Howdy out there Youtube I am PatrickJMT and this is my math channel

What do I do on my math channel?

Well I make math videos

How many math videos?

a lot of math videos (chuckle) over a ton of topics and I plan on expanding on this just like I do every school year

got a decent, certainly a fairly robust I think math background

lots of experience teaching, a masters degree in math, did a lot of private tutoring as well

What's my goal?

Just to make clear and effective videos

I feel like people are probably looking at these at 11 o'clock at night

they're trying to get their homework done

maybe not always 11 o'clock at night, but just some extra supplements for when you're not in the classroom

and you have those questions at night and you don't know where to turn

I appreciate all of the love and support through the years from everybody out there on Youtube

So I hope you guys keep watching and tell your friends and teachers as well

and again thanks for all the support, it means the world to me

Welcome to PatrickJMT's Math Channel!

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Tons of supplemental mathematics videos for your viewing pleasure.  I try and explain the mechanics and technical side of mathematics as clearly as possible.
  Thanks for all of the super nice comments.  I sincerely appreciate them!
Note: These videos are intended to be supplements to your own studies and not stand-alone lectures.


okay in this example I want to express

the

inequalities um X greater than or equal

to -6 and X less than 10 using interval

notation and the way again I like to do

these is I like to make a little number

line so the first condition it says X is

greater than or equal to

-6 okay so I'm just going to put -6 um

on my number line again I could use a

solid dot or we can use a bracket to

indicate that it gets uh used and for

this inequality we would just shade

everything to the right forever and ever

and ever and ever and ever and ever but

that's not the only inequality we're

trying to

satisfy we want to simultaneously

satisfy the

inequality that uh X also has to be less

than

10 so I'm going to put 10 over here in

my number line and we again we could use

an open circle or we can use parentheses

to indicate that we get arbitrarily

close to 10 but we don't include it and

then if you shade to the left for that

one well everything less than 10 would

work for that one but again we're trying

to satisfy both inequalities at the same

time so if you look at the region in the

Middle where they overlap that's going

to be the set of x's that satisfy both

inequalities at the same time so the way

that we can express these two in

inequalities using interval notation we

can say that we include -6 and then we

get up to but not including the number

10 so -6 in Brackets up to positive 10

in parentheses would be an equivalent

way of writing those inequalities using

interval

notation

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


okay here we're going to graph two

periods of the function y = 4 * cosin X

and the period for cosine X is equal to

2 pi so what I like to do when I graph

trig functions is first off I like to

find the period I should say for sign

and cosine I like to find the period and

then what I like to do is to divide that

into four equal parts so 2 Pi / 4 will

Al Pi / 2 and what I do is I I use these

values as my increments when I Mark

values on the x- axis so okay so what I

mean is here's uh you know x equals 0

I'll move over there's pi over two I'll

move over another pi/ 2 which would put

me at

pi and let's see if I go over another

I'll have 3 Pi / 2 and then I'll have 4

Pi / 2 or 2 pi let's see so 1 piun / 2 2

piun over 2 3 piun over 2 4 piun / 2

we'll have 5 piun / 2 6 pi over 2 um

let's see so 6 pi over

2 well 6 pi over 2 is just going to be

equal to 3 Pi then we would have 7 pi

over 2 and then we would have 8 pi over

2 which would be equal to 4 Pi so

that'll give us two complete periods

okay and now you

know what you can always plot a couple

points so if x is 0 cosine of 0 is equal

to 1 so you'll have 4 * 1 or you'll get

four so here we are up here at 1 2 3 4

again uh the X the coordinates on the x-

axis are not going to be the same as the

coordinates on the Y AIS here so not

exactly to scale but uh you know reason

ly

close and then what's going to happen is

so that's going to be the highest point

that cosine touches so the reason why I

like to divide these into four parts is

when I go to the next increment that's

where cosine is going to hit the xaxis

and then at the next value that's when

we'll be at - 1 2 3 we'll be down at -4

so I'm going from positive4 to 0 to

-4 back up to zero and then I'll be back

up at

and that'll roughly be a little sketch

of one period of cosine and now I just

repeat this again so again this is why I

mark them off in fours I kind of get the

the high point the midpoint the low

Point uh back to the midpoint and then

back to the high

point so it'll go at 5 pi/ 2 it'll be

zero at 3 Pi again we'll be back at -4

7 piun / 2 I'm back at zero and then at

4 Pi you'll be back at positive 4 so

that would be two periods

of Y = 4 * cosine X and again you can

always plot points uh do other things to

make sure that the graph is exactly what

you want but again sign and cosine I

take the period divide it into four

that's what I use as my increments and

and then I just start uh you know

putting some points down and again uh

just connecting them with those curvy

curvy coine uh lines

Graphing a Cosine Function  EX 2

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Just another example of graphing a trigonometric function


okay here we're going to solve the

system of equations for X minus y equals

two and negative 12x plus 3y equals

negative six so I'm going to use

elimination by addition and the thing

that I'm going to try to do is to

eliminate the the terms containing X so

I think you know if I just had a

positive 12x in my as my first term in

my first equation well then I could add

I would have 12x minus 12x the terms

involving X would cancel so to make a

12x appear what I can do to my first

equation is I can simply multiply both

sides by 3 well if I do that I'll get

12x minus 3y equals 6 and then we have

negative 12x plus 3y equals negative 6

just our original second equation notice

if I add these two equations

I'll get 12x minus 12x I'll get 0 X

negative 3y plus 3y is 0 Y so we're left

with 0 on the left side on the right

side we have 6 minus 6 which is 0 and

what this tells me is we have infinitely

we have infinitely many solutions

so if you actually want to write your

solutions down to sort of give a a

generic sort of form people sometimes

ask for this what you can do is take any

one of your two equations I started with

the 4x minus y equals 2 I started with

my equation before we you know before we

multiplied everything out 4x minus y

equals 2 I'm going to solve for y so I'm

going to subtract the 2 to the left side

I'm going to add Y over and now the idea

is we can say let's let X equal some

constant K well if we do that then Y is

going to equal 4 K minus 2 so we can say

our solutions that are of the form K

comma 4 K minus 2

so the idea the idea is pick your

favorite value for X and then do four

times that number minus two that will

give you the Y value and that's the way

that you can actually find specific

specific pairs of numbers that satisfy

the original system of equations

A Linear System of Equations with Infinitely Many Solutions: Quick Example

(https://youtu.be/rnyLW2-lL7o)
(https://youtu.be/CsTOUbeMPUo)
(https://youtu.be/cJg2AuSFdjw)
(https://youtu.be/hu6B1d3vvqU)
(https://youtu.be/9LYVi-n-6Jw)
(https://youtu.be/iMQRo0tHORw)
(https://youtu.be/Re1F4d24Fxc)
(https://youtu.be/tquUDsduwys)I

n this video I work a system of two linear equations and show that they are linearly dependent, meaning that there are infinitely many solutions to the system.  For two linear equations, this simply implies that the two lines are exactly the same and would cover each other if you were to graph them.
This is a typical topic seen in algebra, linear algebra, linear programming, and other areas.
For more detailed info about these types of systems, check out my other videos!

