EIT exam

Power Functions – Fundamentals of Engineering Exam Review

December 20, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Power Functions.

The goal is to reinforce both your understanding of the various rules of power functions, but also to put this knowledge to work through a few example problems.

This Fundamentals of Engineering Review of Power Functions is part of the global subject Mathematics.

Having served over 1.5 million lessons in 150 different countries, it’s safe to say we have been in the trenches.

In the trenches working with those who are taking the exam for the first time, either fresh out of college or having taken a few years off from the books.

In the trenches working with those who have already attempted the exam, maybe multiple times, and can’t seem to crack the code.

In the trenches working with those who have been away from the books for ages, from standardized testing for the same amount of time, and can’t fathom taking on this beast of an exam.

And every iteration in between.

Every student’s situation is unique, but the building blocks of what success looks like are always the same.

Click below to register for the FREE myFE Exam Academy Course.

Once you are registered, you’ll get instant access to a proven roadmap, strategies and tactics…

Straight to your email box.

Transcription

Hey what’s gong on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review Of Power Functions.

There are some essential rules of power functions that we must be familiar with in order to greatly simplify the solving of these complex equations.

So let’s start off our Fundamentals of Engineering Exam Review Of Power Functions by defining exactly what a Power Function is.

A power function takes the

form Y is equal to a raised to the X

where a is known as the base and X is

known as the power or the exponent both

interchangeable terms

now when given a power function and that

power function strays away from this

standard form we can dig into our bag of

tricks to simplify using a list of rules

the first rule is the power of products

say we are given a power function where

the base is made up of a product of two

numbers say a times B and that base is

then raised to the X power

given a function in this form it can be

re-written as a raised to the x times B

raised to the X now if we’re given a

function as a product of two power

functions of the same base being raised

to different powers such as a raised to

the x times a raised to the W this can

be rewritten by taking the exponents

adding them together and setting them

against the common base so in other

words we can rewrite this as a x + W now

what if we’re given a function where one

power function is being divided by a

another power function of the same base

similar as before say we have a raised

to the X divided by a raised to the W

well the rule here is that this function

can be rewritten by taking the exponents

and subtracting them from one another

and setting that against the common base

so in other words you can rewrite it as

a raised to the X minus W now if the

base of itself is a power function and

is raised then again by another power

say a raised to the W and that’s raised

to the X this can be rewritten by

multiplying the exponents together and

setting it against the original base so

in other words it can be re-written as a

raised to the W times X now all these

rules right here we’re dealing with

positive exponents but all numbers are

possible in power functions so let’s

look at the special cases when the

exponents are equal to 0 1 are negative

or even non integer rational numbers

such as fractions so first we can note

that any base raised to the power of 0

is 1 so if we the rule here the special

rule the power of 0 if we have any base

and it’s raised to the 0 that is going

to be equal to 1 now let’s talk about

the power of power of 1 rule any base

raised to the power 1 means that that

base is being multiplied only once by

itself so that means that a raised to

the 1 is equal to itself a it’s only

it’s only multiplied once but what about

negative numbers how do we deal with

negative exponents well we determined

that when given a function where one

power function is being divided by

another power function of the same base

if you recall a X divided by aw that the

function can be rewritten by subtracting

the exponents and setting them against

the common base so in other words a X

minus W so we’ve already established

that but what happens when W in the

denominator the power in the denominator

is larger than the power in the

numerator

well since the larger exponent is on the

bottom it shouldn’t be a surprise that

the beta U that the base raised to a

negative exponent can be re-written as a

raised to the say negative W is actually

rewritten as a over so r1 divided by a

raised to the W so in other words we can

quickly illustrate this let’s say that

we have a squared divided by a to the

third if you just expand this out that

will be a times a in the in the

numerator and we got a times a times a

in the denominator and now common terms

cancel out so we got two A’s on the

bottom that cancel out two A’s on the

top one remains on the just the number

