbers. When x = 1,
the polynomial evaluates as 2 + 3 – 30 – 57 – 2 + 24 = –60, so x = 1
isn't a root. But when x = –1, I get –2 + 3 + 30 – 57 + 2 + 24 = 0, so
x = –1 is a root, and I can take care of it right away:

synthetic division by x = -1

This leaves me with the smaller polynomial 2x4 + x3 – 31x2 – 26x
+ 24. (Since I've divided out the factor x + 1, I've reduced the
degree of the polynomial by 1. That's how I know this is a degree-four
polynomial.) Now I'll apply the Rational Roots Test to get a list of
potential zeroes to try:

±1/2, 1, 3/2, 2, 3, 4, 6, 8, 12, 24

From experience, I've learned that most of these exercises have
their zeroes near the middle of the list, rather than at the extremes.
This isn't always true, of course, but it's usually better to stay
away from the larger numbers. In this case, I won't start off by
trying stuff like x = –24 or x = 12. Instead, I'll start out with
smaller values like x = 2. And I can narrow down my options further by
"cheating" and looking at the graph:

graph of y = 2x^4 + x^3 - 31x^2 - 26x + 24

This is a fourth-degree polynomial, so it has, at most, four
x-intercepts, and I can see all four of them on the graph. It looks
like one of the zeroes is around –3.5, but –7/2 isn't on the list that
the Rational Roots Test gave me, so this must be an irrational root.
I'll leave it until last.

It also looks like there may be zeroes near –1.5 and 0.5. But
the clearest solution looks to be at x = 4 and since whole numbers are
easier to work with than fractions, x = 4 would probably be a good
value to try:

synthetic division with x = 4

The zero remainder says that x = 4 is a root. The bottom row of
the synthetic division tells me that I'm now left with factoring 2x3 +
9x2 + 5x – 6. Looking at the constant term "6", I can see that x =
±24, ±12, ±8, and –4 won't work as rational roots (even if I didn't
already know from the graph), so I can cross them off of my list.
(Always check the numbers as you go. The Rational Roots Test can give
a very long list of possibilities, and it can be helpful to notice
that some of those values can be ignored, especially if you don't have
a graphing calculator to "cheat" with.) Comparing the remaining values
on the list with the intercepts on the graph, I'll try x = 1/2:

synthetic division with x = 1/2

The remainder isn't zero, so that test root didn't work. This
means that the zero close to x = 1/2 on the graph must be irrational;
I'll find it when I apply the Quadratic Formula later. For now, I'll
try x = –3/2: Copyright © Elizabeth Stapel 2005-2011 All Rights
Reserved

synthetic division with x = -3/2

The division came out evenly, leaving me with the polynomial 2x2
+ 6x – 4. Since I'm looking for the zeroes of the polynomial, what I
really have here is 2x2 + 6x – 4 = 0. Dividing through by 2 to get
smaller numbers gives me x2 + 3x – 2 = 0, to which I can apply the
Quadratic Formula:

x = [-3 ± sqrt(17)]/2, or about -3.5 and 0.5

Then the complete solution is:

x = -1, [-3 +/- sqrt(17)]/2, -3/2, 4

Asking you to find the zeroes of a polynomial means the same thing as
asking you to find the solutions to a polynomial equation. The zeroes
of a polynomial are the values of x that make the polynomial equal to
zero. So the above problem could have been stated along the lines of
"Find the solutions to 2x5 + 3x4 – 30x3 – 57x2 – 2x + 24 = 0" or "Find
the solutions to 2x5 + 3x4 – 30x3 – 57x2 – 2x = –24", and the answers
would have been the exact same list of x-values.

You can use these same techniques to factor bigger-than-quadratic
polynomials....