#math #algebra #khanacademy #khan #geometry #education #montessori #homeschooling #functions #notation #organicchemistry #organicchemistrytutor
#logarithm #naturallog #exponential #system #linearsystems #linearlyindependent #linearlydependent #linearequations #graphing


all right in this example we're going to

arrange some integers from the least to

the greatest um we've got -7 pos4 -10 0

18 and ne1 and again maybe it's easiest

to think about a number line so the

first thing I do is pick out my negative

values so I know those are going to be

the smallest numbers well let's see if

here if there's zero negative 1 would be

right about there I'm not going to fill

in every digit but there's -7 and 10

would be the furthest one to the left so

that means -10 would be our smallest

number let's see I guess the next one

would be -7 and then we would have ne1

so that takes care of those zero would

be the next number and then last but not

least we've got four and 18 well

certainly four is smaller than

18 and now we've got our numbers

arranged from least to greatest

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Integers : Arranging Integers from Smallest to Largest on Number Line.
Here we look at arranging some integers from smallest to largest on a number line.


all right in this video I want to talk

about functions and the vertical line

test and the basic idea is if you have a

graph and you want to decide if that

graph represents a function the idea is

if you were to draw any vertical line

and if it touches the graph in two or

more places then the graph does not

represent a function uh nothing nothing

worse than

that so here's a three different graphs

um so let's look at the one here on the

left um kind of looks like a parabola or

sort of like a lazy U we want to decide

if that would represent a function okay

well again the idea is to me you know

the arrows it's kind of always opening

to the left and always opening a little

bit and moving to the right well to me

that says no matter where I were you

know no matter where I was to put a

vertical line it would hit the graph in

at most one

place and since it doesn't hit the graph

in two or more places so since it only

hits the graph and at most one place we

would say yes that graph uh is a

function okay so let's look at this

graph in the middle notice on the left

side if we were to draw vertical lines

it would only hit it in one place but

notice for example the Y AIS hits the

graph in two places and there's

definitely some uh some lines to the

right of the y- AIS vertical lines where

it would hit the graph in two places

so in that case we would say no not a

function okay and then we have a graph

over here on the right uh notice it's

not one smooth graph there's a break in

it okay if it's not kind of one smooth

piece we say it's not

continuous okay so that's okay a graph

could be not continuous and be a

function and I think this one would be a

function because no matter where I put a

vertical line to me it's going to hit

the graph in at most one place notice

there's there's some places where the

vertical line won't hit the graph at all

but that's okay as long as vertical

lines don't hit it in two or more places

likewise any vertical line on the right

would hit it in at most one place so we

would say yes this graph um is a

function

The Vertical Line Test

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Vertical Line Test.  Using the Vertical Line Test to determine if a given graph represents a function.


all right in this video we're going to

look at finding some Arc lengths of

parametric curves so in part A we've got

the absolute value of six - t uhal x and

then yal T and we're going to evaluate

this from uh 0 to

3 so we'll have to integrate from 0 to 3

we'll take the derivative of x with

respect to

t^2 plus the derivative of y with

respect to

t^2 DT

so notice uh for x equals the absolute

value of 6 minus

t we can actually just say this is the

same thing as 6 minus

t for t on the interval 0 to three

because uh if you take any value between

0o and three uh the 6 minus t will stay

positive so we can just remove the

absolute value well that says the

derivative of x with respect to T then

is just going to be1

and that tells us the derivative of x

with respect to t^ S is just going to

equal

pos1 the derivative of y with respect to

T that's also just going to be positive

one so the derivative of y with respect

to t^2 will also just be positive one

easy enough so we've got the integral

from 0 to 3 we would just have 1 + one

DT well that's the integral from 0 to

three of just theun of 2 * uh or of 2 DT

so when we integrate we'll get of 2 * T

we'll have to evaluate that from 0 to

three and that's just going to leave us

with 3 < tk2 minus 0 so our Arc Length

will simply be 3 * the < TK of two

Find Arc Length of a Curve Given Parametric Equations , x(t) , y(t) , 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 !! Find Arc Length of a Curve Given Parametric Equations , x(t) , y(t)


all right in this video I'm going to do

some examples of factoring and or

simplifying uh some Expressions so in

part A here we have S S Plus Cent s *

sin s now I think we could certainly

factor a sin squ out maybe even um I

don't know if it's any quicker or any

easier um you could even rewrite

cotangent

squared as cosine squar over sin squar

cancel and use an identity but let's use

this idea of factoring so we can factor

out the sin squ

and then we would have 1

+ S th but recall there's an identity

for 1 + Cent s thet that is cose squ of

theta but what is cose s Theta that's

the same thing as 1

over sin 2 Theta so really uh we have

sin s thet over sin s thet and that's

simply going to equal pos1 so this first

expression simply reduces to positive

one um let's maybe do B here as

well so let's see we've got 2 - cosine 2

over 1 - sin x so um couple different

things that we could do here uh again

the first thing that kind you know we

could try to get common denominators and

group things together uh maybe that'll

make things work out nicely um I also

remember remember though there's an a

relationship between cosine squ and sin

squ so again we can use our identity

that cosine 2 x + sin^2 x = 1 so if we

solve this for cosine s we'll get that

cine 2 x = 1

- sin^2 x that's what I'm going to plug

in the

numerator so we would have 2 - well 1 -

sin^2 x over 1 - sin X and you know why

why does this help well um I recognize

that this numerator is the difference of

perfect squares we can now simply Factor

this numerator um a little

bit you know and kind of think about you

know sort of equivalently if this was

like 1

minus x^2 you've got one minus something

squared how does one minus something

squared Factor it factors as well 1 - x

and 1 + x again it's a difference of

perfect squar

so instead of just plal X we've got sin

x so 1 - sin^2 is going to factor as 1 -

sinx and then 1 +

sinx but hey we've got that 1us sinx

just hanging out in the bottom so really

we can cancel those uh those factors out

and then we'll be left with

2us 1 + sinx but now I can simply

distribute out this negative I will get

2 - 1 and again if we distribute the

negative - sinx well 2 - 1 is one so

we'll simply be left with 1 minus sinx

so you know again a little trickier

again this is what makes these problems

hard you know I I know it has something

to do with an identity um it's got

something to do with an identity but

it's just sort of

uh you know where do you use it and when

do you use it and sometimes once you use

it you know maybe it's not readily uh

maybe you don't see that this factors

immediately but uh you know again you

just kind of have to get your hands

dirty play with things a little bit

certainly as you do more examples I

think they'll start becoming a little

more intuitive and a little bit easier

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


all right in this video I'm going to do

a couple examples of solving logarithmic

equations and anytime you solve

logarithmic equations once you get your

answers you always have to go back and

check for extraneous solutions remember

we can't take logarithms of any

nonpositive number so this is something

you just have to worry about at the end

so uh in in a here we're going to solve

2 * log base 3 of x minus log base 3 of

x + 6 = 1 uh the first thing I'm going

to try to do on the left side is rewrite

this all as a single logarithm so I can

rewrite uh 2 * log base 3 of X as log

base 3 of x squared we can move the

coefficient um up as an exponent not

going to do anything with uh my second

term

here but remember if we have like bases

if we have subtraction this becomes

division so we'll have log base 3 of x^2

over x + 6 thatal = 1 and now what I'm

going to do is just rewrite this using

properties of exponents so it says 3 to

the

1st um would equal

x^2 over x + 6 so okay now we've gotten

rid of the the logarithms and now I'm

just going to multiply both sides by x +

6 on the left side we would have to

distribute our positive3 that would give

us 3x + 18 = x^2 on the right um and now

this is a quadratic I'm going to make

the left side Zero by subtracting 3x

from both sides and by subtracting 18

from both sides and I think this is a

quadratic that's going to factor so I

believe this factors as um how about x -

6 and x + 3 uh that looks okay so if we

said each factor equal to zero we'll get

our Solutions our potential solutions to

be xal 6 and x =

-3 now let's go back and look at the

original

problem um we were trying to solve 2 *

log base 3 of x - log base 3 of x + 6 I

kind of check notice if we

plug if we put in the value six we'll

get a six here which is fine I'll get a

logarithm of a positive

number if I plug six into the other

logarithm I'm going to get a logarithm

of a positive number and so far so good

and you can check that if you plug six

in this is going to work but notice if

we plug -3 in we're going to have a

logarithm of a negative number um on the

second term we would end up still if we

plug in neg3 we' have positive three but

on the first part we would end up with 2

* log base 3 of -3 uh minus some other

stuff but already because we have this

negative inside of a logarithm that is

bad and that means X = three is what's

called an extraneous solution so xal 6

would be our only solution in this

problem

❖ Solving Logarithmic Equations - Example 1 ❖

Solving Logarithmic Equations: Step-by-Step Using Log Properties and Algebra!
In this video, we'll solve a logarithmic equation by applying properties of logarithms and some essential algebra techniques. You'll learn how to use logarithmic properties to combine logarithmic terms and rewrite the equation using exponentials. From there, we'll produce a quadratic equation that needs to be solved. Be careful—we'll also discuss how to identify and handle any extraneous solutions that may arise when solving logarithmic equations.

