in this video we're going to talk about

the harmonic series and if it's

convergent or divergent

so consider the series which starts from

1 and goes to infinity of the sequence 1

over n so if we list out the terms when

n is one it's just going to be one when

n is two it's one half

and then plus a third

plus a fourth

and plus one fifth and so forth

now if you keep adding

these terms you'll realize that the sum

will continue to increase it's going to

get bigger and bigger and bigger and

bigger it's not going to converge to a

single value if you just add it

and so just by doing that it tells you

that this series

is divergent

now if you're thinking about doing the

divergence test

to prove that it's divergent

it's not really going to work out too

well

so if we take the limit as n approaches

infinity for a sub n

so a sub n is one over n

this will equal zero

now for the divergence test to work

if it doesn't equal zero the limit

then

the series diverges but if it equals

zero

well it's inconclusive the series it may

converge

or it may

diverge

so the divergence test won't help us for

this example

but another test that we could use

is the integral test

now if we take the integral let's say

from

one to infinity

of f of x dx

if this integral is convergent

then

the series

the corresponding series

is also

convergent

likewise if the integral test fails if

it's divergent then the series will be

divergent

so the corresponding function for f of x

is going to be 1 over x

so let's integrate it from 1 to infinity

the anti-derivative of one over x is ln

x

and

we can't really use the infinity symbol

so i'm going to use some constant a with

the limit

as

a approaches infinity since we have an

improper integral

and so this is going to be equal to

the limit as a approaches infinity

the natural log of a

minus the natural log of 1.

so the limit as a approaches infinity of

ln a

what is that equal to

well the natural log of infinity is

going to be a large number that's still

going to be infinity

and ln 1 is zero

so the limit i mean not the limit but

the integral

of one over x from one to infinity is

divergent

because

we didn't get a finite answer

our answer is infinity

so because

according to the integral test the

integral from one to infinity is

divergent then the series must also be

divergent so that's another way in which

you could show

that the harmonic series is diverging

you

Harmonic Series

This calculus 2 video provides a basic introduction into the harmonic series.  It explains why the harmonic series diverges using the integral test for series.

Improper Integrals:                                    https://www.youtube.com/watch?v=ND9cEdfCFr0

Converging & Diverging Sequences:
https://www.youtube.com/watch?v=XdkoTb8PEG0

Monotonic & Bounded Sequences:
https://www.youtube.com/watch?v=tHy3TXmZpF0

Absolute Value Theorem - Sequences:
https://www.youtube.com/watch?v=kH5Zqg5j-3M

Squeeze Theorem - Sequences:
https://www.youtube.com/watch?v=21loY9MBK3A

________________________________
Geometric Series & Sequences:
https://www.youtube.com/watch?v=zRKZ0-kOUZM

Introduction to Series - Convergence:
https://www.youtube.com/watch?v=zg9N2gAf6a4

Divergence Test For Series:
https://www.youtube.com/watch?v=nC_IU1IzzS4

Harmonic Series:
https://www.youtube.com/watch?v=OqBZCycIYfw

Telescoping Series:
https://www.youtube.com/watch?v=XVkdhU6nJbo

__________________________________
Integral Test For Divergence:
https://www.youtube.com/watch?v=f9SJz4-UaQQ

Remainder Estimate - Integral Test:
https://www.youtube.com/watch?v=vaU5DDcTfZk

P-Series:
https://www.youtube.com/watch?v=x621rtIwb_4

Direct Comparison Test:
https://www.youtube.com/watch?v=oZtAgihok5s

Limit Comparison Test:
https://www.youtube.com/watch?v=LBxYQ0TJxYM

___________________________________
Calculus Final Exam and Video Playlists:
https://www.video-tutor.net/

Full-Length Videos and Worksheets:
https://www.patreon.com/MathScienceTutor/collections


now let's say if we have a sphere

with a radius of

what is the volume of the sphere

the volume of a sphere is four thirds

pi r cubed

so let's say this is ten inches it's

going to be four over three times pi

times 10 inches raised to the third

power

10 to the third that's 10 times 10 times

10 which is a thousand

and four times a thousand is four

thousand

so it's going to be four thousand over

three times pi

cubic inches

and so that's the volume of this

particular sphere

and let's see what the answer is as a

decimal

so this is equal to four thousand

hundred

cubic inches

so that's the rounded answer for volume

you

Volume of a Sphere

This geometry video tutorial explains how to calculate the volume of a sphere.

3D Shapes - Faces, Edges, & Vertices:     https://www.youtube.com/watch?v=Wmwe1fmR1SM

Rectangular Prism - Volume:
https://www.youtube.com/watch?v=rNIKbT0-ywU

Rectangular Prism - Surface Area:
https://www.youtube.com/watch?v=BsAQbJyEF9g

Rectangular Prism - Diagonal Length:
https://www.youtube.com/watch?v=iO7vb3HLHqU

Triangular Prism - Volume:
https://www.youtube.com/watch?v=HfDHYFgTbxk

Triangular Prism - Surface Area:
https://www.youtube.com/watch?v=152tLGR0lms

Triangular Prism - Lateral Area:
https://www.youtube.com/watch?v=PljECBkHQpg

_______________________________
Hexagonal Prism - Surface Area:
https://www.youtube.com/watch?v=JKyF2r-bIC0

Cube - Volume:
https://www.youtube.com/watch?v=bpRsuccb6dI

Cube - Surface Area:
https://www.youtube.com/watch?v=ZMhkvFwLOR4

Cube - Diagonal Length:
https://www.youtube.com/watch?v=wr3EvQZzoRI

Pyramid - Volume:
https://www.youtube.com/watch?v=8onylJ72ZPQ

Pyramid - Surface Area:
https://www.youtube.com/watch?v=2g6QuOizzac

________________________________
Square Pyramids - Volume:
https://www.youtube.com/watch?v=x8wEnG4GURQ

Cylinder - Surface Area:
https://www.youtube.com/watch?v=LdztqPczYiI

Sphere - Volume:
https://www.youtube.com/watch?v=eAwc8rwMRnI

Sphere - Surface Area:
https://www.youtube.com/watch?v=aXEcU-M_qNA

Cone - Volume:
https://www.youtube.com/watch?v=LypmeZ0tnJA

Cone - Surface Area:
https://www.youtube.com/watch?v=MeWoIcQNqRA

_________________________________
Final Exams and Video Playlists:
https://www.video-tutor.net/


now what if we have an absolute value

expression

with an inequality

what do we need to do

well we still need to write two

equations

the first one is going to remain the

same

now for the second one we need to do two

things

we need to change the positive five to a

negative five and

we need to change the direction of the

inequality

so that's important make sure you

understand that

and then after that just it

five plus one is six and then let's

divide by three

so x is greater than two

now for the other one negative five plus

one

is negative four and if we divided by

three we're going to get a fraction x is

less than negative four thirds

now because we're dealing with

inequalities we need to plot the

solution on a number line

so we need two negative four thirds is

about negative one point three which is

between negative one and two

so x

is greater than two

which means that we have an open circle

at two shaded towards the right and x is

less than negative four thirds which is

about negative one point three

so it's an open circle

shaded towards the left

now the solution

using interval

let's write it up here it's going to be

negative infinity

to negative 4 over 3

and then union

2 to infinity

so here's the 2 to infinity part

and here's the negative infinity to

negative four thirds

the open circle is at negative four

thirds just keep that in mind

let's try another example

feel free to pause the video and work on

it

now right now we can't

write two equations yet

we need to get the absolute value

expression on one side of the equation

first

so first let's subtract both sides by

so what we're going to have is negative

2 times the absolute value of 2x minus 1

is less than or equal to negative 10.

next

let's divide both sides

by negative two

so this will change the direction of the

inequality just keep that in mind

so the absolute value of two x minus one

is equal to or greater than

positive five

now since we have the absolute value

expression on one side by itself

with no other numbers outside of it

this is when we can write two equations

2x minus 1 is greater than or equal to 5

and 2x minus one is less than or equal

to negative five

don't forget to change the direction of

the inequality

so now let's add one

so two x is greater than or equal to six

and next let's divide by two so x is

equal to or greater than 3.

here we're going to add 1

so 2x is less than or equal to negative

4 and then divide by 2.

so x is less than or equal to negative

so we have x is less than or equal to

negative two and x is equal to a greater

than three

so here's zero

here's negative two

and here's three

so x is equal to or greater than three

so we have a closed circle by three

shaded towards the right

x is equal to or less than negative two

so we have a closed circle shaded

towards left

so therefore the solution

is negative infinity to negative 2 with

a bracket

union

3 to infinity

How To Solve Absolute Value Inequalities, Basic Introduction, Algebra

This Algebra video tutorial provides a basic introduction into absolute value inequalities.  it explains how to solve absolute value inequalities and graphing the solution using a number line and using interval notation.  This tutorial contains many examples and practice problems.