one remains on the top and so now we

have 1 over a which is also equivalent

to a to the negative one so now lastly

let’s look at let’s look at the dreaded

fractions when a base is raised to a

fraction let’s say 1 to the N for

definition purposes that function can be

rewritten as the nth root of a

such as say if we have won a raised to

one to the end can be re-written as the

nth root of a but the fraction doesn’t

have to be one over some variable it can

be anything such as two thirds or five

six

so let’s generalize this by saying that

if we are given a function raised to M

in a raised to the M over n then that

function can be re-written as the nth

root of a raised to the M so that’s

going to be our general equation so for

given any function let’s just look at

one right real quick let’s say we have 5

raised to the one-sixth well that can

simply be re-written as 6 the sixth root

of 5 raised to the 1 and once again that

wine could be 3 could be 4 or 5 whatever

number is on top it doesn’t have to be

one that’s just M so lastly I want to

touch on one particular base which is of

singular importance in math and that is

the transcendental number or e which is

written as E and that’s equal to two

point seven one eight two and and change

it keeps going this base shows up

naturally and calculates because the

power of the base is equal to its own

derivative so in other words if we’re

given the function e to the X the

derivative of that dy/dx is also equal

to e to the X the base power a function

either the X is called the exponential

function and is sometimes abbreviated as

exp such that e each of X is equal to e

XP X so you might see the exponential

function

written in either of these two ways so

that will wrap up the review of the

power functions if you guys have any

Logarithms – Fundamentals of Engineering Exam Review

December 20, 2011 by Justin Leave a Comment

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In this episode of Engineer In Training Exam TV will reinforce your understanding of Logarithms. We will be defining different rules of Logarithms and run through a few example problems using these ratios.

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Enjoy!!

Trig Ratios – Fundamentals of Engineering Exam Review

December 20, 2011 by Justin Leave a Comment

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In this episode of Engineer In Training Exam TV will reinforce your understanding of the Trig Ratios. We will begin by defining SOH CAH TOA and then run through a few example problems using these ratios.

CLICK HERE to get immediate access to the FREE “EIT PREPARATION BOOT CAMP” and get yourself started on the road to DOMINATING the EIT Exam NOW.

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Right Triangles – Fundamentals of Engineering Exam Review

December 20, 2011 by Justin Leave a Comment

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In this episode of Engineer In Training Exam TV will reinforce your understanding of the Right Triangle. We will begin by defining the common right triangles and then run through a few example problems.

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Hyperbola – Fundamentals of Engineering Exam Review

December 19, 2011 by Justin Leave a Comment

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In this episode of Engineer In Training Exam TV will reinforce your understanding of the Hyperbola. We will begin by defining the general equation for a Hyperbola and then run through a few example problems.

CLICK HERE to get immediate access to the FREE “EIT PREPARATION BOOT CAMP” and get yourself started on the road to DOMINATING the EIT Exam NOW.

Enjoy!!

Ellipse – Fundamentals of Engineering Exam Review

December 19, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Ellipses.

We will begin by defining the general equation for an ellipse and then run through a few example problems.

This Fundamentals of Engineering Exam Review of Ellipses is part of the global subject Mathematics.

Having served over 1.5 million lessons in 150 different countries, it’s safe to say we have been in the trenches.

In the trenches working with those who are taking the exam for the first time, either fresh out of college or having taken a few years off from the books.

In the trenches working with those who have already attempted the exam, maybe multiple times, and can’t seem to crack the code.

In the trenches working with those who have been away from the books for ages, from standardized testing for the same amount of time, and can’t fathom taking on this beast of an exam.

And every iteration in between.

Every student’s situation is unique, but the building blocks of what success looks like are always the same.

Click below to register for the FREE myFE Exam Academy Course.

Once you are registered, you’ll get instant access to a proven roadmap, strategies and tactics…

Straight to your email box.

Transcription

Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review of Ellipses.

We’ll start off this tutorial discussing the theory and then get in to working a problem.

So let’s start off our Fundamentals of Engineering Exam Review of Ellipses by reviewing the General Formula.