What You Will Learn:

How to use properties of logarithms to combine and simplify terms.
Techniques for rewriting logarithmic equations using exponentials.
Solving the resulting quadratic equations.
Tips for identifying and avoiding extraneous solutions.
Step-by-step examples to deepen your understanding.
📚 Check out my book: 1001 Calculus Problems for Dummies for more practice!

👍 **If you find this video helpful, please like, share, and subscribe for more math tutorials!

Support My Work:

If you'd like to support the creation of more math content, consider becoming a patron on Patreon: https://www.patreon.com/PatrickJMT

#Logarithms #Algebra #LogarithmicEquations #MathTutorial #PatrickJMT #SolvingEquations #QuadraticEquations #Exponentials #ExtraneousSolutions #Mathematics #MathHelp #EquationSolving #MathProblems #LogProperties #AlgebraTechniques


all right so one more example here using

the direct comparison test part see

we've got the series from N equals 1 to

infinity arctangent of n over N cubed so

in this one again I think you know I'm

not sure I think a lot of people though

would be thrown off by the arctangent of

n you know sort of sort of what's going

on in you know in this case as in

increases

well let's recall the graph of

arctangent of X so recall arctangent of

X looks roughly like my little graph

here and again this has an asymptote at

PI over 2 and at negative PI over 2 well

you know we're plugging in you know N

equals 1 N equals 2 N equals 3 or

equivalently x equals 1 x equals 2 x

equals 3 x equals 4 etc and those are

going to be the values we get in the

numerator but you know again if you

think about it our tangent of n is

always going to be well it's always

smaller than this asymptote PI over 2

and it's again always greater than or

equal to 0 again for values of n that

are greater than or equal to 1 well if

this is true we can simply say well our

tangent of n divided by what do we have

here in cubed

well if divided by n cubed that's going

to be less than or equal to PI over N

excuse-me PI over 2 divided by n cubed

and again these are all positive so I'm

kind of leaving out the zeros so that

means the series from N equals 1 to

infinity of arctangent of n over n cubed

that would be less than or equal to the

series you know PI over 2 divided by n

cubed I'm just going to pull the PI over

2 out front though because that's a

constant and we are certainly allowed to

do that well if you think about the

series N equals 1 to infinity again this

is our familiar you know p series and

again since our p-value is greater than

our power is greater than one in this

case we have a convergent P series and

again since our series that we're trying

to decide whether it converges or

diverges since it's smaller than this

convergent series and again we always

know that this is greater than or equal

to zero by the direct comparison test we

can say that our series must also

converge so again nothing too crazy and

again you know if I just looked at this

problem and sort of had no clue you know

I tell people you know again maybe get

rid of the stuff you don't like maybe

you don't like the arctangent of n if

you're just left with 1 over N cubed

hopefully you remember you know you say

oh well that's a convergent P series

again at that point you know I'm

thinking maybe this is also a convergent

series and somehow I can use you know

some sort of comparison test with that

series 1 over N cubed and you know

that's exactly what we did we kind of

had to justify you know the correct

inequalities a little bit but to do that

we just had to really remember the graph

of arctangent and the fact that has an

asymptote and I think if you can

remember this part you know the rest of

it isn't isn't too terribly bad

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Direct Comparison Test - Another Example 5. In this video I show that another series converges or diverges using the direct comparison theorem. I do not run over the formula/theory in this video (but do in another video, so look around if that is what you need!).


okay here I'm going to graph the

function cotangent of 2x and also we'll

pick out any vertical asymptotes and

zeros so this still has the same basic

basic resemblance to cotangent of X the

only thing that's going to change here

is going to be the period so in this

case the period will take PI and divide

it by the absolute value of B which is 2

it'll give us PI over 2 cotangent has a

vertical asymptote at x equals 0 so to

get to the next vertical asymptote we'll

go over here to x equals PI over 2 and

likewise if we want to do maybe a couple

two periods we can put an asymptote

there as well at x equals negative PI

over 2 and now cotangent if we take the

point in between the asymptotes so this

would be PI over 4 it still has that

same basic shape as cotangent so it'll

increase as we get close to the y axis

and then it will start decreasing down

to negative infinity as we approach PI

over 2 likewise this is the point

negative PI over 4 it's going to

increase and then decrease so I say

increasing as we move to the left the

graph is definitely decreasing

everywhere so just to hopefully clear up

any confusion there so that's all there

is to it it's just the graph of

cotangent the only thing that has

changed is the period so the zeros the

zeros the first positive zeros at PI

over 4 and then we said the period is PI

over 2

so we'll move over multiples of PI over

2 to get the zeros and the vertical

asymptotes those occur at multiples of

PI over 2 so we can write that as PI

over 2 times n

again wherein is an integer

Graphing a Cotangent Function, EX 3

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 Cotangent Function, EX 3


all right in this video we're going to

look at some examples of adding and

subtracting fractions and to add or

subtract fractions we have to have

common denominators we need to have the

same number in the bottom part of the

fraction and if that happens well then

we just do the the arithmetic in the

numerator so a over C plus b over C we

just do the arithmetic a plus b all over

C and likewise if it's subtraction we

just do a minus B all over C if you

don't have a common denominator

ators you'll have to make common

denominators appear but in these first

examples here I think everything should

work out without too much trouble so we

have 4 over 11us 3 over 11 again we have

the same number in the denominator uh so

we just do the arithmetic in the

numerator so 4 Min - 3 will leave us

with 1 over

11 the same thing happens if you have

multiple fractions again as long as you

have the same denominator so in this

case we would get 7 + 4 + 3 again all

being divided by 10 so let's see 7 + 4

is 11 + 3 is going to be 14 over 10 and

I can reduce that so I'm going to

they're both even so if we divide top

and bottom by two that'll leave us with

7 over five and I'm going to leave it uh

as a fraction I'm not going to write it

as a mixed

number here we have 8 and 34 - 3 and 1/4

different ways you can do this um you

could just subtract the whole numbers

and subtract the fractions um in general

uh I tend to you know and this is

probably a bit um you know too much for

this problem again we could just do 8 -

3 which would be 5 34s - 1/4 well that

would be 3 - 1 or

24s so that would be one way to do it I

have a tendency to always just turn them

back and into fractions so 4 * 8 is 32 +

3 is 35 over 4 uh 4 * 3 is 12 + 1 is

13 and now we can do the subtraction uh

35 minus 13 that's going to leave us

with 22 over

4 which again reduces they're both even

into 11 over two which if we made it

into a mixed number that is five and2

so um in general um I do this method

just because uh you know some weird

things can happen with the fractions

I'll put it that way um you know you may

end up with a fraction over one and then

you have to carry a digit to the whole

number um things of this nature so in

general um you're going to see me do

this method although in some cases you

can

just well really in all cases you can

just do this subtraction or addition

normally but you have to be a little

careful I guess what I'm trying to say

is maybe if you had eight and 34s

plus you know five and another

34s in this case you know if you add the

whole numbers eight and five is going to

give you 13 and then you would do 34s

and 34s which would give you um 64s so

that would be you know so you would have

13 and really You' have another 64s

which you'd have to turn into one and a

half Etc okay so maybe more than you

wanted to hear but I'm going to uh to uh

kind of deal with these situations

you'll see me just write things um as

fractions from the beginning it may take

an extra step or two but not the end of

the world

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Adding and Subtracting Fractions / Mixed Numbers: Example 1