Plotting Inequalities on a Number Line:
https://www.youtube.com/watch?v=KkiYqww4eg0

How To Solve Linear Inequalities:
https://www.youtube.com/watch?v=rIl2USa8XPY

Solving Absolute Value Inequalities:
https://www.youtube.com/watch?v=ZJvOGkykJ44

Graphing Linear Inequalities In 2 Variables:
https://www.youtube.com/watch?v=cskHdgwB5k8

Graphing Systems of Linear Inequalities:
https://www.youtube.com/watch?v=FWbcKade3rw

__________________________________
How To Solve Compound Inequalities:
https://www.youtube.com/watch?v=msiceXvwriE

Quadratic Inequalities:
https://www.youtube.com/watch?v=_gWjLKsFOPE

Polynomial Inequalities:
https://www.youtube.com/watch?v=Fd5ys4PQ-aM

Rational Inequalities:
https://www.youtube.com/watch?v=gfnVHwhEe6U

SAT Math - Inequalities:

_______________________________________
Final Exams and Video Playlists:
https://www.video-tutor.net/

Full-Length Videos and Worksheets:
https://www.patreon.com/MathScienceTutor/collections


in this lesson

instead of expanding a single log into

multiple logarithms

we're going to condense

multiple logs into a single log

so this is the opposite of the last

lesson

so hopefully you've watched the last

lesson before this one

so we're going to write this as a single

log

now the ones that contain a positive

sign

will go on top so that's a and c so it's

going to be a times c

on top

now b has a negative in front of it so

it's going to go on the bottom

so it's going to be log

ac over b

let's try another example

2 log a

plus

5 log b

minus 7 log c

so how can we express

this expression as a single log

the first thing we need to do

is move the coefficient

to the exponent position

so this is going to be log

a squared plus

log b to the fifth

minus

log c to the seventh

now we can write a single log

a and b both have a positive sign in

front so they're going to be on top a

squared b to the fifth

c has a negative sign in front of it

so c to the seventh is going to be on

the bottom

therefore this is the answer

try this one

one third

log a

minus two thirds

log b

plus

one fourth

log c

minus four over five

log d

so first so let's move all of the

coefficients

to the exponent position

so this is equal to log a

raised to the 1 3

minus log b

to the two thirds plus

log c

to the one fourth

minus log d

to the four fifths

and now let's write it as a single log

so a and c both have a positive sign

so it's going to be a to the one-third

c to the one-fourth on top

now b and d both contain a negative sign

in front of the log

so they're going to go on the bottom b

to the two thirds

and then d to the four fifths

so you can leave your answer like this

or you can convert it back

into radical form

so you can write it like this as well

the cube root of a

times the fourth root of c

divided by the cube root of b squared

times the fifth root of d to the fourth

you

Condensing Logarithmic Expressions

This algebra video tutorial explains how to condense logarithmic expressions into a single logarithm using properties of logarithmic functions.

Logarithms - Free Formula Sheet:  https://bit.ly/4fEwqv0

_______________________________
Logarithms - The Easy Way!
https://www.youtube.com/watch?v=kqVpPSzkTYA

Log to Exponential Form:
https://www.youtube.com/watch?v=f0C1KL7GkqY

Change of Base Formula:
https://www.youtube.com/watch?v=FFm-zaFW_X4

Change of Base Log Problem:
https://www.youtube.com/watch?v=p7hD9VdXv9U

Properties of Logarithms:
https://www.youtube.com/watch?v=Jtv9Lnf7Zw8

____________________________________
Expanding Logarithmic Expressions:
https://www.youtube.com/watch?v=OIz-5MyJA3g

Condensing Logarithmic Expressions:
https://www.youtube.com/watch?v=luRrOlsB4cY

Natural Logarithms:
https://www.youtube.com/watch?v=daUlTsnCNRQ

Solving Exponential Equations:
https://www.youtube.com/watch?v=9tutJ5xrRwg

Exponential Equations - Quadratic Form:
https://www.youtube.com/watch?v=yNgmVu0R_T8

_______________________________________
Solving Logarithmic Equations:
https://www.youtube.com/watch?v=fnhFneOz6n8

Graphing Logarithmic Functions:
https://www.youtube.com/watch?v=-nptxS9rZNA

Graphing Exponential Functions:
https://www.youtube.com/watch?v=DASfP8KAyvs

Compound Interest Word Problems:
https://www.youtube.com/watch?v=Hn0eLcOSQGw

Logarithms Practice Problems:
https://www.youtube.com/watch?v=7DVbQKI600k

_______________________________________
Final Exams and Video Playlists:
https://www.video-tutor.net/

Full-Length Videos and Worksheets:
https://www.patreon.com/MathScienceTutor/collections


I would say it would move from here to

here you're right is that right okay so

yes the tail does have to go from

whatever electrons we're moving and then

from there that's going to move towards

the bond a lot of times students want to

point this to the positive charge but

it's not going there necessarily like we

want to point this to then show that

there's going to now be a double bond at

this end that's correct next let's draw

the resonance structure and then my

positive charge would be here now yes

that is exactly correct

Drawing resonance structures when there’s a positive charge #organicchemistry #resonance


alright here I'm going to do another

example of solving a compound inequality

and then we'll write the solution and

interval notation so here it says

negative 4 has to be less than 3w plus 5

which has to be less than 12 when we

write this in one fell swoop really what

they mean is negative 4 has to be less

than 3w plus 5 and also 3w plus 5 has to

be less than 12 okay so it's really just

a shorthand way of writing both of these

statements at once ok so let me go back

to the original problem here so just

something that you should think about

what the heck do they mean here so again

another way to think about it is you

know we're going to pick some number W

multiplied by 3 add 5 at the end of the

day it's got to be between negative 4

and 12 those are the numbers that we are

looking for the way that you do this

again you know imagine if there was only

one inequality here you know negative 4

less than 3w + 5 I mean to me the first

thing I would do would be to subtract 5

from both sides I would get negative 4

less than 5 which is negative 9 and then

I would have the 3w left over likewise

if we just had you know 3w plus 5 less

than 12 you know to solve that

inequality I would do the exact same

thing which is subtract 5 so typically

you know what you'll I'm going to do it

all at once I think you know I've got

this negative for less than 3w plus 5

less than 12 all when I see that I'm

trying to get the W by itself in the

middle so to me it says just subtract 5

from everybody so on the left we'll have

negative 9 we'll be left with 3w in the

middle let's see 12 minus 5 would be 7

and again the same thing I would do here

if only one of the inequalities were

there I would divide everything by 3 and

again that's what I'm going to do

everywhere I'm just going to divide

everything by 3 so negative 9 over 3 is

going to be negative 3

so it says we'll get W in the middle

just one W and then less than 7/3 so

basically it says the numbers that work

it says W has to be strictly greater

than negative 3 but less than 7/3 so if

we want to write that again I'm going to

make my little number line I always like

to do that first so negative 3 we don't

use it so we make an open circle and it

says 7/3 that's something positive over

there I'm going to put an open circle

sometimes people always have a tendency

to want to put 0 in the number line I

think is just a point of reference I

just put the numbers that I'm using

because if I put 0 down there to me that

thinks maybe I you know I have to use

that in my interval somewhere and I

don't so I just put these numbers down

it says it has to be bigger than

negative 3 but smaller than 7/3 so the

numbers in between there would be the

numbers that work so to write our

solution and interval notation okay so

if you kind of think about the left

number that's going to be the number

that goes on the left with interval

notation since it's less than we don't

include it so we just use our

parentheses we put a comma and then it

says we go up to 7/3 and again we use a

parenthesis because we don't include it

okay so this is the interval notation to

describe this set of solutions

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving Linear Compound Inequalities - Example 2


all right so here we're going to look at

one more example using the order of

operations

and here we're going to evaluate the

expression negative x minus y squared

all divided by x times y minus the

quantity 3 minus x

but what we're going to do is we're

going to substitute in x equals negative

and y equals positive 3. so

anytime i substitute things in i usually

put

it in parentheses okay so what i mean is

we've got

a negative well x is negative 2. so

notice i'm

replacing i'm putting the x inside of

parentheses

minus the same thing i've got y squared

but again y had the value of positive

three

x times y so x is negative two

times three minus three

minus and again another set of

parentheses negative two

and now i've replaced everything

and this looks uh certainly

a little obnoxious and i guess it is but

it shouldn't be too bad

so i've got multiplication i've got

negative one times negative two that's

going to give me a positive two

but next i'm gonna have to do my

exponents okay so we've got

minus three squared well we do exponents

first so

i'll just leave the minus sign in there

3 squared is 3 times 3 which is

9. in the denominator i see

multiplication

i see subtraction i see some parentheses

i'm going to do negative 2 times 3 which

will be negative 6.

there's my minus sign inside the

parentheses we have three

minus negative two well that's the same

thing as three

plus two which is going to give us five

and now we're getting pretty close 2

minus 9

that's going to be negative 7. so again

i'm just thinking 9 minus 2

but changing you know the changing it to

negative

so 2 minus 9 is negative 7 i see

negative 6

minus five negative six minus

five is going to be negative eleven well

the last thing we can do

a negative divided by a negative is a

positive

so we'll be left with the value of seven

over eleven

The Order of Operations, Example 3

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Order of Operations, Example 3.  Here we look at using the order of operations to evaluate some expressions!