The general equation of an ellipse that takes the form of X minus H squared

divided by a squared plus y minus K

squared divided by B squared is equal to

1 now please note that in order for the

ellipse to be in the standard form that

the right side must equal 1 the point H

K on the other hand as it was in all the

other general forms such as the circle

on the parabola that represents this

center of the ellipse now a is the the

square root of the term that is below

the X term in the general form equation

and a is the distance that the that the

ellipse moves from the center right and

left so if we have the term say we have

the center is located at H K then the

rightmost point on the graph is going to

be H plus a ok

and the left most point this is the

right point and the leftmost point is

going to be H minus a ok so that’s just

how far the ellipse extends out from the

center to left and right now the the B

term on the other hand is the square

root of the term that is below the Y

term in the general equation and the B

term represents how much the lip

stretches in the up-and-down direction

now once again if we’re given HK is the

center then the top most the top most

point

is going to be h k+ b and the bottom

most point is going to be equal to H K

minus B so if we’re given these terms

the a square terms for given the a

squared term in the B squared term if

we’re given H and K then that’s all we

really need to graph the ellipse and

determine all the points on the ellipse

so let’s quickly look at an example here

let’s say that we were given an equation

we’re given in eclipse with the general

in the general form of X plus 2 squared

divided by 9 plus y minus 4 squared

divided by 25 is equal to 1 so what we

want to do is find the rightmost the

left-most the top and the bottom most

points of this ellipse so we’re lucky

because this is just in the standard

form we easily can see what we’re H and

K is or where the center is and in this

case H is going to be equal to negative

2 and K is going to be equal to 4 so the

center of the Klipsch HK is located at

negative 2 4 and so what we want to do

now is determine what a and B is now we

know this term at the bottom below the X

term is a squared and we know this term

below the Y term is B squared and so all

we need to do is take the square roots

to find out that a is going to equal to

3 and B is going to be equal to 5 so

let’s go ahead and write this down at

the right point we want to find the left

point we want to find the top point and

the bottom point of this ellipse

so all we need to do is know that the

right point is going to be H plus a okay

the left point is going to be H minus a

plus K the top point is going to be H K

plus B and the bottom is going to be H K

minus B so all we need to do is plug in

our information and we find that H plus

a H is negative 2 a is 3 we’re going to

have 1 and we know K is 4 so that the

right point is going to be 1/4 we have h

minus a H again is negative 2 and a is 3

so we got negative 2 minus 3 we’re going

to get negative 5/4 and then now we got

to find the top and now we’re dealing

with the B term and we know that H is

going to be negative 2 and that K is 4

and we want to add 5 to that so it’s

going to be negative 2 9 and now we want

the bottom turn once again H is going to

be negative 2 and we got B is equal to 5

and we – that from K so we get negative

1 we get that the rightmost point is 1

for the leftmost point is negative 5 4

we got the top point at negative 2 9 and

we have the bottom point at negative 2

negative 1 so that’s it guys just a

quick review on how we can determine

some of the information if given an

equation of ellipse if we’re asked to

find Center if we’re asked to find some

certain point on that ellipse we know

the general equation and we know how to

find that information now so if you guys

have any more questions don’t hesitate

to visit the site at engineering

training exam comm where I continue to

put up more information more content to

help you guys out I could definitely use

suggestions though I

I know that you know I’ve been through

this process but the only way I can help

you is if you tell me what you guys need

you know I know I know it’s hard

sometimes to you know just give you guys

the information that you need and that’s

all I want to do so head on over there

and contact me through the contact page

or go ahead and sign up for my free EIT

preparation boot camp and you can

contact me through that mailer

I’m always responding to emails every

day to every single question I get so

don’t hesitate to do that all right guys

so take care tonight and we’ll talk soon

all right

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Circles – Fundamentals of Engineering Exam Review

December 19, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Circles.

He will first start with defining the general equation of a circle and how it relates to the distance Distance formula, then run through a few example problems.

This Fundamentals of Engineering Exam Review of Circles is part of the global subject Mathematics.

Having served over 1.5 million lessons in 150 different countries, it’s safe to say we have been in the trenches.

In the trenches working with those who are taking the exam for the first time, either fresh out of college or having taken a few years off from the books.

In the trenches working with those who have already attempted the exam, maybe multiple times, and can’t seem to crack the code.

In the trenches working with those who have been away from the books for ages, from standardized testing for the same amount of time, and can’t fathom taking on this beast of an exam.

And every iteration in between.

Every student’s situation is unique, but the building blocks of what success looks like are always the same.

Click below to register for the FREE myFE Exam Academy Course.

Once you are registered, you’ll get instant access to a proven roadmap, strategies and tactics…

Straight to your email box.

Transcription

Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review of Circles.

We’ll start off this tutorial discussing the theory and then get in to working a problem.

So let’s start off our Fundamentals of Engineering Exam Review of Circles by revisiting the Distance Formula.