okay so another example here of using a

double angle identity to solve a trig

equation so here we're gonna solve sine

of 2t equals sine of T and again we're

gonna find inner solutions on the

interval 0 to 2pi this one will actually

be I think a little less involved than

the first one sine of 2t we can simply

write that as 2 sine T times cosine T

equals sine T and again be careful you

don't want to divide both sides by sine

T you would end up throwing some

solutions out the window so what we're

gonna do is we're simply gonna subtract

sine T from both sides set that equal to

zero and now we can factor the sine of T

out that'll leave us with two cosine t

minus 1 equals 0 and now all we have to

do is simply just take you know each

factor to set it equal to zero so we'll

set sine of T equal to zero

we'll take two cosine t minus 1 and set

that equal to zero well we can add 1 and

divide by 2 so now we're just thinking

about solutions I'm thinking okay where

does sine of T equals zero on the

interval 0 to 2pi well that would happen

at the angle zero at the angle PI and at

the angle 2 pi

cosine of T equals 1/2 let's see cosine

of the angle let's see I believe PI over

3 cosine of PI over 3 that gives us 1/2

and down in Quadrant 4 will also find

the angle 5 PI over 3 as our other

solution so in this case we've got 5

solutions to our original equation sine

of 2t equals sine of T get 0 PI over 3

PI 5 PI over 3 and our last solution we

found was 2 pi

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Using Double Angle Identities to Solve Equations, Example 2.
In this video, I use some double angle identities for sine and/or cosine to solve some equations. Great fun!!  For more free math videos, visit http://PatrickJMT.com


okay so one last example here using

these sum and difference identities so

suppose we want to find the exact value

of tangent of 75 degrees minus tangent

at 15 degrees divided by 1 plus tangent

of 75 degrees times tangent of 15

degrees

well okay certainly this looks like you

know our tangent identity here and again

it says if there's a negative sign in

the numerator it says we're going to use

the the difference identity

it says we'll take tangent of X minus y

and in this case our X is just going to

be the 75 degrees and the Y is just

gonna be our 15 degrees here so it says

simply we'll get tangent of the first

one 75 degrees again if there's a

negative in the numerator we we preserve

that sign in between so we're going to

keep it negative in between here so

really we're just evaluating tangent of

75 degrees minus 15 degrees which is

going to be tangent of 60 degrees but

hey that's just sine of 60 degrees

over cosine of 60 degrees sine of 60

degrees is gonna be root 3 over 2 cosine

of 60 degrees is gonna be 1/2 you can

always simplify this a little bit we can

leave the numerator alone we can flip

the denominator and multiply so in this

case the twos will simply cancel out and

we'll be left with our solution of

square root of 3

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Sum and Difference Identities to Simplify an Expression, Example 3.  In this video, I simply show how to simplify some expressions by using sum and difference identities.  Nothing crazy, just showing how one must know identities to make things more manageable!  For more free math videos, visit http://PatrickJMT.com


all right in this video I'm going to do

another example of solving a quadratic

equation by completing the square so in

this case we're going to solve x^2 + 4x

- 20 = 12 and again there's probably you

know slightly easier ways to to solve

this quadratic other than completing the

square but again we're just trying to

illustrate the procedure here um so in

this case notice we have you know x's on

one side we've got a number and then we

also have a number floating around on

the right side as well a lot there's

sort of different ways that you can do

these they're all equivalent but one way

to go about doing this would be to get

all the numbers on the right side so the

first thing I'm going to do is add 20 to

both sides of this

equation so that's going to leave me

with

X2 + 4x =

32 and now kind of the the completing

the square step uh the idea is we're

going to make the left side into a

perfect square by doing this following

procedure and you'll you'll see that

happen so whatever numbers in front of

the X the coefficient on the X what we

do is we take 1/2 always 1/2 of that

number so 1/2 of four is simply going to

give us two and then what we do with

that number we take that number and we

Square

it so 2^2 is going to equal

pos4 what you do is on the left side

we're going to write this as x^2 +

4x and there's our equal 32 equals 32

that was there from before what we're

going to do is whatever this whenever

you take 1/2 of the number and square it

whatever that new number is we're just

going to add that to the left side but

if we add it to the left side you also

have to add it to the right side to keep

everything uh equal and the I is now x^2

+ 4x + 4 factors it's a perfect square

it factors as x +

2 * x +

2 on the right side we have 32 + 4 which

is

36 and again if you take x + 2 um we can

X+ 2 * X+ 2 we can simply rewrite this

as x + 2 quantity squared equal 36 okay

so this is kind of a to me sort of phase

one this is the first thing you want to

get done is you want to make the left

side into a perfect square and yet

you'll have some number hanging around

on hanging around on the other side

that's fine so now the idea is to get

rid of the square we're going to take

the square root of both sides okay so I

should be a little careful here with my

notation because when you take um square

roots you get positive and negative the

square root of that number so I should

put a little plus minus you know in the

previous step since I already threw my

square root in

there

um what we're going to do is well we can

take the square root of 36 that's just

going to be 6 so we have um x + 2 =

positive or -6 is what we're trying to

say here we basically break this off

into two little equations now we're

going to make the equation x + 2 = POS 6

and then we're going to make the

equation x + 2 = -6 and we'll solve

those individual equations and that will

give us our solution to the original

quadratic equation so in the first one

we can simply subtract two from both

sides we would get X = pos4 we could

subtract two from both sides and we

would get x = 8 and now we've got our

two solutions to our quadratic equation

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Completing the Square to Solve Quadratic Equations: More Examples - 1


all right so just one more example here

finding uh components of of vectors

along other vectors and projections of

vectors onto other

vectors so we did part A here before so

we're going to do the same thing we'll

find the component and the projection of

b along a so in this one uh Vector a is

given by 1 0 0 Vector B is -6 comma 2

comma 1 so the first thing we'll do is

just find the the component

so again to find the component of b

along

a we take a. B and we divide that by the

magnitude of vector

a so the dot product is going to be

pretty easy we'll get 1 time ne6 which

will be

-6 um and then we'll just simply

multiply so we'll have 0 * uh 2 which is

0 we have 0 * 1 which is

0 if we take the magnitude of vector um

of vector a we'll do 1 2 plus well 0 2 +

0^ 2quared so that's simply going to

leave us with -6 over

1 or a value

of6 to figure out the

projection of vector b along a again we

just do a b we can divide that by the

magnitude of a s and then we multip

multiply that by Vector a so again we've

already figured out the dot product we

figured that was

-6 um if we take the original magnitude

which was one and we Square it well we

still get one and then we just multiply

that by Vector a which again just had

components of 1 0 0 so if we multiply

we'll be left with -6 comma 0 comma 0 as

our Vector projection

Orthogonal Projections - Scalar and Vector Projections - Example 2

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Orthogonal Projections - Scalar and Vector Projections. In this video, we look at the idea of a scalar and vector projection of one vector onto another.