okay in this video we're just going to

do some more derivatives involving the

power rule so in Part A here we've got P

of Q equals negative 3 over pi times the

square root of Q to the 4.1 power and

all I'm gonna do in Part A is I'm just

gonna rewrite this first I'm gonna

rewrite it and again any time I see a

square root or a radical in there the

first thing I always do is just try to

rewrite this with no radical just as a

power so we can write this as P of Q

equals negative 3 over PI and don't be

thrown off by the negative 3 over pi

that's just a constant so nothing's

really gonna happen to it when we take

the derivative we can just kind of take

the derivative of the stuff involving

the variable and then multiply that at

the end so we've got Q raised to the 4

point 1 recall a square root is

equivalent to raising something to the

1/2 power

well again I'm just going to keep

simplifying this so we have negative 3

over pi recall if things are in

parentheses we just multiply the

exponents so we would have Q to the 4.1

times one-half or equivalently 4.1 over

2 so that's negative 3 over PI and then

if we simplify we would have well 4.1

divided by 2 is going to be two point

zero five and now I'm kind of in a good

place where I think well now I'm finally

going to take the derivative so P prime

of Q the negative 3 over pi just kind of

gets carried along and we'll just do the

same thing as before we've got a

variable to a number okay so that in

that number that exponent just comes out

front times our variable Q and then we

do the same thing as before we just take

one away from this number so if we

subtract one from 2.05 we'll just get

one point zero five and if you wanted to

I mean you could always multiply you

could think about this as being over 1 I

guess we could take negative three times

two point zero five we could make that

negative six point one five the

you race to the 1.05 would also go in

the numerator and then we would have our

pie in the denominator so if you wanted

to kind of write it all back together it

certainly would be okay to write it like

this but that's all there is to it so

again you'll see derivative problems all

the time that involve square roots cube

roots fourth roots other types of roots

always first thing I do is just

immediately rewrite those as exponents

and if I can I simplify it before I

start taking the derivative

Power Rule and Derivatives, A Basic Example #1

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Power Rule and Derivatives, A Basic Example. In this video I use the power rule to find the derivative of a function.  There is a touch of algebra involved, but nothing too heavy! : )


all right so one more example here using

alternating series so here we've got1 to

the N

+1 all divided by the square < TK of n^2

+ 6 N -

2 so in this case uh again if we get rid

of the the alternating the 1 to the n +

1 you can think about there as being a

times one on top so my B subn is going

to be 1 over the square Ro T of n^ 2 + 6

N - 2 so again same two questions um is

it

decreasing and does the limit go to

zero I think this question is

uh easy to address because as n goes to

Infinity we're certainly going to get a

very large number in the denominator

we'll have one over a large

number and that will get arbitrarily

close to zero so that condition

satisfied to do decreasing again you

know our series I think we started it at

nals

one you can always again just check

values so let's see if we plug in one I

think you know this one it's pretty

clear if you plug in one you'll get 1 +

6 that's 7 7 - 2 is going to be

5 when we plug in two well 2^ 2 is 4 6 *

4 or excuse me 6 * 2 is 12 so we have 12

+ 4 12 + 4 is 16 - 2 that's going to

give us the sare < TK of 14 and then you

know you can keep plugging in values and

you'll see that you know n^2 + 6 nus 2

larger and larger and larger so

absolutely it looks like our our our B

subn is decreasing so since both

conditions are satisfied we would say

that this alternating series converges

Alternating Series - Another Example 3

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Alternating Series - Another Example 3.  Just another example showing an alternating series converges or diverges.


okay in this example we're going to find

the domain of the function x / x^2 - 10

x -

24 so the only restrictions the only

values of X that we'll have to Omit will

be those values of X that make the

denominator equal to zero so I'm going

to find those values so I'm going to

take the denominator x^2 - 10 x - 24 and

set that equal to zero and to solve this

we've got a quadratic equation I'm going

to try to

factor so in this case again we want two

numbers that multiply to

-4 but add up to -10 so let's see uh

let's see if any if we can find any that

work um I think some combination of 2

and

12 right that's going to multiply to 24

and I think let's see if we make the the

12 negative and the two positive those

two numbers will multiply to -4 but add

up to -10 so that's a correct

factorization and then we just take each

factor set it equal to Z and solve so

from the first Factor we'll get that x =

-2 is a solution to our quadratic and

then we'll get x - 12 also equal 0 well

we can add 12 to both

sides so the domain

it'll be all values of X such that X

does not equal -2 and also X does not

equal

pos2 and you could also certainly write

this using interval notation um but

however you prefer to write it

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding the Domain of a Rational Function / Fraction with Variable in Denominator.
In this example, I find the domain of a rational function by factoring the denominator.


okay in an act of uh Shameless

self-promotion here uh I'm thinking

about making some T-shirts after people

can consistently keep uh asking well not

consistently periodically often enough

so um basically got a couple t-shirt

designs uh I'm just kind of wondering if

anybody's actually interested so um I've

thought about doing this for a while so

I've got a couple different ideas you

can vote for your favorite t-shirt idea

I'll put a link too somewhere in the

comments at just math tutoring

blogspot.com so a couple different ideas

so the first idea is iart Patrick JMT

right a little less than three uh so

iHeart Patrick JMT um okay so again very

uh self-gratifying

here um the classic smart equals sexy

okay so would you wear a shirt that says

that I would um I would wear either one

of them um number three girls love a guy

with a big dot dot dot brain okay so hey

a little racy right how exciting um last

but not least hey the classic the Geeks

shell inherit the earth um so if you go

there I think I've got those listed if

you've got a great idea for a shirt feel

free to post it as a comment uh maybe

I'll add it into the survey I think I

can still add things into the survey um

and uh who knows maybe your your uh

T-shirt design will be will be picked if

it is and I make them I'll definitely

send you one for free so

um you can even go on there and say hey

I would never ever buy one of your

shirts so that's useful information to

you so uh just a little idea tell me

what you think um be uh brutally honest

it's not going to hurt my feelings so

all right everybody um I know this is

the last week of uh finals finals week

for a lot of you um those of you in high

school still have a couple weeks I know

so hang in there it's almost over hope

your semester's going well it's been

busy for me got a few more days be happy

when it's over so all right um take care

I appreciate your feedback out there

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okay suppose we know that the sum of

three consecutive even integers is 180

we want to find the largest of those

three

integers so the idea is we've got three

even integers we don't know what they

are so I'm going to let X denote the

smallest even

integer well to get to the next even

integer we would just add two to the

original number well to get to the next

even integer after that we would have to

add yet another two so we would get x +

2 + 2 or x + 4 well we know that if we

add these

up if we add these three numbers

up we get

180 and that's now going to be the

equation that we have to solve you you

can certainly just remove the

parenthesis here and combine like terms

so we would have X Plus X Plus X which

is 3x we would have 2 +

4 which is

six that equals 180 we can subtract six

from both

sides we'll have 3x =

174 and now the last thing we have to do

is divide both sides by

three so 3 divid 3 will will just leave

us with

1X let's see uh 174 / 3 that looks to me

like that's going to give us uh

58 okay so x equals 58 so we have to be

careful right that's not our the

solution to the question uh but we now

know what the smallest integer is uh we

know that it is equal to 58 well that

means the next one would have to be 60

and then the one after that would have

to be

62 so the solution to our our question

here when we want to find the largest of

the three integers we would say we would

get

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Linear Equation, Word Problem #2 : Consecutive Even Integers.
 Here, we are given that three consecutive even integers add up to 180.  We find the largest of these three integers using a linear equation.


okay I mentioned this in the uh last

part of one of my videos but uh I was

going to see if uh anyone was interested

in doing a little video response problem

and uh what this has to do with changing

the order of Integrations for triple

integral so if you want to uh uh show

how to work this problem I would love if

you would uh like to post a video

response I'll absolutely uh let people

see it this way too again you know I

would encourage if you're having issues

with this stuff practice this you know

hopefully you watched the other videos

they were helped a little bit practice

this example on your own um you know

again what we want to do is rewrite

it using the five other orders of

integration so my advice to you um if

you're having issues with it you know

practice it on your own check out the

video responses see where you went wrong

if uh if you feel like it's finally

starting to click for you help others I

think you know know again um there's no

better way to learn something than to

try to explain it to someone I think uh

it really uh verbalizing things makes

you really um

uh I guess uh uh it makes you really

again I look I can't say at it I'm

having a hard time it makes you really

clarify your thoughts I guess right um

you know if you can explain something

simply you probably don't know it well

enough I think uh Einstein said

something like that that so sorry to

throw the Einstein quotes out there but

uh I think it's a good one so anyway

what we're doing is we've got this

little uh three-dimensional object that

we're integrating over kind of like a

little almost like a little uh wedge of

cheese here or something so kind of the

uh sort of the roof and the front part

of this uh um of this little wedge of

cheese is the equation zal 1 - x^2 the

side of it can be represented by the

plane yal 1 - x um

the back of it looks like the YZ plane

the side of it is just the XZ plane they

uh give you one correct way of setting

up the uh the orders of integration so

again uh if you're interested try to

show people how to switch these orders

of integration do Dy dxdz and then do

the other ones first do X first and then

uh do Z first so if you uh if you're

interested that'd be great if uh if

people like it maybe I can uh send you a

free shirt or uh you know do something

to uh uh um I don't know to uh to reward

your hard work U definitely would love

to put it up on my channel as a featured

video for for at least a few days so if

you're interested great if you need some

uh pointers or some hints um post

comments and uh you know try to help

each other out so all right um good luck

and uh I don't know hopefully uh

hopefully somebody will uh will will

post some good some good Solutions

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Video Response Challenge for Triple Integrals!
Want to show your stuff? Or maybe you are just bored?! :)
Post a video response about how to change the order of integration on the given problem!  Show how to set up the other 5 orders of integration!