Let’s take at the point say X Y and x1 y1 the

distance formula tells us that the

distance between these two points D D is

equal to x1 minus x squared plus y 1

minus y squared all under square root

now a circle or the equation in the

circle is just a reiteration of the

distance formula all the way around it

is telling us that each point on the

outside of a circle is a fixed distance

R from a central point H K so let’s call

X Y here our central point HK let’s call

our D our set radius so D is equal to R

and we can rewrite this equation to say

that are the fixed radius distance

between H K and any point on a circle is

actually at x1 minus H squared plus y1

minus K squared all under the square

root to see this in the standard form or

in a form that were more used to seeing

all we need to do is square both sides

so by doing that we get R squared is

equal to let’s get rid of the x1 and y1

let’s just call them XY representing any

point XY on the outside of a circle so R

squared is equal

to X minus H squared plus y minus K

squared so this right here is going to

be our lovely equation standard form

equation for circle so this is the

standard form in this form we very

quickly can see what the radius is and

we can also see what the center of the

circle is which is going to be located

at HK whatever those two numbers are so

let’s look at a quick example here and

just illustrate how we can go about

defining the equation of a circle say

we’re given a circle with the radius of

8 and let’s say that we’re told that the

center of that circle is located at

negative 4 7 so let’s just recall that

this term negative 4 7 the point

negative 4 7 is our HK in our standard

equation and this of course is our R so

once again restating the standard form

it’s R squared is equal to X minus H

squared plus y minus K squared and this

is very simple all we need to do is plug

in R all the given information we got 8

squared is equal to X minus negative 4

squared plus y minus 7 squared so this

ultimately plays out to be let’s see X

plus 4 squared plus y minus 7 squared is

equal to 64 so this is an equation

that’s telling us that the circle is

located at negative 4

and has a radius of a square root of 64

which is 8 so let’s look at another

example here let’s say on the other hand

were given the equation of a circle and

we’re asked to find the center and also

the radius of that circle let’s say that

we’re given the equation let’s go with x

squared plus y squared plus 8x plus 7

now when we first glance at this

equation we obviously hope that it’s

going to over already be in the standard

form so we can quickly just pick out the

necessary information that we need which

is the radius and the center of the

circle but in this case obviously it is

not so we do have a little work to do so

the first step or the first 2 steps is

to first get the constant on the right

side of the equation and also group the

terms together group the X and the y

terms together so to do that we get x

squared plus 8x plus y squared is equal

to negative 7 so the next step is for

each variable with two terms we need to

complete the square of those terms and

we look on the left side here and see

that X is the only term that or the only

the only variable that has two terms so

that is the only variable that we need

to complete the square of so let’s

recall to complete the square what we do

is we take half the coefficient of x and

square it so we’ll go 8 divided by 2

squared which is equal to 4 squared

which is equal to 16 so now what we do

is add that we add this term to both

sides of the equation

so we get x squared plus 8x

plus 16 plus y squared is equal to

negative 7 plus 16 so that what that is

if we just lets see now we can go ahead

on this right side and just do this

calculation real quick we got 9 and on

the left side here now we can factor the

X term such that it’s equal to X plus 4

squared plus y squared which is also

equal to Y minus 0 squared is equal to 9

so now we have our equation the same

equation here now we have it in the

standard form of a circle and we can

quickly determine what HK is and we can

also determine what the radius is so

looking at this we know that HK the

center of the circle is going to be

located at negative 4 0 and that the

radius of the circle is the square root

of 9 which is also equal to 3 so the

answer to this this this problem is if

we’re asked given the N equation we’re

asked to determine the center as well as

the radius we can just simply put it

into the standard form pick out the

necessary information and answer the

question so that’s it guys that’s all I

got for you today quick recap of circles

if you guys have any more questions or

suggestions or any other tutorials you

want to see go ahead and visit my site

at engineer and training exam com either

subscribe and get the free AIT

preparation bootcamp that I got or go

ahead and contact me through the contact

form either or to shoot me your

suggestions your feedback just let me

know how everything’s going here if you

guys are getting what you need my

concern is that you guys are at least

getting a little bit of value out of

each one of these videos and you know

that’s what I aim to do is just help you

guys out in your preparation so anyways

that’s it

for today you guys take care and we’ll

be talking soon all right bye

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Distance Formula – Fundamentals of Engineering Exam Review

December 19, 2011 by Justin Leave a Comment

In this episode of Engineer In Training Exam TV, Justin walks you through a Fundamentals of Engineering Exam Review of Distance Formula.

He will start with defining the Pythagorean Theorem and how the Distance formula relates, then finish off with a quick example of how to find the distance between any two given points.