all right in this video we're going to

find a power series representation by

doing a littleit of integration so we're

given the power series for cosine X is

x^2 um plus x 4 over 4 factorial - x 6

over 6 factorial Etc we want to find

what function is represented by the

series x^2 - x 4 over 3 factorial + x 6

over 5 factorial - x 8 over 7 factorial

Etc so what I'm going to do is integrate

both sides of our of our function

here okay so on the left side the

anti-derivative of cosine that's just

going to be

sinx on the right side um the

anti-derivative of one is just X then

we'll get x to the 3 notice we'll get 3

* 2 factorial um so I guess let's even

write that and then we'll get x 5 over 5

* 4 factorial minus X 7 over 7 * 6

factorial

Etc now when we integrate um again we

pick up this arbitrary constant I'm

going to call it C so to figure out the

C value uh we can plug in any value that

we want for x uh but it needs to be

something that's easy to compute so

notice if we plug in x equals z we'll

get S of 0 equal C + well we're plugging

in zeros for all the X's so those would

all just be

zeros but sign of zero we know that that

is zero so in this case our C value is

just going to be zero so what we have is

we have sin x is going to be x - x 3r

over well 3 * 2 factorial that's the

same thing as 3

factorial um 5 * 4 factorial faial

that's 5 factorial X 7th 7 factorial

Etc okay so let's let's what we're

actually trying to find a series

for notice we've got um you know X squ

notice the exponents are sort of one

larger um than the uh you know the

factorials in the bottom so the

exponents are a little bit larger than

the factorials okay so one thing we

could try to do is I guess we could

integrate again um but then we're going

to have like x^2 over two and X 4th over

4 factorial and then the factorials are

kind of messed up so I think the easiest

way just to increase the

exponents uh by one we can just simply

multiply both sides by

X and I think if we do that that's going

to give us exactly what we

needed because when we multiply by X

we'll just have well X2

- x 4 3 factorial + x 6 5 factorial - x

to 8 over 7 factorial so we're just

increasing all the exponents by one and

again that's actually the uh Power

series that we wanted to produce so

we've now got it so this original uh

function uh that we were trying to

figure out what well when we were trying

to decide what function is represented

by this power series we've now got it

this is just going to equal x *

sinx

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Power Series Representation Using Integration.
In this video, I find a power series representation of a new function and identify that function by integrating a known power series.


حسناً , هنا سنقوم بحل مثالين سريعين

متعلقان بإختبار الحد النوني للمتسلسلة المتباعدة

وبالنسبة  لي , هئا الإختبار مفيد جداَ  لمعرفة إذا ما كانت المتسلسلة  متقاربة أم متباعدة

إنها غالبا أول ما أفكر بة

غالباً أول ما أفكر بة

إذا , ما يقوله إختبار الحد النوني للسلسة المتباعدة هو

أن لديك متسلسلةِ ما بصيغةٍ ما

أ ن (ألف نون)

يقول الإختبار إذا أخذت نهاية عندما تقترب ال(نون) لما لا نهاية

ل (أ ن)

إذا لم تساوي تلك النهاية صفراً

عندها تكون المتسلسلة تلقائياً متباعدة

وذلك هو الأمر , قد إنتهيت  من هذه المشكلة فتنتقل للمشكلة التالية

أما إذا كانت النهاية تساوي صفراً

سيكون ذلك غير مؤدٍ لنتيجة

حيث أن السلسلة قد تكون متقاربة أو متباعدة , نحن لانعلم شيئاَ

إذاً , كن حذراَ لأنة خطأ شائع

المتسلسلات مربكة

لكثير من الناس كما أعتقد

بالتأكيد كانت مربكة لي

عندما بدأت بالتعامل معها لأول مرة

و مازالت المتسلسلات مربكة كما أعتقد في كثير من الأوقات

كما تعلمون , الناس يقعون في الفخ

ويقولون إذا لم تساوي النهاية صفراً

فتذكر إنها متباعدة

وبعدها سيقولون بتفكير خاطئ

إذا كانت النهاية تساوي صفرا . فأعتقد أن السلسة متقاربة

لو كانت الأمور بتلك البساطة وكان ذلك صحيحاً

لم نكن لنزعج أنفسنا في عمل كل هذة الإختبارات والمطاليب المختلفة التي نقوم بها

إذا مجدداَ , الإفادة الأولى  هي

هي ما سنستخدمة هنا

إذاً هنا نريد ان نكتشف ما إذا كانت المتسلسلة التالية

متقاربة أم متباعدة

إذاَ

إذاً  للجزء (أ) هنا

لدينا النهاية.. أقصد لدينا  المتسلسة  من ن= 1 إلى مالا نهاية

ل ل ن^2+3ن+1  على 2ن^2+4

حسنا,  إذا أخذنا النهاية حيث (ن) تتجة لمالانهاية

ل ن^2+3ن+1  على 2ن^2+4

لقد رأينا هذه الأنواع من النهايات

إذا كانت درجة البسط مساوية لدرجة المقام

حيث يتساويان في هذه الحالة

نستطيع أن نستخدم النسبة للمعامِلات الرئيسية

إذا في هذه الحالة ,  النهاية تساوي نصف

والذي قطعا لا يساوي صفراً

وهذا يخبرنا بأن المتسلسلة متباعدة

حسناً , سهل بما فية الكفاية

إذا في الجزء (أ) نقول ببساطة أن المتسلسلة متباعدة

في الجزء (ب)  إذا لاحظنا

كما تعلمون ,  لدينا المتسلسلة من ن=1 إلى مالانهاية

ل اللوغاريتم الطبيعي

لِ جتا ن (جيب التمام ل ن)

حسنا , نأخذ النهاية حيث تتجة النون لمالانهاية

ل الوغاريتم الطبيعي ل جتا ن

و قانون النهايات يقول بأنة

يمكننا  ببساطة أن نضع النهاية في الداخل

حيث يكون لدينا اللوغاريتم الطبيعي للنهاية حيث (ن) تتجة لمالانهاية

ل جتا ن

لكن ريثما تتجة النون لمالانهاية

جتا تبقى متذبذبة

إذاً هذه النهاية ببساطة غير موجودة

وإذا لم تكن موجودة

هي قطعاً لا تساوي صفراً

إذا في تلك الحالة مجدداً

يمكننا القول بأن السلسلة متباعدة

طبعاَ , كما تعلمون

إذا كانت النهاية تساوي صفراً

مجدداَ في تلك الحالة لم نكن لنستطيع قول أي شيئ

لا نستطيع أن نقول أن السلسلة متقاربة أو متباعدة

بعد ذلك , يجب أن تبدأ بالتفكير بمختلف الإختبارات التي تعلمتها

إختبار التكامل

المقارنة المباشرة , مقارنة النهايات , إختبار النسبة , إختبار الجذر

سوف تكون هناك كل هذه الأنواع المختلفة من الأختبارات للتجربة

إذا كانت النهاية تساوي صفراً

إذاً, إذا كانت النهاية لاتساوي صفراً فإن السلسلة متباعدة

ولقد انتهيت!

إذا كانت النهاية تساوي صفراً

سوف تحتاج لعمل أكثر نوعاً ما

لأتك ستحتاج لتكتشف طريقة لتتعامل مع السؤال

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Test for Divergence for Series, Two Examples.  In this video, I discuss the test for divergence and show two examples of series who diverge by using the test for divergence.


all right in this video I'm going to do

another problem involving the

Pythagorean theorem

so in this one suppose we have two

cyclists and they start from the same

location

suppose one cyclist is going due north

and the other is going due east and

suppose somehow magically they're going

exactly the same speed the whole time

suppose after two hours we know that

there are 17 times the square root of

two models apart and we want to know how

fast each is going in miles per hour

okay so we want the account for this

fact that they've done this two hours so

that'll come into it but at the moment

you know so sometime later the cyclist

going north will be here the cyclist

going east will be here we know that the

distance between them is 17 times the

square root of 2 since they were going

the same speed they went the exact same

distance so we can call each side of the

right triangle we can say that each one

of those distances is X and now we can

just use the Pythagorean theorem it says

x squared plus x squared would have to

equal 17 square root of 2 squared so

we've got x squared plus x squared

that's going to give us 2x squared we've

got we would have to take the square

root of 17 times the square root of 17

excuse me right we would have to take

simply 17 times 17 I'm looking at the

square root part there

so that's 289 so 17 times 17 is 289 we

do have to take the square root of 2

times the square root of 2 which would

be 2

and then we get simply divide both sides

by two and that would give us x squared

equals 289 and again we know well I mean

we squared it to get it so if you square

root both sides again normally we get

positives negatives but these clearly

have to be positive values the square

root of 80 to 89 is just 17 so what it

says again this is in two hours they've

each gone a distance of 17 miles so that

means the speed of each is 17 divided by

2 which would be 8 point 5 miles per

hour

so each cyclist is going 8.5 miles per

hour over 2 hours if they do that though

end up being 17 times the square root of

2 miles apart so nice leisurely bike

ride

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Word Problems Using the Pythagorean Theorem - Example 2