By now, you should be familiar with variables and exponents, and you
may have dealt with expressions like 3x4 or 6x. Polynomials are sums
of these "variables and exponents" expressions. Each piece of the
polynomial, each part that is being added, is called a "term".
Polynomial terms have variables which are raised to whole-number
exponents (or else the terms are just plain numbers); there are no
square roots of variables, no fractional powers, and no variables in
the denominator of any fractions. Here are some examples:
6x –2 This is NOT
a polynomial term... ...because the variable has a negative exponent.
1/x2 This is NOT
a polynomial term... ...because the variable is in the denominator.
sqrt(x) This is NOT
a polynomial term... ...because the variable is inside a radical.
4x2 This IS a polynomial term... ...because it obeys all the rules.

Here is a typical polynomial:

terms

Notice the exponents on the terms. The first term has an exponent of
2; the second term has an "understood" exponent of 1; and the last
term doesn't have any variable at all. Polynomials are usually written
this way, with the terms written in "decreasing" order; that is, with
the largest exponent first, the next highest next, and so forth, until
you get down to the plain old number.

Any term that doesn't have a variable in it is called a "constant"
term because, no matter what value you may put in for the variable x,
that constant term will never change. In the picture above, no matter
what x might be, 7 will always be just 7.

The first term in the polynomial, when it is written in decreasing
order, is also the term with the biggest exponent, and is called the
"leading term".

The exponent on a term tells you the "degree" of the term. For
instance, the leading term in the above polynomial is a "second-degree
term" or "a term of degree two". The second term is a "first degree"
term. The degree of the leading term tells you the degree of the whole
polynomial; the polynomial above is a "second-degree polynomial". Here
are a couple more examples:

* Give the degree of the following polynomial: 2x5 – 5x3 – 10x + 9

This polynomial has four terms, including a fifth-degree term, a
third-degree term, a first-degree term, and a constant term.

This is a fifth-degree polynomial.

* Give the degree of the following polynomial: 7x4 + 6x2 + x

This polynomial has three terms, including a fourth-degree term,
a second-degree term, and a first-degree term. There is no constant
term.

This is a fourth-degree polynomial.

When a term contains both a number and a variable part, the number
part is called the "coefficient". The coefficient on the leading term
is called the "leading" coefficient.

terms

In the above example, the coefficient of the leading term is 4; the
coefficient of the second term is 3; the constant term doesn't have a
coefficient. Copyright © Elizabeth Stapel 2000-2011 All Rights
Reserved

The "poly" in "polynomial" means "many". I suppose, technically, the
term "polynomial" should only refer to sums of many terms, but the
term is used to refer to anything from one term to the sum of a
zillion terms. However, the shorter polynomials do have their own
names:

* a one-term polynomial, such as 2x or 4x2, may also be called a
"monomial" ("mono" meaning "one")
* a two-term polynomial, such as 2x + y or x2 – 4, may also be
called a "binomial" ("bi" meaning "two")
* a three-term polynomial, such as 2x + y + z or x4 + 4x2 – 4, may
also be called a "trinomial" ("tri" meaning "three")

I don't know if there are names for polynomials with a greater numbers
of terms; I've never heard of any names other than what I've listed.

Polynomials are also sometimes named for their degree:

* a second-degree polynomial, such as 4x2, x2 – 9, or ax2 + bx +
c, is also called a "quadratic"
* a third-degree polynomial, such as –6x3 or x3 – 27, is also
called a "cubic"
* a fourth-degree polynomial, such as x4 or 2x4 – 3x2 + 9, is
sometimes called a "quartic"
* a fifth-degree polynomial, such as 2x5 or x5 – 4x3 – x + 7, is
sometimes called a "quintic"

There are names for some of the higher degrees, but I've never heard
of any names being used other than the ones I've listed.

By the way, yes, "quad" generally refers to "four", as when an ATV is
referred to as a "quad bike". For polynomials, however, the "quad"
from "quadratic" is derived from the Latin for "making square". As in,
if you multiply length by width (of, say, a room) to find the area in
"square" units, the units will be raised to the second power. The area
of a room that is 6 meters by 8 meters is 48 m2. So the "quad" refers
to the four corners of a square, from the geometrical origins of
parabolas and early polynomials.