This Fundamentals of Engineering Exam Review of Distance Formula is part of the global subject Mathematics.

Having served over 1.5 million lessons in 150 different countries, it’s safe to say we have been in the trenches.

In the trenches working with those who are taking the exam for the first time, either fresh out of college or having taken a few years off from the books.

In the trenches working with those who have already attempted the exam, maybe multiple times, and can’t seem to crack the code.

In the trenches working with those who have been away from the books for ages, from standardized testing for the same amount of time, and can’t fathom taking on this beast of an exam.

And every iteration in between.

Every student’s situation is unique, but the building blocks of what success looks like are always the same.

Click below to register for the FREE myFE Exam Academy Course.

Once you are registered, you’ll get instant access to a proven roadmap, strategies and tactics…

Straight to your email box.

Transcription

Hey what’s going on everyone, it’s Justin Dickmeyer from EngineerInTrainingExam.com.

In today’s video we are going to present a Fundamentals of Engineering Exam Review of Distance Formula.

We’ll start off this tutorial discussing the theory and then get in to working a problem.

So let’s start off our Fundamentals of Engineering Exam Review of Distance Formula by defining exactly what the Distance Formula is.

So what is the distance formula?

Well essentially
the distance formula is a restatement of
the Pythagorean theorem packaged up into
a simple equation that will allow us to
find the distance between two given
points so let’s consider the two points
negative 2 1 and 1 5 now it’s the
standard process to find the distance to
the of these two points we can simply
graph them and then draw a straight line
between them and then develop a triangle
as such would this be inside a this side
be inside being the hypotenuse which is
the distance between the two points be
inside C so graphically once again it’s
easy to find what C is just using the
Pythagorean theorem which is a squared
plus B squared is equal to C squared so
using this process we can simply find
out that if this is a point 1 5 and this
is point negative 2 1 that we can see
that a is going to be 3 and B is going
to be 4 so if we just plug that into the
Pythagorean theorem here we find that 3
squared plus 4 squared is equal to C
squared and just calculating that out we
find that it is equal to 9 plus 16 which
is equal to 25 which is equal to C
squared and we find that the hypotenuse
or the distance between these two points
is the square root of 25 which is equal
to
five so this this method obviously holds
true in any case however the the
distance formula
like I said repackages this formula into
a simple a simple equation that we can
just quickly plug in any two given
points in a coordinate system to figure
out what the distance is between those
two points
so the distance formula states that D is
equal to the square root of x2 minus x1
squared plus y squared and all that is
under the square root now this is of
course if we’re given x1 y1 and x2 y2 so
all we need to do is plug this given
information into the distance formula
and we will be able to determine the
distance between those two points so
let’s look at a quick example here let’s
say that we’re given the points negative
3 negative 2 and we’re given the point 5
2 now we can go ahead and graph this
make a triangle determine the sides a
and B and solve for the hypotenuse C or
we can go ahead and just take our
distance formula and plug in the
information that we are given which we
will do in this case so D is equal to
the square root of x2 which in this case
is 5 minus x1 which is negative 3 all
that squared plus y2 which is 2 minus y1
which is negative 2 all squared so
working that out we get we get the
square root of 8 squared plus 4 squared
which comes down to the square root of
80 and just thrown that into our
calculator real quick
the distance between these two given
points is eight point nine four so
that’s essentially it guys not much to
it it just once again is a reach the
distance formula is once again a
restatement of the Pythagorean theorem
it just allows us a quick concise way to
plug in two given points and define real
quickly what the distance is between
those two points so if you guys have any
questions go ahead and visit my site at
engineer in training exam comm check out
the other videos I got posted there as
well as here on YouTube leave comments
suggestions you know email contact me
I’m always open to help you guys out in
any way that I’m able once again
everything’s free for you guys to use so
check it out and look forward to talking
to you guys soon all right take care

—

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Power Functions – Fundamentals of Engineering Exam Review

Transcription

Logarithms – Fundamentals of Engineering Exam Review

Trig Ratios – Fundamentals of Engineering Exam Review

Right Triangles – Fundamentals of Engineering Exam Review

Hyperbola – Fundamentals of Engineering Exam Review

Ellipse – Fundamentals of Engineering Exam Review

Transcription

Circles – Fundamentals of Engineering Exam Review

Transcription

Distance Formula – Fundamentals of Engineering Exam Review

Transcription

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