all right in this video I want to talk a

little bit about graphing sine and

cosine functions with phase shifts and

if you remember horizontal shifts from

you know from just regular algebra I

mean it's the exact same thing all we're

doing is just shifting graphs

horizontally except now assigning cosine

we call them phase shifts so these

definitely have some applications and

physics at least I know and I'm sure in

other places as well so basically what

we're going to have to do is you know we

could have some amplitude out front

whatever we've either got sine or cosine

inside the parenthesis if your variable

X or R is originally has a coefficient

you kind of have to factor that out

within the parenthesis but then whatever

is left over this plus K that's going to

help us figure out the phase shift and

remember if it's on the inside if it's

actually positive K that moves you K

units to the left and if the inside is

actually a negative K its kind of

opposite and we go K units to the right

so let's talk about a couple of these so

we're just going to figure out the phase

shift for each one of these following

functions so all right I've got cosine

of X minus 4 okay I see a negative 4 so

negative normally makes me think kind of

to the left in terms of a number line

but actually this is just going to be

the regular graph of cosine shifted

four units to the to the right again

kind of the opposite direction so notice

the next one we've got sine of two times

the quantity x plus one again I'm just

looking at this number

notice the twos outside the parentheses

and that's what we want we're going to

have to for example modify our next

example here down here but since I see a

plus one again normally positive the

positive makes me think to the right but

in this case again it's sort of opposite

so we're going to shift it will be

shifted in this case one unit

well since it's positive we're going to

go one unit to the left last but not

least here we've got cosine of ten X

plus thirty all I'm going to do is I'm

going to factor the ten out okay it

still has to stay inside the parentheses

though I can't like pull it out front

that's definitely incorrect so be

careful about things like that well if i

factor the ten out we would have so ten

out front and then it looks like X plus

three in the parentheses and again the

same thing I see a positive three but

now I'm really just shifting it's going

to be shifted

the original graph of cosine would be

shifted three units to the in this case

the left okay so again I said the

original graph of cosine actually it

would be different than plain old cosine

because of this this ten in there that

would change the period but yeah this is

the basic idea to figure out just the

horizontal or the phase shift

just make sure within the you know sort

of within the sine or cosine you know if

there's any coefficient factored out and

then just remember this trick that's

sort of if it's on the inside of the

parentheses you kind of have to do the

opposite so I see positive but it

actually moves in the negative direction

which would be to the left

Graphing Sine and Cosine with Phase (Horizontal) Shifts, 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 !! In this video, I show how to find the phase shift (aka, the horizontal shift) of a couple of trig functions.  In the next video, I actually graph a trig function by thinking about phase shifts.


all right in this video we're going to

look at solving uh another uh

differential equation by using

integrating

factors so here we've got dy over DX +

2x over 1 + x^2 * y = 4 1 + x^2 squared

so the first thing we'll do is calculate

our integrating Factor so we'll do e to

the integral of 2x over 1 + x^2

DX so kind of off to the side here I'm

going to integrate 2x over 1 +

x^2 and to integrate this I think we can

just do a nice little U substitution we

can let u = 1 +

x^2 du is going to give us 2x

DX so after we do our U substitution

we're really just integrating one over U

du and that's going to give us the

natural logarithm of U plus

C so so that's going to be the natural

logarithm of we can replace the uh

absolute value with uh just parentheses

1 plus X2 is always positive plus

C okay so really when we integrate we'll

get e to the natural logarithm of 1 + X2

so again we're trying to find a

particular um integrating Factor doesn't

have to be the most General so again we

don't have to worry about the plus C

here all right so let's see we' got e to

the natural logarithm of 1 + x^2 well

that's just going to leave us with 1 +

x^2 so that'll be our integrating Factor

so we're going to multiply both sides by

1 +

x^2 so we'll have dy over DX + 2x over 1

+ x^2 * y we're going to multiply all of

that by 1 + x^2 and we'll do the same

thing on the right side as well

so again the left side we can write as

the derivative with respect to X of 1 +

X2 all multiplied by

y on the right side notice the uh 1 plus

x^2 will cancel out with one of the 1

plus X SS in the denominator so that's

going to leave us with four time or

excuse me four over the quantity 1 +

and again what we're going to do is just

integrate simply both sides with respect

to

X all right so again on the left side

we'll just be left with 1 + x^2 * y on

the right side well the anti-derivative

of uh you know so I've pulled the four

out the anti-derivative of 1 1 + x^2

that's just going to be ar tangent of X

I'm going to add on the plus C and now

we can get y by itself by just dividing

so we could write our solution as for

arct tangent of x + C all divided by 1 +

❖ Integrating Factor to Solve a Differential Equation ❖

Integrating Factor to Solve a Differential Equation.
In this video, I look at an example of using an integrating factor to help us solve a differential equation.
In this video our differential equation is already in the proper form so we simply need to multiply both sides by the integrating factor and then integrate.
Notice that the left side of the equation simplifies nicely as y multiplied by the integrating factor after integrating both sides (this is the magic of the integrating factor!).

#differentialequation #maths #integratingfactor #exactdifferentialequation #exactdifferentialequations #ordinarydifferentialequations #patrickjmt


all right in this example we're going to

sketch the graph of the ellipse y

squared over 4 plus X minus 5 squared

equals 1 so what we'll do is basically

we're going to find the center and we're

going to find some the vertices and also

the co vertices and we'll just make a

rough sketch based on that so we could

write this as Y minus 0 quantity squared

we can write 4 as 2 squared then we have

X minus 5 squared we could put that over

1 but again we could even just write 1

as well 1 squared and again I'm just

putting this you know in the standard

form of an ellipse so the center of our

ellipse is going to be positive 5 comma

0 so we can now read that off from our

standard form and to get the vertices

well I'm even going to go ahead and

start making a rough little sketch based

on this and point out some other things

as well so 1 2 3 4 5 so there's going to

be the center the center of our ellipse

now to get the vertices we can basically

look at our a value and our B value

again a is greater than B so in this

case our a value is going to equal

positive 2 so the way I always again

just kind of thought about things so I'm

going to go in to put 5 0 there okay the

number that's being squared is 2 again

that's our a value and you know that's

underneath the y squared term the term

that involves Y squared that tells me

how many units we go up and down from

our Center so I'm going to go up 1 2

units and then I'm going to go down 1 2

units and those will give us the

vertices of our ellipse so we've got 5

comma 2 and then our other point is

going to be here at 5 comma negative 2

well likewise if we look at our beef

you are B value is going to equal

positive one and that tells me how far I

move left and right from where it's

centered to get my co vertices so from

the center it says move left one unit

okay and then it says move right one

unit as well so the co vertices will be

at 4 comma 0 and also at 6 comma 0 and

now again I'm just going to make a rough

little sketch of my ellipse and not

quite to scale but not too terrible so

all right so again that's what I like to

do I pick out the center and then again

I look at my a value or my B value and

again if it's underneath a you know I

look at the I look at the numerator that

kind of tells me again hey that's how

many units I either go up or down and

again for example if you look underneath

the term involving X the B values 1 well

X I think about the x-axis that's kind

of to the left and to the right so that

helps me remember sort of what direction

to go to get my other points

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, Ellipse, Shifted: Sketch Graph Given Equation.  In this example, we sketch the graph of a shifted ellipse.