all right in this video I want to solve

another quadratic equation by factoring

so here we want to solve the quantity y

plus 6 squared equals 81 and for those

of you who have seen the square root

property and you're saying hey Patrick

the square root property is much easier

to use I say I absolutely agree but

again you know we're just trying to

illustrate factoring I don't you know

the procedure is what's important here

so okay in this case the first thing I'm

going to do since I'm going to try to

factor it again we want one side of our

quadratic equation to equal zero

so I'm going to subtract 81 from both

sides so well we've just got the

quantity y plus 6 squared now minus 81

equals 0 there's two things you could do

here you could now oil out the Y plus 6

you know take Y plus 6 times y plus 6

collect your like terms

subtract 81 and then kind of try to

refactor that but to me when I see this

this is really a difference of perfect

squares I see something squared and

there's my minus sign and then I see

something else squared namely 9 squared

ok and remember a squared minus B

squared that factors nicely so a squared

minus B squared is just a minus B a plus

B so that's what I'm gonna do in this

case so to me it says you're gonna get

okay so my a is the Y plus 6 so you get

y plus 6 minus B and again the number

being squared is 9

and then we multiply that by basically

the exact same thing but we just put a

plus sign in the middle so y plus six

plus nine equals zero and now inside the

parentheses we could really think about

there's being a positive one so if you

distribute that it would just get rid of

the parentheses on the y plus six so we

have y plus six minus 9 which is going

to be Y minus three and then we have Y

plus six plus nine which is going to be

Y plus 15 equals zero and now that we

have it nice and factored we just do the

same thing we set the first factor equal

to zero we set the second factor equal

to zero I'll add three to both sides on

the first equation and get y equals

three and I'll subtract fifteen on my

second one to get y equals negative

fifteen and now we've got our two

solutions and notice it's pretty easy to

check these if you plug three in you

would get three plus six or nine squared

which is certainly 81 if you plug

negative fifteen and you would get

negative fifteen plus six which would

give you negative nine

but that squared is still equal to 81

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving Quadratic Equations by Factoring - Another Example


okay so this is another SAT question

taken from the Princeton Review says

which of the following points is not on

the graph of absolute value of negative

x minus the absolute value of negative y

equals one and then they give you a

bunch of points this again is another

very typical you know don't don't try to

graph this that's not at all what you

want to do I mean you could I guess you

could do a couple simplifications you

know basically just start plugging

things in that's what I would do so

let's see it says again we have to make

sure we're finding which one is not on

the graph so if we plug negative 2 in

the absolute if we take negative

negative 2 will get positive 2 minus if

we plug in negative 3 we'll get positive

3 well does that equal 1 well we get 2

minus 3 which is negative 1 so hey 1

negative 1 does not equal 1 so tada

there's our answer negative 2 negative 3

would be the solution you can check the

others make sure that they work but this

is one that will not be on the graph

SAT Math Question #7

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! A sample math SAT question taken from the Princeton Review.


okay in this video I just want to do a

few examples of simplifying some

Expressions involving absolute value so

I've got three examples um and let's do

uh the first one first um so we're going

to take the absolute value of the of 8 -

12 and again on these problems typically

you know I just try to work inside out

so um if we do 8 - 12 we get -4 and then

remember if we take the absolute value

of uh a number negative a negative

number it simply turns into the positive

of that number so the absolute value of

-4 is simply four and again be careful

um because again a very one of the

classic mistakes you don't want to do

this um it would be easy to make this

the absolute value of 8 minus the

absolute value of 12 which would be 8 -

12 which would be -4 which is not

correct okay so do not do any of these

steps here in Red so don't you can't

break up the absolute value when you're

adding and subtracting you have to be

real careful about that okay so just a

little uh word of warning I'm sure I've

done it myself uh more than

once okay so in the next one here we

have the absolute value of -4 time the

absolute value of three and again I

think well what's the absolute value of

-4 that's simply pos4 the absolute value

of positive3 is just itself positive3

and then we get 4 * 3 and that gives us

our solution of

12 and in the last one here the same

thing we have the absolute value of -25

over 5 uh we could simplify inside First

- 255 over 5 and make that ne5 and the

absolute value of5 is pos5 another way

to do this uh just to illustrate some of

the properties remember if you have a

fraction inside of absolute value we can

write that as the absolute value of the

numerator over the absolute value of the

denominator and the absolute value of -

255 would simply be posi 25 the absolute

value of five is just five and if we

divide 25 over 5 again gives us our

solution of positive five so um and just

one little you know one little thing to

point out if you accidentally did this

mistake you know back in the first

example where you got a negative number

to me you know I'm taking the absolute

value of something uh for example in

this first problem the absolute value of

anything either has to be zero or a

positive number so at the end if I did

get some negative number out I I I I

would just recognize that something had

to be wrong there and uh maybe I would

you know figure out that this was the

incorrect step so just think about that

too when you're doing absolute value so

um all right I hope these few uh

examples help

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Evaluating Expressions Involving Absolute Value - Example 1
Just a few very basic problems involving simplifications with absolute value.


okay here we're going to solve one last

equation where we have variables on both

sides here we're going to solve 6 minus

x over 5 equals 3 times the quantity X

plus 2 over 10 so we could start you

know distributing out in the numerator

we could break up the fraction we could

do lots of things here typically I think

the easiest thing to do if you have

fractions a lot of times it's it's

easiest just to try to multiply both

sides of the equation by something that

will get rid of the fractions so in this

case if we think about the least common

multiple of 1 5 and 10 the least common

multiple of 1 5 and 10 would be 10 so

what's the smallest number that's

divisible by 10 5 and 1 well I'm going

to multiply both sides by that number

again 10 in this case so on the left

side if we distribute well 10 times 6

will be 60 I'll write it out this is 10

over 1 we would get 10 x over 5 but

again 10 x over 5 that's just going to

be 2x right so we're getting rid of the

fractions by multiplying by 10 so on the

right side the tens will just cancel out

and we'll have 3 times the quantity X

plus 2 I'm going to go ahead and

distribute the 3 so we'll have 3 times X

which is 3x 3 times 2 will be 6 and now

it's a matter of putting all the terms

involving X on one side and all the

constants on the other so the first

thing I'm going to do to both sides is

add 2x again you could move the

constants around first as well it

doesn't matter on the left side we'll be

left with 60 3x plus 2x is 5x and again

I chose to add 2x I could have

subtracted 3x but again I just kind of

observe that by adding 2x I'm going to

keep everything positive

to me I like just having things positive

it keeps me from making sign mistakes

but again nothing says you have to do it

that way

so 60 equals 5x plus 6 will subtract 6

from both sides 60 minus 6 will be 54

equals 5x and the last thing we'll do

here since we're multiplying by five is

we'll divide both sides by five and then

we'll have our solution x equals 54 over

five that doesn't reduce so I'm going to

leave our solution just like that as a

fraction

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving a Linear Equation with Variables on Both Sides of the Equation, Ex 3


okay suppose that the graph of a linear

equation has the following values it

says if we put in x equals zero we get y

equals one if we put in x equals one we

get y equals negative one and if we

substitute in x equals three we get y

equals negative five we want to know

what's the slope of the line is it

negative 4 negative 2 positive 2 or

positive 4 so this again is just kind of

an application of using that slope

formula M equals y2 minus y1 over x2

minus x1 and to use this formula all you

need you only need two points notice

we've actually been given three points

here we've got the point zero comma one

we've got the point 1 comma negative one

and then we also have the point 3 comma

negative 5 we can use any two of these

that we that we wish so I don't know it

doesn't really matter I'll use the 1

comma negative 1 and the 3 comma

negative 5 for no particular reason just

picking two at random so okay so there's

going to be my x1 there that's going to

be my value of y1

there's x2 and there is y2 okay so let's

do the change in Y so we'll do negative

5 minus negative 1 so I'm going to put

that in parentheses just to make it look

a little cleaner here and then we've got

3 minus positive 1 well negative 5 minus

negative 1 a negative and a negative

will make a positive well in the

denominator 3 minus 1 that's going to be

positive 2 negative 5 plus 1 that's

negative 4 and a negative divided by a

positive that's going to be a negative

number 4 divided by 2 is equal to 2

that's going to be the slope of our line

negative two and you can check it

doesn't matter you know you could have

used the first point in the second point

or the first point in the third point

and have went through the same you know

calculation and you would have found in

all cases that the slope is equal to

negative two so it doesn't matter you

know which which of those two points

that you pick

GED 'Most Missed' Math Practice Problems , Code A.5.b,  #3

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The GED Testing Service keeps a list of the most missed math topics and can be seen here: http://www.gedtestingservice.com/uploads/files/126de0283c94ff6323b3b5b3f310da5d.pdf
This video shows how to work a sample problem using their coding system that is shown on the left using the link above.


okay here we're going to evaluate some

factorials uh just to get a feel for

factorials since you have to use

factorials when you calculate

permutations and

combinations well five factorial that's

just shorthand for 5 * 4 * 3 * 2 * 1 so

let's see 5 * 4 is going to be 20 3 * 2

is going to be 6 so if we multiply 20 *

6 we get 1 120 so 5 factorial is equal

to

120 um factorials get very large very

quick you know five factorial is already

bounced up to

120 suppose we want to do four time five

factorial well you need to do the

factorial first or you know it

definitely doesn't hurt to write it all

the way out so this is four * again well

5 * 4 * 3 * 2 *

well we've already seen that 5 factorial

is

120 so 4 * 120 if we simplify that we

get

480 so 4 * 5 factorial is 480 uh do not

multiply you know something uh that that

might be tempting to do that's

definitely not

correct don't say it's 20 factorial you

know don't somehow multiply the four and

the five 20 Factor is going to be a huge

big number definitely not going to be

480 so do not do it that way if you're

not sure whether or not you can sort of

do a process just write it out and then

you'll sort of see everything you know

written out and you can just

multiply okay so let's see um in part C

we've got six factorial so that's 6 * 5

* 4 * 3 * 2 *

1us we've got 7 - 3 factorial so I'm

going to going to uh simplify 7 - 3 is 4

factorial over 3 factorial well let's

see we can always uh simplify this we've

seen that 5 * 4 * 3 * 2 * 1 that's equal

to

120 so if we take 6 * 120 uh 6 * 120 is

going to be 720 so again if you multiply

those all out you'll get 720 you could

calculate four factorial calculate 3

factorial and then uh you know simpli it

uh notice though 4 factorial is 4 * 3 *

2 *

1 3 factorial is just 3 * 2 * one so if

we cancel out those common factors

really we're just going to be left with

a four in the numerator so we have

720 minus 4 if we simplify that we get

seven uh

Factorials ; Evaluating Expressions With Factorials

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Just a few basic examples of evaluating numerical expressions involving factorials.