Evaluation

"Evaluating" a polynomial is the same as evaluating anything else: you
plug in the given value of x, and figure out what y is supposed to be.
For instance:

* Evaluate 2x3 – x2 – 4x + 2 at x = –3

I need to plug in "–3" for the "x", remembering to be careful
with my parentheses and the negatives:

2(–3)3 – (–3)2 – 4(–3) + 2
= 2(–27) – (9) + 12 + 2
= –54 – 9 + 14
= –63 + 14
= –49

Always remember to be careful with the minus signs!

The Factor Theorem

The Factor Theorem is a result of the Remainder Theorem, and is based
on the same reasoning. If you haven't read the lesson on the Remainder
Theorem, review that topic first, and then return here.

As the Remainder Theorem points out, if you divide a polynomial p(x)
by a factor x – a of that polynomial, then you will get a zero
remainder. Let's look again at that Division Algorithm expression of
the polynomial:

p(x) = (x – a)q(x) + r(x)

If x – a is indeed a factor of p(x), then the remainder after division
by x – a will be zero. That is:

p(x) = (x – a)q(x)

In terms of the Remainder Theorem, this means that, if x – a is a
factor of p(x), then the remainder, when we do synthetic division by x
= a, will be zero.

The point of the Factor Theorem is the reverse of the Remainder
Theorem: If you synthetic-divide a polynomial by x = a and get a zero
remainder, then, not only is x = a a zero of the polynomial (courtesy
of the Remainder Theorem), but x – a is also a factor of the
polynomial (courtesy of the Factor Theorem).

Just as with the Remainder Theorem, the point here is not to do the
long division of a given polynomial by a given factor. This Theorem
isn't repeating what you already know, but is instead trying to make
your life simpler. When faced with a Factor Theorem exercise, you will
apply synthetic division and then check for a zero remainder.

* Use the Factor Theorem to determine whether x – 1 is a factor of
f (x) = 2x4 + 3x2 – 5x + 7.

For x – 1 to be a factor of f (x) = 2x4 + 3x2 – 5x + 7, the
Factor Theorem says that x = 1 must be a zero of f (x). To test
whether x – 1 is a factor, I will first set x – 1 equal to zero and
solve to find the proposed zero, x = 1. Then I will use synthetic
division to divide f (x) by x = 1. Since there is no cubed term, I
will be careful to remember to insert a "0" into the first line of the
synthetic division to represent the omitted power of x in 2x4 + 3x2 –
5x + 7:

completed division: 2 2 5 0 7

Since the remainder is not zero, then the Factor Theorem says that:

x – 1 is not a factor of f (x).

* Using the Factor Theorem, verify that x + 4 is a factor of
f (x) = 5x4 + 16x3 – 15x2 + 8x + 16.

If x + 4 is a factor, then (setting this factor equal to zero
and solving) x = –4 is a root. To do the required verification, I need
to check that, when I use synthetic division on f (x), with x = –4, I
get a zero remainder:

completed division: 5 –4 1 4 0

The remainder is zero, so the Factor Theorem says that:

x + 4 is a factor of 5x4 + 16x3 – 15x2 + 8x + 16.

In practice, the Factor Theorem is used when factoring polynomials
"completely". Rather than trying various factors by using long
division, you will use synthetic division and the Factor Theorem. Any
time you divide by a number (being a potential root of the polynomial)
and get a zero remainder in the synthetic division, this means that
the number is indeed a root, and thus "x minus the number" is a
factor. Then you will continue the division with the resulting smaller
polynomial, continuing until you arrive at a linear factor (so you've
found all the factors) or a quadratic (to which you can apply the
Quadratic Formula).