okay in this video I want to do some

slightly harder uh examples of

evaluating exponential

Expressions so in the first one we have

2 + the quantity 1 - 5^ 2 + 3^ 2

typically what I like to do is just work

inside the parentheses first if there's

any other clear simplifications I'll do

it the two I'm just going to leave him

hanging out front inside we have 1 - 5 1

- 5 is going to give us -4 again that's

being

squared and I'll think I'll go ahead and

simplify we have 3 squar which means 3 *

3 3 * 3 is going to give us the number

n and couple more steps here so ne4 * -4

will give us a positive 16 and then we

still have our 9 hanging out

there well 2 + 16 is going to give us 18

18 + 9 gives us the number 27 so our

first expression will reduce to the

value

27 let's look at our next one here we

have 7 minus the quantity 4 - 2 cubed +

-2 cubed I'm going to do the same thing

I'm going to leave the 7 out front I'm

going to take 4 - 2 which is 2

cubed likewise we have -2 to the 3 power

and again that's -2 * -2 which is which

is pos4 * another -2 will give us

A8 okay

so we then have 7 - 2 the 3 power we can

replace that simply with 8 okay so 2 the

3 power is 8 we have a positive and a

negative so a positive and a negative

remember that turns into a

negative and there's our eight still

just hanging

out and now we're almost there okay so 7

- 8 7 - 8 is going to be -1 we still

have our other - 8 there and -1 - 8 is

going to give us

-9 as our final

solution

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Evaluating Exponential Expressions (Numbers Only, No Variables) - Ex 2


all right in this video we're just going

to look at a couple examples of adding

whole numbers so in part A here we've

got 61 + 14 and what I like to do uh you

know when I add these is I like to just

write them vertically and I've got

everything lined up in the sense that I

have all the ones and a column and all

the tens in another column and when we

add these you know if we have to carry a

digit to the next column we'll do so but

other than that hopefully you know

relatively straightforward so um well

you know if we add the stuff in the ones

column 1 + 4 is five so I'll just drop

that down and then I just go to the next

column 6 + 1 is 7 so it says Hey 61 + uh

14 that's going to equal

75 we'll do the same thing for our next

example 77 +

18 now here we have to be a little more

careful um you know if we do 7 +

8 well 7 + 8 is 15

so what we're going to do you know 15

would be the uh ones digit so the 15 is

the ones digit that's the digit we're

going to write down so I'm going to

write the one's digit in the ones

column um and then our one would be our

10's digit so that's what I'm going to

carry to the next column so that one is

getting carried to the next column um so

then I'm just going to add again

everything in my column so 1 + 7 is 8

plus 1 is 9 and it says 77 + 18 that's

going to give us 95

Adding Whole Numbers 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 !! Adding Whole Numbers ; Adding Two Digit Numbers.  In this video, we look at two examples of adding two digit numbers.


okay so here we'll have one more example

of solving an equation

that'll take a few steps to uh to

simplify

so we have two times the quantity r

minus three

plus four minus five r equals sixteen

so again my goal is you know basically i

want to get rid of the parentheses

combine all my like terms get the term

involving

r on one side a number on the other

and then divide by whatever the

coefficient

is so it's kind of a rough outline

quickly of what we're going to do

so to start off to get rid of the

parentheses i'm going to distribute so 2

times r will just be 2r

2 times negative 3 will be negative 6.

plus 4 minus 5 r

equals 16. and now i'm going to combine

like terms

so i see a term involving r and a term

involving r

what i'm going to do is just take the

coefficients so we have a 2 and a

negative 5.

so 2r minus 5r will be negative

3r and then we have constants we have

our negative 6

plus 4 well negative 6 plus 4 will give

us

negative 2. and again i'm going to leave

the 16

on the right side nothing really to do

with it for the moment

i want to get the term involving r by

itself

so what i'm going to do is add 2 to both

sides

since i since i'm subtracting 2 we'll

add 2

we'll have well negative 3r equals

18. and to get the r by itself since

we're multiplying

by negative three what we'll do

we'll just divide uh both sides

by negative three so negative three over

negative three is one

we're left with r on the left and a

positive over a negative will give us a

negative

eighteen over three is six

so our solution will be r equals

negative

six so again kind of the basic outline

get rid of the parentheses

combine like terms get the term

involving your variable isolated

and then just divide by the coefficient

Solving Multi - Step Linear Equations , Another Example

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving Multi - Step Linear Equations.  Here we solve some linear equations that require a few steps to solve.


all right in this video I'm going to do

another example of finding the center

radius form of a circle so again the

equation of a circle with Center h k and

radius

R is going to be the equation x - h^2 +

y - k^2 = r^ 2 so we basically have to

fill in the value of H the value of K

and we'll have to take R and square it

as what we have to replace in this

formula so in this example suppose we

want to find the equation of a circle

that's centered at -4a 5 if we know that

the point 1A 7 is also a point on the

circle well already when we do our x -

the x coordinate squar + y - the y

coordinate squared equals the radius

squared I already know the center so I

could simply drop in here -4 so if I

plug in my -4 two negatives will make a

positive4 likewise if I plug in my

positive 5 I'm going to have a y - 5^ 2

the only thing we don't really know

still is the

radius but since we know this 17 is on

the circle and it goes from the circle

to the center well the distance between

these two points would uh would be the

the the length of the radius so we need

to find the

distance between our two given

points

-45 and

the17 and the distance formula simply is

just the square root again we take the x

coordinates and subtract them so -4 -1 2

plus then we subtract the Y coordinates

5 - 7^

SAR and if we simplify

here -4 - 1 is -55 * 5 is POS 255 and

then if we have 5 - 7 that's going to be

-2 -2 * -2 is pos4 so the distance

between those two points is going to be

the square < TK of 29 but again we said

in this particular case this this

distance between these two points that's

actually going to be the value of the

radius so now we're pretty much there

all we have to do is just simply replace

our r with the sare < TK of

29 so it says our final answer it says

the equation of the circle will be x +

2ar + y -

5^ 2 and then it says the radius is the

square < TK of

29 so we'll have the square < TK of

29 quantity squared but obviously uh we

can simplify the right side pretty

easily thek of 29 s is just posi 29 and

now we've got our solution x + 4 2 + y -

5 2 = 29 is going to give us the

equation of the circle that satisfies

the original given

conditions

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Center-Radius Form for a Circle - A few Basic Questions, Example 2


okay so one more example here finding an

interest rate to match certain

conditions so

suppose you deposit nine grand the

account

compounds quarterly at the end of 10

years the balance has doubled we want to

know what the interest rate is

okay so i'm going to start off just by

filling in what i know

so okay so we know the starting

principle is nine

thousand we know that the final amount

has to be double

which would make that eighteen thousand

one plus the rate well we don't know

that it compounds

quarterly so i'm going to divide by four

four times per year

and then 4 times 10 because we're doing

this for 10 years

so a couple things i'm going to do i'm

going to divide both sides by 9000

so 18 000 over nine thousand you can

cancel out the zeros eighteen over nine

is two

the nine thousands cancel then i have

one

plus r over four well four times ten

will be to the fortieth

power and what i want to do now again is

i want to get the

get rid of this exponent of 40.

so what i can do is i can raise both

sides to the

1 over 40 power or equivalently i'm

taking a 40th root

so i'm going to have to take 2 to the 1

over 40th power

let's see so

so 1 divided by 40 that looks about

right

on the right side the if we multiply

we'll just get to the first power so

we'll just have 1

plus r over 4. and now i'm going to take

2 and raise it

to the point .025

so i'm getting uh 1.017479692

that's going to equal 1 plus r

over four okay well we can simply

subtract one from both sides in this

case

so if we subtract one from both sides

we'll just be left with zero point

zero one seven four seven nine

six nine two equals r over four

now i'm simply going to multiply both

sides by four

and let's see here so if we multiply

all of that by 4 i'm getting the number

on the right side the 4s would cancel so

that's going to be our interest rate as

a decimal and that tells me as a

percentage

it says that the interest rate is

roughly so we move it over twice

percent we'll round it off to two more

spots so

it says if you basically have almost a

seven percent interest rate

under those given uh under these given

conditions

you'll end up having a

an accumulated amount of 18 000

if you deposit an initial amount of 9

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding an Interest Rate to Match Certain Financial Goals, Ex 3.  In this video, I do an example of where one has to find the correct interest rate when given other requirements on time, amount earned, etc.!