all right here we're going to look at

some examples related to subtraction of

integers and what we're going to do is

we're just going to make some number

lines to illustrate these subtraction

equations so in part A we've got 2 - 3 =

-1 well in terms of our number line the

first number we have is positive two so

I'm going to go two units to the right

and then we're subtracting away positive

three units so you can think about again

subtracting a positive as going to the

left so we'll go left one 2 three units

and again if we if we subtract three

units um we'll end up at the value of

negative one so that could be our number

line two to the right three to the

left part B is a little tricky here a

little bit trickier just because of uh

you there's a lot of minus signs

floating around so the first thing I'm

going to do is -2 so that's going to

move us two units to the

left so there's our two units to the

left now if we were subtracting away a

positive five we would keep going to the

left but since we're subtracting a Nega

five what we're going to do is now move

to the right so let's see we're at -2

then we move one unit two units three

units four units and five units and

again you can count this would 0 1 2 and

this would put us at posi3 so that would

be a diagram that goes with or a number

line that goes with Part B

Subtracting / Subtraction of Integers

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Subtracting / Subtraction of Integers.
Here we look at some examples of subtracting integers. In these examples, we create a subtraction statement from a number line.


suppose a fountain has a circular base

with a diameter of 12 ft we want to know

which of the following Expressions could

be used to find the area of the bottom

surface of the

fountain so this is just an area

question here so again let me draw a

picture there's our Fountain if we were

looking at it from above we said that it

has a diameter of 12T okay so a diameter

of 12 ft that means the distance across

is

12 or equivalently

if the diameter is 12T that means that

it has a radius of 6t so we're just

wanting to find that that area of the

fountain well the area of a circle

recall that that's Pi * the radius

squared well we're going to approximate

the radius or excuse me we're going to

approximate Pi by using

3.14 okay so we'll approximate Pi by

using 3.14 we know the radius exactly

right we said the diameter is 12 so half

of that is going to be six so our radius

is going to have a value of six so we'll

just take

3.14 and we would multiply that by 6^

squared and that would give us our

solution so this is just one of those

questions it's just a matter of really

knowing uh knowing the formula so in

this case our answer let's see so 3.14 *

6^ 2 it looks like answer Choice B

would be our solution we've got 3.14 *

6^ s and again if I saw a question like

this on the test and I wasn't really

sure if you at least you know I don't

know I mean if you knew the area formula

again the thing that gets squared in the

area formula is the radius so if I saw

this 3.14 squ and this 3.14 squ I would

know that C and D are not correct um and

it would be one of the two choices A and

B so at least at that point I would have

it narrowed down to two possible

solutions

GED 'Most Missed' Math Practice Problems , Code Q.4.b,  #3


all right in this example we're going to

start uh looking at some derivatives of

the inverse trig functions so part A

here we're going to find the derivative

of x * AR cosine of < TK X so recall the

derivative of Arc cosine of some

function what we get is we get negative

uh 1 over theun of 1 minus the

function squar then we have to do the

chain Rule and take the derivative of

that function so uh we'll definitely

have to use uh this formula and we're

also going to have to use the product uh

the product rule because hey well we've

got a product so uh we'll definitely

come back to this formula so I'm going

to use it in just a

second so maybe we can rewrite this real

quick we've got y = x * AR cosine of x

to the

12 well okay the derivative let's see so

the derivative of 1x that's easy we'll

just get one so we'll leave the uh Arc

cosine of the X to the 12 or the square

otk of X we'll leave that all alone plus

and now we'll leave the X alone when we

take the derivative of AR cosine of x to

the 12 this is where we'll use our

formula so we'll get -1 over the square

< TK of 1 minus the function which is x^

12 all of that's going to get squared

and then we have to multiply by the

derivative of the inside

function so that'll give us 12 x raised

to thega

one2 so I'm going to try to clean this

up a little bit so we've got AR cosine

of squ < TK of X let's see we've got a

positive and a negative so that would

give us a negative um the X could go on

top um the two would be in the

denominator um underneath the square

root we would just be left with 1us x uh

when we take x to the 1/2 and square it

and then our x to the -

one2 um you know or excuse me I think

it's our x to the negative one2 uh you

know we could even take our x to the 1

and multiply that by x to the one2 and

that would leave us with X to the

positive 1/2 again we're just adding the

exponents so really this would be an x

to the positive2 which we can rewrite

simply as the square root of x so you

know I don't see a lot lot of other good

meaningful ways to combine this you

could write this as a single square root

the square root of x over 1 - x but I

say let's leave it right there and that

would be our derivative so you know all

the same rules still apply chain rule

quotient rule product rule you just have

kind of these new formulas now you know

just these new formulas now that you're

going to have to remember and uh just

have to have these uh you know kind of

ready uh just to uh they're kind of easy

to forget they're very similar to some

of you know like a the inverse sign is

very similar to this um so again just

kind of new formulas you got to remember

and uh be ready to use

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Derivatives of Inverse Trigonometric Functions - Another Example!.  Just another example involving a derivative of an inverse trig function.


okay in this video we're gonna do

another example related to the law of

cosines and what we want to do here is

just find the measure of the smallest

angle in our given triangle here so we

have side lengths of four six and nine

the first observation is the smallest

angle is going to be opposite the side

of the triangle that has shortest length

so the side with shortest length would

be our side here a B that has length 4

so that means again we're trying to

figure out the angle here the the

measure for angle C well we can use law

of cosines so 4 squared that would equal

9 squared plus 6 squared minus 2 times

well 9 and 6 and then that's going to

help us bring in our cosine of the angle

C so now just a little bit of rithmetic

4 squared 16 9 squared - 81 81 plus 36

is going to give us 117 2 times 9 times

6 let's see so 9 times 6 is 54 if you

double that you'll get 108 which will be

cosine of C

we can subtract the 117 from both sides

that's gonna leave us with negative what

I want on the left equals negative 108

times cosine of C we can divide both

sides by the negative 108 so that'll

just leave us with positive 101 over 108

equals cosine of C squeeze this in here

101 over 108 I'm getting that to be

point 9 3 5 equals cosine of C and now

if we just take the inverse cosine of

both sides that's going to give us our

angle C and let's see here I think this

should be let me double check one more

time so point nine three five if we do

the inverse cosine of that I'm getting

this to be twenty point seven seven

degrees after rounding here a little bit

so that's all there is to it

and again you know we need the four

isolated at the beginning on the left

side again because that's what's going

to give us the cosine of this angle that

we want here so nothing worse than that

just again just noticing that the

smallest angle will be opposite that

side of length 4 using law of cosines

just being a little careful with the

arithmetic and that's all there is to it

❖ The Law of Cosines - SSS Example ❖

In this video we look at an example involving the Law of Cosines.
In this triangle we are given the 3 sides of the triangle and asked to find the measure of the smallest angle.  In order to solve the question we merely use the Law of Cosines.
This is just a quick example to refresh or quiz yourself!


okay in this example we've got a right

triangle with side lengths Three Square

7 and then a hypotenuse with length four

we're going to find the S cosine tangent

cosecant secant and cent of that angle

Theta so I've jotted down the

definitions quickly here um so you can

take a look at them and again notice my

triangle the Theta is in a uh a

different corner so be careful about

that when you read off the angles or the

side lengths from the angle and again s

of theta is defined to be the the length

of the opposite side divided by the

length of the hypotenuse cosin's the

adjacent over the hypotenuse tangent the

opposite over adjacent um Etc there's

the other one so really to compute these

it's just a matter it's just a matter of

knowing those definitions

and maybe you have to rationalize a

denominator at some point so s of theta

okay that's the opposite side divided by

the hypotenuse again the length of the

opposite side well in this case the

opposite side it's it's the

side furthest away from our angle Theta

the opposite side is going to be theun

of

7 and then we divide that by the length

of the hypotenuse which in this case

will be

pos4 cosine of

theta well that's the adjacent side and

in this case three is going to be the

adjacent

side divided by the hypotenuse again

which is

four and tangent of theta that's the

opposite side squ of

7 over the adjacent side which is

three now we could go through here

cosecant secant cotangent you know you

could read everything off notice that

cosecant you just take the value you got

for sign and you just take the

reciprocal just flip the fraction over

so the same thing with cosine and secant

and tangent and

Cent so let's see so cosecant of

theta that's going to be 4 / the < TK of

7 here we're going to rationalize the

denominator so multiply top and bottom

by thek of

7 that'll leave you with 4 * the of 7 in

the numerator the 7 * the 7 is just 7

the fraction 4 7 doesn't reduce so

that'll be our

solution then we have secant of theta

same thing I'm just going to flip the

fraction that'll be

4/3 okay so we don't have to rationalize

anything there it's already

rationalized and also cotangent that'll

be 3 / the < TK

7 and you can multiply top and bottom by

the < TK of 7 that'll give us 3un 7 over

7 and now we've got all our Solutions

again you can also so of course you can

think about cosecant secant and

cotangent in terms of ratios of of side

lengths of the triangle but to me it's

much easier just if I can remember

cosecant goes with s uh secant goes with

cosine and cotangent goes with Tangent

it to me it's easier just to take the

flip of the fraction and simplify it or

rationalize the denominator if I need to

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! All the Trigonometric Values Given a Right Triangle.  Just a quick little example of finding all of the trigonometric values given a right triangle.