* Using the fact that –2 and 1/3 are zeroes of f (x) = 3x4 + 5x3
+ x2 + 5x – 2, factor the polynomial completely. Copyright ©
Elizabeth Stapel 2002-2011 All Rights Reserved

If x = –2 is a zero, then x + 2 = 0, so x + 2 is a factor.
Similarly, if x = 1/3 is a zero, then x – 1/3 = 0, so x – 1/3 is a
factor. By giving me two of the zeroes, they have also given me two
factors: x + 2 and x – 1/3.

Since I started with a fourth-degree polynomial, then I'll be
left with a quadratic once I divide out these two given factors. I can
solve that quadratic by using the Quadratic Formula or some other
method.

The Factor Theorem says that I don't have to do the long
division with the known factors of x + 2 and x – 1/3. Instead, I can
use synthetic division with the associated zeroes –2 and 1/3. Here is
what I get when I do the first division with x = –2:

completed divison: bottom row: 3 –1 3 –1 0

The remainder is zero, which is expected because they'd told me
at the start that –2 was a known zero of the polynomial. Rather than
starting over again with the original polynomial, I'll now work on the
remaining polynomial factor of 3x3 – x2 + 3x – 1 (from the bottom line
of the synthetic division). I will divide this by the other given
zero, x = 1/3:

completed division: bottom row: 3 0 3 0

This leaves me with the quadratic 3x2 + 3, which I can solve:

3x2 + 3 = 0
3(x2 + 1) = 0
x2 + 1 = 0
x2 = –1
x = ± i

If the zeroes are x = –i and x = i, then the factors are x –
(–i) and x – (i), or x + i and x – i. I need to remember that I
divided off a "3" when I solved the quadratic; it is still part of the
polynomial, and needs to be included as a factor. Then the
fully-factored form is:

3x4 + 5x3 + x2 + 5x – 2 = 3(x + 2)(x – 1/3)(x + i)(x – i)

REMAINDER THEOREM

The Remainder Theorem
The Remainder Theorem is useful for evaluating polynomials at a given
value of x, though it might not seem so, at least at first blush. This
is because the tool is presented as a theorem with a proof, and you
probably don't feel ready for proofs at this stage in your studies.
Fortunately, you don't "have" to understand the proof of the Theorem;
you just need to understand how to use the Theorem.

The Remainder Theorem starts with an unnamed polynomial p(x), where
"p(x)" just means "some polynomial p whose variable is x". Then the
Theorem talks about dividing that polynomial by some linear factor x –
a, where a is just some number. Then, as a result of the long
polynomial division, you end up with some polynomial answer q(x) (the
"q" standing for "the quotient polynomial") and some polynomial
remainder r(x).

As a concrete example of p, a, q, and r, let's look at the polynomial
p(x) = x3 – 7x – 6, and let's divide by the linear factor x – 4 (so a
= 4):

completed division: quotient x^2 + 4x + 9, remainder 30

So we get a quotient of q(x) = x2 + 4x + 9 on top, with a remainder of
r(x) = 30.

You know, from long division of regular numbers, that your remainder
(if there is one) has to be smaller than whatever you divided by. In
polynomial terms, since we're dividing by a linear factor (that is, a
factor in which the degree on x is just an understood "1"), then the
remainder must be a constant value. That is, when you divide by "x –
a", your remainder will just be some number.

The Remainder Theorem then points out the connection between division
and multiplication. For instance, since 12 ÷ 3 = 4, then 4 × 3 = 12.
If you get a remainder, you do the multiplication and then add the
remainder back in. For instance, since 13 ÷ 5 = 2 R 3, then 13 = 5 × 2
+ 3. This process works the same way with polynomials. That is:

If p(x) / (x – a) = q(x) with remainder r(x),

then p(x) = (x – a) q(x) + r(x).

(Technically, this "if - then" statement is the "Division Algorithm
for Polynomials". But the Algorithm is the basis for the Remainder
Theorem.)