all right in this video we're going to

do another example of permutations

involving repeated symbols and we saw in

the first example how many different

linear Arrangements there are using the

letters from the word Mississippi so

slight variation on this problem at

first may not seem like it's related but

then I think it's pretty clear it's

really the exact same type of problem so

suppose we want to look at the number of

staircase paths in the XY plane that

start from the origin 0 0 and so we're

going to start at the origin and move

over to the point 5 comma 3 and each

path is made of steps moving one unit to

the right or one unit upwards ok so

again that's how our paths are made we

can't go diagonally or we're you know

other things like that so for example

one little path you know you could go

over you know maybe I go over twice and

then I go up once and then I go over one

so then I go up once then I go over once

and I go up once and hey then I go over

again ok that would be an example of a

path where I'm either moving to the

right or moving upwards so we want to

count all of those different paths ok

well if you think about it you know each

path is going to be made up of well to

go all the way over we're going to have

to go to the right 1 2 3 for a total of

5 times ok so we're going to really move

right 5 times and notice we would have

to go up 1 2 3 times and if you think

about the path I just did I could really

write it so maybe you know we go right

one unit so I'm going to list that as

you know move right and then we moved

right again and then we moved up and

then we moved right and then we moved up

and then we move to the right and then

we moved up and hey then we move to the

right

really what what's going on to make our

our paths you can really think about it

as being a word kind of like our

Mississippi example where we've got five

R's to two

from we've got a you know go right five

times and we have three use okay so the

total number of you know letters in our

word if you think about this as being a

word we've got eight objects total again

there's five hours three use so now

we're just using it's very much like the

Mississippi example so again our

original formula it says we look at the

total number of objects in this case

there's eight factorial objects and then

we just multiply so we've got eight

factorial on top and then in the

denominator we would have five factorial

because eventually we have to go right

five times so our word would have five

hours and it's somewhere and then we

have to go up three times so our word

would have three use in it and this is

going to be the total number of little

staircase patterns that we can make so

again this is just 8 times 7 times 6

times 5 times 4 times 3 times 2 times 1

5 factorials 5 times 4 times 3 times 2

times 1 3 factorial 3 times 2 times 1 so

let's see we can cancel out the 5 4 3 2

1 from the top and bottom 3 + 2 is 6 so

it looks like we're left with 8 times 7

or it says there's a total of 56

different paths to go from the origin to

that point 5 comma 3

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Permutations Involving Repeated Symbols - Example 2. In this example, we count how many ' stair case ' paths there are from the origin to the point (5,3).


all right in this example we're gonna

write the standard form equation of a

parabola with a vertex at the origin and

a focus at the point 1 8 comma zero so

we don't have to sketch anything at all

but I always like to have a little

sketch it helps it helps me just to

visualize and to kind of a double check

and make sure that I'm doing things

correctly so we know the focus here is

at the point 1 8 comma 0 that's given

and we're assuming that the vertex is at

the origin so I know that my parabola in

this case is gonna open to the right

well recall if our parabola opens to the

right it's gonna be of the form Y

squared equals 4p times X so all we

really have to somehow is we just need

to deduce the value of P do the

arithmetic and hey we'll have the

standard form recall though that our

focus our focus is gonna be of the form

P comma 0 but we know what our focus is

is 1/8 comma 0 so we're actually already

being told our value of P in this case

we know that P in this case is going to

equal 1/8 so that's all we have to do is

just substitute in the 1/8 and 4p so we

have Y squared equals 4 times our p

value of 1/8 simply multiply that by X

and we get Y squared equals well for you

can think about this as being 4 over 1

so I have 4 over 8 which will reduce to

1/2 times X or if you wanted to you

could write that as Y squared equals x

over 2 either one looks good to me and

now we've got the standard form of our

parabola

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Conic Sections, Parabola : Find Equation of Parabola Given the Focus.  In this example, we are given that a parabola has a vertex at the origin and we are given the coordinates of the focus.  From that, we find the equation of the parabola.


okay here we're going to graph the

function y = -23 * tangent X we'll pick

out the vertical asmp tootes and the

zeros as well so tangent of X normally

has ASM tootes its first ASM toote

occurs

at the first positive ASM toote occurs

at Pi /

[Music]

and I guess you could say the large just

negative ASM toote is atunk / 2

and the the graph of tangent tangent has

a period of Pi so every time we move

over Pi units we're going to get another

ASM toote so let's see uh 1 2 3 4 so 1 2

3 4 1 2 3 4 let me try to keep at the

scale here that would be another ASM

toote that would be at 3 Pi / 2 and

likewise I'm not going to count exactly

but over here at x

= 3i / 2 again we'll have another

vertical ASM toote so tangent normally

if will go through 0 0 and this graph is

certainly doing the same as you move to

the right it increases and it decreases

but since we have -2/3 it's going to

reflect that so the graph is now going

to go

down and we'll go go

up if you move over Pi

units so here's xal Pi again this graph

is just going to repeat itself and

likewise at NE

Pi okay so let's see so the zeros in

this

case the zeros

those are going to occur at multiples of

Pi so we'll say Pi * n where n is an

integer and the vertical ASM tootes

those are going to occur

at xal Pi / 2 plus multiples of Pi so

we'll say Pi / 2 plus pi n again where n

is an

integer if you wanted to you could plot

maybe even an extra point so tangent at

Pi over4 normally goes through positive

one but in this case it's going to go

through at Pi over4 due to the

reflection it'll go through

-2/3 and

at Pi over4 so let's see if I can fit

that in there attive Pi over4 it would

be going through the POS 23 so likewise

piun over4 comma 23 would be a point on

the graph and this just keeps repeating

forever and ever and ever

Graphing a Tangent Function - EX 1

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Graphing a Tangent Function - EX 1


all right here's another example of

finding derivatives using the product

rule along with the chain rule so again

I see two separate pieces here being

multiplied so the derivative of this if

I take the derivative of x 4th I'll get

4X 3r you leave the other piece

alone and then from that I will leave

the X to the 4th alone and I'll take the

derivative of the other piece

remembering I have to use the chain rule

so the three will come out front I'll

leave the inside part alone so 5x minus

one I'll take one away and get to the

second power and then multiply that by

five so at this point we have to

basically clean things up a little bit

and this can be the tricky part so

there's really nothing to simplify in

the first part so I'm going to leave

that part alone

and I'm going to multiply things out

over on the right side so notice I have

a three and a five that's going to give

me the 15 I have x to the 4th still

hanging out and then I have a 5x -1

sared now at this point we can keep

factoring even a little

further okay so I look at my numbers

there's a four and a 15 well the largest

number that goes into four and 15 is

just simply one so you can't really

factor a number out notice I have an x

to the 3 thir power and an x to the 4th

power well I can factor out an x to the

3 power notice also I have a 5x -1 I

have another 5x - one just like with the

X's you have to pull out the smaller

power I'll pull out the smaller power so

I can pull out 5x

-1 squared and then I'm just going to

fill things back in on the

parentheses so to get the first term

I'll need my four because I'm missing

the four I'm going to need an extra 5x -

one term because again I've got two of

them outside I need an extra one on the

inside so notice notice if I multiply

the first thing to the first thing here

I will get my first term well I'm going

to need a

15 and an X to get back the second

term and now you can

simply reduce things down inside of the

parentheses so if you distribute you'll

get 20x + 15x that's

35x you'll get 4 * -1 which is -4 and

that's now simplified and factored so

these problems are a little confusing I

think people know how to do the

derivatives some of the factoring can

get a little tricky but here's a nice

basic example on what you need

❖ Product Rule and Chain Rule Example ❖

📚 Mastering the Product Rule and Chain Rule: Essential Derivative Techniques with Algebra Simplification 🧮

In this video, we dive into two fundamental derivative techniques: the Product Rule and the Chain Rule. These rules are crucial for differentiating functions that are products of other functions, especially when one or both involve compositions. You’ll learn how to apply both rules together effectively, followed by simplifying the result using algebra.

🚀 What you’ll learn:

The formulas for both the Product Rule and Chain Rule, and how they work together.
Step-by-step examples applying these rules to solve real derivative problems.
A walk-through of algebraic simplification at the end of the process to tidy up your final answer.
Key tips to avoid common mistakes when using both rules on complex functions.
Whether you're prepping for AP Calculus, refreshing your derivative techniques, or looking for clear examples, this video will help you gain confidence in using these powerful rules. Stick around for the final step, where I simplify the derivative using algebra for a cleaner solution!

For more advanced calculus tutorials, be sure to check out my other videos on implicit differentiation, quotient rule, and more.

👍 Don’t forget to like, subscribe, and hit the notification bell for more math tutorials! 🔔✨

Watch now to master both the Product Rule and Chain Rule and take your calculus skills to the next level!

#ProductRule #ChainRule #Calculus #Derivatives #MathTutorial #LearningMath #APCalculus #Mathematics #StudyTips #DerivativeTechniques #AlgebraSimplification #Differentiation #ImplicitDifferentiation #MathHelp #STEM #STEMEducation #MathLessons #MathConcepts #DerivativesExplained #MathSkills #CompositeFunctions #DerivativeExamples #ChainRuleExplained


okay here we're going to do a limit

problem involving the absolute value

theorem so here we're going to find the

limit of the sequence x sub n = -1 n +

n^2 over n^2 and again intuitively to me

what's going to happen you know if you

plug in larger and larger values for n

let's forget about the first part for a

moment as we plug in larger and larger

values for n you know we're going to get

some big number squared over some big

number squared it's the same number so

this ratio you know if it wasn't for the

negative 1 to the end we would just get

one well what's going to happen um you

know the negative 1 to the N all that's

either going to do to the numerator is

either subtract one or add one so again

to me for large values of n the the

numerator and the denominator are almost

going to be exactly the same value which

makes me think this limit should equal

zero to justify this again we're going

to use the absolute value theorem

so remember that says if the

limit of the absolute value of a subn

equals 0 or in this case X ofn if that

happens that tells us that the sequence

uh without the absolute value also

equals zero so we're going to use this

little result and the way that we're

going to do this is well we're going to

take the limit as n goes to Infinity of

-1 n + n 2 all over n 2 and all we're

going to do is just break this up using

some algebra so we can write this as the

limit as goes the limit as n goes to

Infinity of 1 the N over n

2 plus n^ 2 over n

2 so we'll have the limit as n goes to

Infinity again of -1 to the N over n^2

well n^2 over n^2 that's just going to

give us

one

so again we can kind of look at these

individually and then the limit as n

goes to Infinity of just one well the

limit as n goes to an Infinity of one is

just one the limit of a constant is the

constant so the only thing we really

have to think about now is What's

Happening Here well again I think it's

pretty easy to argue the top is always

going to bounce back and forth between

negative one and positive one the

denominator is going to get arbitrarily

large so to me this this first limit

should certainly go to zero but again to

bring in this absolute value

theorem notice

um so notice that the limit as n goes to

Infinity if we take the absolute value

of1 to the N over n^2 so I'm just

looking at the first part well that's

just going to equal the limit as in goes

to Infinity if we take the absolute

value of the numerator U again you know

without the absolute abute value we're

either getting positive one or negative

1 uh when we raise it when we take the

absolute value of it it'll just become

postive one over N2 but again uh this

limit as n goes to Infinity the

denominator will get arbitrarily large

one over a big number is going to be

zero so hey in fact by the absolute

value theorem this first limit is

actually going to be zero so really

we're left with 0 + one so again in this

case our limit will simp converge to one

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Squeeze Theorem and Absolute Value Theorem, #2. Just another example of finding the limit of a sequence by using the squeeze and / or absolute value theorem.


all right in this video I just want to

start looking at some examples of just

simplifying trig functions involving uh

fractions so nothing too crazy here so

uh maybe we can do at least one or two

so uh in part A we've got s time secant

times tangent a lot of times what I will

do um you know there's really not much

going on here there's nothing to add or

subtract uh a lot of times what I'll do

is I'll just rewrite everything in terms

of s and cosine and see see what if

anything I can cancel out so I'll leave

the sin x alone see of X that's the same

thing as 1 cosine of

x well tangent of X that's sin x over

cosine X well um I think you'll notice

in the top you can really put sinx over

one so really in the numerator we have

sinx over cosine

X and again that's simply being divided

sin x over cosine X well you have S you

have sin / cosine over sinid by cosine

or equivalently we have tangent of X

over tangent of X however you want to

look at it or you know last but not

least you could even think about it just

in terms of fractions uh we've got a

fraction divided by a fraction flip the

denominator and make it into

multiplication and again basically

everything's just going to cancel out um

we still have you know you can think

about there as being ones in the

numerator so this first expression is

simply going to simp is just going to uh

just reduce down to the number one so s

* secant over tangent is just another

way of writing

one um let's look at Part B we've got

cosecant time cotangent divided tangent

time secant so I think I'm going to do

the same thing in this example um let's

see so cosecant that's going to be 1/

sin Theta

coent is cine

thet over sin

thet let's see tangent of theta that's

going to give us sin Theta over cosine

Theta and then secant of theta is going

to be 1 over cosine of theta okay so

let's try to uh clean all this stuff up

here a little bit so again you

know couple different uh sort of ways

can regroup things we we really have

cosine of theta on top um we have sin

squar in the

denominator of the numerator and then in

the denominator we have S of theta over

cosine squar of theta and what I'm going

to do now is just simply take this

fraction these two fractions and we can

multiply so I'm going to take cosine

over sin squ

I'm going to multiply that by the

reciprocal of the denominator so uh

we're just going to flip the the bottom

fraction so we'll get cosine s Theta

over s of

theta well cosine * cosine squ is going

to be cosine cubed of theta over sin

cubed of theta but again we could really

just factor out the uh the exponent we

could write this as cosine over s all

cubed and remember cosine over s that's

simply

coent so really we have OFA cubed or

again we can rewrite that asent cubed

Theta

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Simplifying Trigonometric Expressions Involving Fractions, Ex 1


So in this question it says,

For what value of n is the absolute value of the quantity n minus one

plus one equal to zero?

So the answer choices as are zero, positive one, positive two, or there is no such value of n.

So, just recall that the absolute,

I mean, to me there is not much to do in this problem as long as you remember absolute values.

So remember if you take the absolute value of a number

you know, so the absolute value of zero is zero.

You can think about this being a distance from a number,

you know, the distance from zero in the number line,

that's someway people think about it.

I always just think of it, you know, if the number's already positive,

you leave it alone.

So, for example the absolute value of two would be two.

If the number's negative, the absolute value of negative four, again we just drop the negative sign.

So the absolute value of negative four is positive four.

The main thing you remember is that the absolute value is always gonna give you, either,

if you simplify, its either gonna be a number that's zero or larger.

So, just based on that, I know that the absolute value n minus one,

no matter what value I substitute in for n,

this absolute value is always gonna give me a number

that is greater than or equal to zero.

Okay, so I'm gonna get something that is zero or larger.

Well, if I add one to that, again if I add one to that value,

it's gonna be, its gonna give me a number

that's greater than or equal to one.

So in this case, there would be no such value of n.

And again, you could check, you know,

if you substitute in zero, you would have zero minus one which would be negative one,

the absolute value of negative one is positive one,

so you have positive one plus one.

If you substitute in zero, you actually get the value of two.

If you substitute in n equals one, one minus one is zero,

the absolute value of zero is zero, plus one would give you one as value.

And if you substitute in the value of two, well, two minus one is one,

the absolute value is still one, plus one, it wll actually gonna give you the value of two.

So if you substitute in zero, you get two.

If you substitute in one, you get one.

If you substitute in two, you get two.

Again,

we're trying to get zero out, so there is no such value of n in this case.

You know, they could give you some really wonky problem,

you know, they could make it like

n squared plus n raised to the four hundredth power minus the square root of pi

plus, you know, eight to the eighth to the eighth

plus one.

You know, what value of n, they could ask you the same question,

for what value of n is this quantity equal to zero?

Same thing, it doesn't, they'll be no solution because

no matter what you substitute in for n,

the absolute value, again your gonna get something that's zero or larger.

If you add one to that, it's gonna give you someting that's one or larger.

So, that's all there is to it, just making sure that you remember absolute values.

So, you know, hopefully if you saw this in the SAT,

this would be pretty quick question and you could just

you know,

circle answer choice D and keep on truckin'.

SAT Math - Question 8, Section 2, 2016 Edition and Beyond

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! SAT Math - Question 8, 2016 Edition and Beyond.
Section 2 - A Calculator IS allowed!


all right so just one more example here

of solving a basic trig equation suppose

we want to solve see of theta equal

2 well all I would do to solve secant

theta equals

2 um you know I have values memorized

about s and cosine so I'm going to turn

secant back into one/

cosine and now I'm just going to do a

little bit of algebra we can multiply

both sides by cosine Theta

well the cosine thetas uh would cancel

out we would have 1 = 2 * cosine Theta

and now I'm simply just going to divide

both sides by two by two and that'll

give us cosine Theta equal

12 all right so I do know angles where

cosine equals 12 let's see where would

those be um I think actually again at

the angle pi

3 again that's where the x coordinat is

12 and the y coordinate is < tk3 over2

well so that means that Theta = pi over

3 would be a solution um if we kind of

make one revolution 0 to 2 pi I think

the next place where the x coordinate is

going to be positive uh2 would

be down here in Quadrant 4 that's where

we would get 12 < tk3 over 2

let's see again if we go all the way

around that's 2 pi radians equivalently

if we multiply top and bottom by three

that'll be 6 pi over 3 radians but we

didn't go 6 pi over 3 we went one less

so if you subtract 1 pi over 3 we'll get

5 pi over 3 so that would be another

uh another solution that works um and

again you know sometimes there's

definitely uh you know more uh maybe

more succinct ways of writing it I'm

just going to say our solutions would be

Theta = pi over 3 again to end back up

at the same place we can add or subtract

multiples of 2 pi so I'm going to write

plus 2 pi *

n same thing for our angle 5 pi over 3

we can add or subtract multiples of 2 pi

so I'm going to write plus 2 pi * n and

again in both cases um n is an integer

for

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving a Basic Trigonometric Equation, Example 3


okay and these two examples I just want

to just do some evaluating of these

exponential expressions and we've got

two problems here so in the first one we

have negative 2 raised to the third

power times 3 raised to the second power

and again you can just you know you can

simply write it all out so negative 2

times negative 2 times negative 2 and

then we have 3 squared which means we

have 3 times 3 well negative 2 times

negative 2 will give us positive for

positive 4 times negative 2 is going to

give us negative 8 if we do 3 times 3

that's going to give us 9 we have a

negative times a positive which is going

to give us a negative and then 8 times 9

will simply give us 72 so negative 2 to

the third power times 3 to the second

power will equal the number negative 72

and our last one here on the second one

we have negative 3 squared and negative

2 squared so again that's just negative

3 times negative 3

and then we have two negative twos so we

have negative 2 times negative 2 well

negative 3 times negative 3 it's going

to give us a positive 9 negative 2 times

negative 2 is going to give us a

positive 4 and 9 times 4 will give us 36

so negative 3 to the second power times

negative 2 to the second power will

equal 36

Evaluating Exponential Expressions (Numbers Only, No Variables) - Ex 1

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okay in this video I just want to talk