In terms of our concrete example: Copyright © Elizabeth Stapel
2002-2011 All Rights Reserved

Since (x^3 – 7x – 6) / (x – 4) = x2 + 4x + 9 with remainder 30,

then x3 – 7x – 6 = (x – 4) (x2 + 4x + 9) + 30.

The Remainder Theorem says that we can restate the polynomial in terms
of the divisor, and then evaluate the polynomial at x = a. But when x
= a, the factor "x – a" is just zero! Then evaluating the polynomial
at x = a gives us:

p(a) = (a – a)q(a) + r(a)
= (0)q(a) + r(a)
= 0 + r(a)
= r(a)

But remember that the remainder term r(a) is just a number! So the
value of the polynomial p(x) at
x = a is the same as the remainder you get when you divide that
polynomial p(x) by x – a. In terms of our concrete example:

p(4) = (4 – 4)((4)2 + 4(4) + 9) + 30
= (0)(16 + 16 + 9) + 30
= 0 + 30
= 30

But you gotta think: Okay, fine; the value of the polynomial p(x) at x
= a is the remainder r(a) when you divide by x – a, but who wants to
do the long division each time you have to evaluate a polynomial at a
given value of x?!? You're right; this would be overkill. Fortunately,
that's not what they really want you to do.

When you are dividing by a linear factor, you don't "have" to use long
polynomial division; instead, you can use synthetic division, which is
much quicker. In our example, we would get:

completed division: bottom row: 1 4 9 30

Note that the last entry in the bottom row is 30, the remainder from
the long division (as expected) and also the value of p(x) = x3 – 7x
– 6 at x = 4. And that is the point of the Remainder Theorem: There is
a simpler, quicker way to evaluate a polynomial p(x) at a given value
of x, and this simpler way is not to evaluate p(x) at all, but to
instead do the synthetic division at that same value of x. Here are
some examples:

* Use the Remainder Theorem to evaluate f (x) = 6x3 – 5x2 + 4x –
17 at x = 3.

First off, even though the Remainder Theorem refers to the
polynomial and to long division and to restating the polynomial in
terms of a quotient, a divisor, and a remainder, that's not actually
what I'm meant to be doing. Instead, I'm supposed to be doing
synthetic division, using "3" as the divisor:

completed division: bottom row: 6 13 43 112

Since the remainder (the last entry in the bottom row) is 112,
then the Remainder Theorem says that:

f (3) = 112.

* Using the Remainder Theorem, find the value of f (–5), for f
(x) = 3x4 + 2x3 + 4x.

I need to do the synthetic division, remembering to put zeroes
in for the powers of x that are not included in the polynomial:

completed division: bottom row: 3 -13 65 -321 1605

Since the remainder is 1605, then, thanks to the Remainder
Theorem, I know that:

f (–5) = 1605.

* Use the Remainder Theorem to determine whether x = 2 is a zero of
f (x) = 3x7 – x4 + 2x3 – 5x2 – 4

For x = 2 to be a zero of f (x), then f (2) must evaluate to
zero. In the context of the Remainder Theorem, this means that my
remainder, when dividing by x = 2, must be zero:

completed division: bottom row: 3 6 12 23 48 91 182 360

The remainder is not zero. Then x = 2 is not a zero of f (x).

* Use the Remainder Theorem to determine whether x = –4 is a solution of
x6 + 5x5 + 5x4 + 5x3 + 2x2 – 10x – 8 = 0

For x = –4 to be a solution of f (x) = x6 + 5x5 + 5x4 + 5x3 +
2x2 – 10x – 8 = 0, it must be that f (–4) = 0. In the context of the
Remainder Theorem, this means that the remainder, when dividing by x =
–4, must be zero:

completed division: bottom row: 1 1 1 1 -2 -2 0

The remainder is zero. Then x = –4 is a solution of the given equation.