about the midpoint formula and all the

midpoint formula says it says that the

midpoint between two points x sub 1

comma y sub 1 and x sub 2 comma y sub 2

it says all you do is add the x

coordinates divide by two add the

y-coordinates and also divide by two um

I've uh in my experience I've noticed

some people sometimes forget you know

the signs they'll sometimes subtract and

one way to maybe think about it is you

know take a kind of a concrete example

suppose you took two tests and you made

a 50% on the first one and 100% on the

second one and I asked you I said well

if those are your test scores what's

your test average most people would say

your test average is

75% and I normally when I ask them how

you get it they say well to average you

add the things together and you divide

by the number of them so you have 150

over two or simply 75 um% so you know

notice a mid point is just an average of

two things so if you can remember it in

that context maybe it'll help you

remember that um you should be adding

the x coordinates divide by two adding

the y-coordinates divide by two so in

this example um I'm just going to Simply

find the midpoint between these two

points I've got the point -5 comma -6

and then I've got the point um 5 comma

8 so let me try to you know maybe

connect those with with a line if I can

so not the straightest of lines but a

line nonetheless so to me intuitively

the midpoint should be you know

somewhere around there so you know based

on my solution at least I can even get a

quick little conceptual guess about what

it should be so it simply says again to

find the

midpoint piece of cake all we have to do

is add the x coordinates so maybe we can

label them as well so this will be X

sub1 this will be the point Y sub one

again these are kind of generic labels

it doesn't matter um which one you call

X1 y1 which one you call X2

Y2 so it says if we add the x

coordinates we'll get -5 + 5 / by two

and then it says uh the midpoint for the

y-coordinates will get -6 + 8 and divide

that by 2 and if we simplify that down

-5 + 5 will be 0 / 2

um -6 + 8 is 2 / 2 and that simply

reduces to the point 0 comma 1 and to me

that seems reasonable 01 hey that would

be a point right on the y axis so that

seems correct and notice in this case as

well things are kind of symmetric um

we're at NE five we're at positive five

so we're five units to the right of the

y- axis five units to the left of the y

axis the midpoint should be you know

zero units away from the y- axis which

would correspond to an x coordinate of

zero um you know so

again it's good to know formulas but I

think if you sketch a little picture on

these two if you're taking an exam or

something like that um you can quickly

verify that whether it's at least in the

ballpark or not so um all right I hope

this video makes some sense and helps

you out if you have any questions about

the midpoint or any questions or

comments about this video as always

always please feel free to post them

hopefully me or somebody else can point

you in the right direction

The Midpoint Formula

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Midpoint Formula - In this video I give the formula for finding the midpoint of two points and do one simple example to find the midpoint.
For more free math videos, visit http://PatrickJMT.com


all right in this video I'm going to do

one more example of graphing a

piece-wise defined function so here in

in this particular case notice we've got

a function that sort of has three parts

it looks like the line y = -2 if x is

less than or equal to3 looks like 4 - x

for X values in between -3 and 0 not

including either and it looks like the

line y equals x if x is greater than or

equal to zero

okay so here is uh here's my little

graph so I'm just going to do them kind

of in order here uh if you think about

the first equation y = -2 y = -2 is just

a horizontal line at the yalue of -2 so

let me drop down here -1 -2

units but it says it only looks like uh

this line Y = 2 for x coordinates that

are less than or equal to -3 okay so it

looks like that line but we don't get

the whole line -1 -2 -3 it says it only

looks like that for x coordinates uh

starting at -3 and then it says it has

uh it has that shape forever and ever

and ever for all x coordinates less than

or equal to -3 so it says we're getting

that that segment of the line up to an

including 3 but then it

stops okay I recognize the second

equation Y = 4 -

x that's simply a line we could actually

rewrite this

ASX +

4 that's a line with a Y intercept of

positive4 so 1 2 3 4 um and a slope of

negative 1 so I know it's pointing

downwards um Let me give myself a little

more room here five six we'll say sevens

up here again what I do I always just

plug in the end points so if I were to

plug -3 into this I would get four - -3

or 4 + 3 which would give me 7 but since

it's not including -3 I'm going to put

an open circle up

there okay if I plug in the value if I

plug in the other end point 0 I would

get 4 minus 0 which would give me four

but again since it doesn't include the

value zero I'm going to make an open

circle there so I'm kind of finding the

end points and again since this is a

line and it it it does look like the

line for all x coordinates in between I

can just play Connect the Dots uh with a

straight line and that'll now be the the

portion of the graph y = 4 - x so again

here was a y = -2 down here and last but

not least if we graph the last little

segment y equal x

X well again I just plug in the value if

I plug zero in well I get zero out so

that means 0 Z's on the graph and yal X

is just our kind of a line that goes

through the origin kind of makes it like

a 45° angle um but again it only looks

like this line yal X for X values

greater than or equal to zero so to me

that's only the right portion of the

graph so now I'll just make it extend on

forever and ever and ever and there's

our line Y = X and hey now we're in

business we've got our nice piecewise

defined function

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Graphing a Piece-Wise Defined Function - Another Example


all right uh in this video that I'm not

really going to do anything uh other

than just put up a table of llao

transforms because we'll use this in

some of the other examples so just as a

reference so um on the left we have our

function f of T and then um on the right

we have the Lao transform associated

with that so uh the Lao transform of the

function one is just going to be 1 / s

um if you have t to a power n you end up

getting in factorial over s the n + 1 if

it's s of a we get a/ s² + a 2 again a

lot of these are just Justified uh

straight using the definition um if you

have cosine of a t you get S over s^2 +

a 2 if it's e to the a t we get 1 / s -

a e to the a t * sin of BT we get b/ S -

a^ 2 + b^ 2 if it's e to the a t cosine

of BT we get S - A over s - a^ 2 + b^ 2

and if we have t to the power of n time

e to the a t we get n factorial over the

quantity s minus a raised to the n + one

so we'll use these in some other

examples to actually calculate Lelo

transforms of functions

Table of Laplace Transforms

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Table of Laplace Transforms.  In this video, I show the Laplace transform for a few common functions.  No theory or derivations here, just a quick reference that I will refer to in other videos.


okay so just one more uh example here

finding the derivative of an inverse

function so uh in this problem we have

FX = cosine x + 3x and we're given that

F inverse of 3 piun / 2 = piun / 2 we

want to find the derivative of f inverse

at x = 3i / 2 um again notice that this

function does have an inverse uh the

derivative of cosine is just going to be

Nega

sinx uh and then the derivative of 3 X

will be positive 3 um if you think about

negative sinx you know s normally looks

basically sort of like that um if we

flip it you know it's just going to

instead of going up it'll go down

um but what I'm trying to say is uh

negative s of x if you think about the

range of negative sinx for any value of

x this is going to be between negative

one and positive one well if we add

three to it so sinx + 3 if we add three

everywhere we'll get two on the left

side and four on the right side it says

the derivative is always between two and

four so the main thing is this

derivative is always greater than zero

and again that tells us that our

original function f is always strictly

increasing so it passes the horizontal

line test so it does have a

derivative well again if we use our

original

formula it says we basically have to do

is take the derivative and plug in F

inverse of our

value well again we're given that F

inverse of 3 pi/ 2 is um going to equal

pi/ 2 so really we're just left

evaluating so when we kind of fill in

our

formula our F inverse of 3 pi/ 2 uh

we're going to plug in our value Pi / 2

so now again uh all we have to do is

just take the derivative and plug in Pi

/ 2 but that's what we did just a second

ago so again we said frime of X was

going to equal sinx and then we had

positive3 left over well F Prime of pi/

2 that's going to give us negative S pi/

2 +

3 S of pi over 2 is equal to 1 so we've

got -1 + 3 which will give us pos2 and

now it's just a matter of plugging all

of that back again so it says uh our

solution in this case will be positive

Derivative of an Inverse Function , Ex 2

In this video I work a quick example of where we need to evaluate the derivative of an inverse function but we are only given the original function, not the inverse.
This is a typical topic in a calculus course.  In order to evaluate the derivative at a point we have a clever work around that uses the derivative of the original function, evaluated at the inverse.

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all right so one more example here of

finding a linear approximation so here

we're going to find the linear

approximation for cosine X at x = pi

over 4 all right so again if we plug pi

over 4 into our function well we get

cosine of pi over 4 and that's just <

tk2 over 2 so I know when we do y minus

the y coordinate the Y coordinat is

going to be < tk2 over two our slope of

our line x minus the x coordinate well

the x coordinate is given to us as Pi

over4 to get our slope we'll have to

take the derivative of our function well

the derivative of cosine would be

sinx if we plug in pi over 4 well we'll

get ne s of pi over

4 and that's going to be < tk2 over 2 so

now all we have to do is just plug that

in so we have Yus < tk2 over2 the slope

of our tangent line our linearization

will be < tk2 / 2 then we'll have x - <

over 4 and now I'm just going to you

know uh just multiply things out a

little bit just uh get y by itself so if

we distribute we'll have < tk2 over

2x um I guess we'll get a positive uh

when we multiply so we have positive <

tk2 * piun over 2 * 4 which would be 8

we can add add the uh positive we can

add over < tk2 over two and again you

can get common denominators whatever you

want to do if you want to write your

constant as a single term but again this

is now our line um so this would be our

linearization

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding a Linear Approximation (Linearization, Tangent Line Approx), Another Ex 2.  Just another example of finding a linear approximation for a function at a particular point.