okay in this video I just want to look

at an example of finding the inverse

lass transform of a function involving a

factor of the form e to the

Cs so here we we're going to be given

that capital Y of s = e to3 * s / S2

sorry the little two kind of got

squeezed in there we want to compute the

inverse Lelo transform of e to3s /

S2 so I've jotted down a couple things

here so again just the the formula which

we proved in a different video the Lelo

transform of f of T minus C multiplied

Capital H of T minus C we said that

that's equal to e the Cs multiplied by

capital F of s and again H is our he

aide

function so okay so what have we got

here if you take the inverse lass

transform of the right side so that's

what I'm doing I'm just saying hey you

know we've got this original expression

let's take the inverse lla transform if

you take the inverse lass transform of

the left side as well they just undo

each other and you get back with that

that original function that you started

with so just to again point that

out so we've got the form so we had e to

the

ne you know e to the --3s over s^2 I'm

just breaking that up to make it crystal

clear so we were trying to find the

inverse plus transform of E-3 s * 1 s^2

well that means that our C value is

going to be equal to 3 so c =

3 so I can already write you know we've

got h of T minus 3 and then we're going

to have okay we've got this F of T minus

3 and this is the part that we still you

know have to have to figure out so once

we can figure out F of T minus 3 we're

done okay that's the last thing we have

to do well recall I mean to get F of

t f of T is the inverse lass transform

of f of

s so to get F of T we just take the

inverse lass transform of f of s well F

of s is our function here 1 / s^2 so

we're just using our inverse lass

transform of 1 /

s^2 and again just by using a table of

values so if you've got a table of llao

transforms the inverse Lao transform of

1 / s^2 that's just going to be equal to

T so again to get this I'm just using a

table of lus

transforms okay so it says we're really

almost done we're pretty much there

these ones uh aren't too bad so our fun

function f of

t f of T is just actually equal to T

well if F of T equals to

T if we replace our t with tus 3 well F

of T minus 3 is just going to equal T

minus 3 and now we have our solution it

says that

the inverse lla

transform of that original function e to

the

3 S / s^ 2 it says that's going to be

equal to F of tus 3 which we said in

this case turned out to be exactly just

T minus 3 and then we multiply that by h

of T minus C so we would have H of T

minus 3 because again our C value is

equal to 3 and that would be our answer

in this

case

Inverse Laplace Transform Involving Heaviside Functions

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! In this video I find an inverse Laplace transform of a function involving the term e^(-cs) using Heaviside functions.


okay here we're going to use our

compound interest formula to find the

the starting or the principal amount

invested suppose we know that the okay

so we're going to find the principal

amount invested if we know the amount in

a semiannually compounded account with a

4.25% interest rate is

$6,000 after 90 months so you you've

accumulated a total of

$6,000 after 90 months compounded

semiannual at

well again we can just fill in our

formula the accumulated amount is going

to be

6,000 the principle is what we don't

know that's what we're going to figure

out one plus our rate so our rate is

going to be

zero 0.0425

we're going to divide that by the number

of compoundings per year which is

two and then we'll do two multiplied by

the number of years well 90 months if we

take 90 and divide that by 12 let's see

that's going to give us so 12 will go

into 97 times that'll leave us with six

months out of 12 left over or 7.5

that'll be the number of years

so now I'm going to simplify the right

side down a little bit we've got

6,000 equals the

principal if we

do if we do

0425 divided by two if we add that to

one we'll get

one25 well 2 * 7.5 is going to be 15

so again I'm going to simplify we've

got we've got 1.02 one25 I'm going to

raise that to the 15th power again I'm

just using a calculator here off to the

side I'm getting

08 22 after rounding so to solve for p

the principal or starting amount I'm

just going to divide both sides by 1.37

0822 1.

37822 and on the left side let's see so

6,000 divided by

822 I'm getting that to be

4,376 and 94 cents after rounding so it

says the starting amount was close to

but not quite equal to $4,400

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Compound Interest Example - Find Starting Principal.
  Here we are told that an account accrues interest semiannually at a rate of 4.25%.  If the account has a total of $6,000 after 90 months, what was the starting principal?


all right here we're going to look at a

little question related to radioactive

decay so suppose a radioactive material

is known to Decay at a yearly rate of

0.2 times the amount of each

moment and suppose uh so in part A there

are a th000 grams of the material right

now we want to know what's the amount

after 10

years so again when things change at a

rate proportional to the amount uh that

tells say that the the original the

amount at uh sometime T equals a this is

going to represent the initial

amount times e to the K T so in part A

here we're given that the initial amount

is

1,000 it's uh decaying at a rate of

0.2 uh K represents the K rate or the uh

growth rate we'll even call it

that sometimes called a growth rate

sometimes also called a Decay rate the

main thing is since it's decaying uh

we'll use

0.2 * T and uh that's that's going to

describe the amount at any time T so in

part A we want to know how much of the

material is there after 10 years well

all we have to do is just plug 10 into

this

so we have 1,00 * e to the

0.2 *

10 so that's going to give us 1,00 * e

to the ne uh I guess just

-2 and now we can evaluate

that so let's

see I'm getting e to the -2 to be

roughly

135 so if we multiply that by ,000 we'll

have well 135 so it says the amount uh

left after 10 years will be roughly 135

I don't know if we had grams at the

beginning yes it will be roughly 135

gram of this material left over

Radioactive Decay and Exponential Growth: Quick Example Involving Exponential Decay

Question: A radioactive substance decays at a yearly rate of 0.2 times the amount each moment.  If we start with 1000 grams, how much remains after 10 years?
This is a basic question related to exponential growth / decay.   Here we know how quickly the material decays and want to know how much remains after ten years.
We use the relative decay rate, the initial amount, and time to compute the solution.

#math #algebra #khanacademy #khan #geometry #education #montessori #homeschooling #functions #notation #organicchemistry #organicchemistrytutor
#logarithm #naturallog #exponential #decay #growth #exponentialdecayproblem


all right in this video I want to do one

more example of finding the x and

y-intercepts of an equation so in the

other video I did the first part we

found the X and y intercepts of this

function now we're going to do the same

thing we'll just find the x and y

intercept of this function y equals x

squared minus 3x minus 18 again the idea

is to find the y-intercept we simply let

X equal 0 and we just do the arithmetic

so we would get y equals well 0 squared

minus 3 times 0 minus 18 which just

simplifies to negative 18 so it says the

y-intercept is at negative 18 or we can

say that 0 comma negative 18 is on the

graph ok that would be our y-intercept

now to find the x-intercept again or

intercepts I should say if there are any

well now we substitute in y equals zero

and we simply solve for X so we have

zero now on the left side equals x

squared minus 3x minus 18 so again now

basically I think well I've got a

quadratic equation I have to solve

let's see hopefully this factors so I

need two numbers that multiply to

negative 18 but add up to negative 3

I think negative 6 and positive 3 would

be two numbers that have the two numbers

that have that property so now we've got

a factored so it says we just said X

minus 6 equal to 0 X plus 3 equal to 0

if we solve the first equation we just

add 6x equals 6 is one of them cracked 3

and we'll get x equals negative 3 is our

other solution so now we found our two x

intercepts at the point 6 comma 0 and

there's a second x intercept at negative

3 comma 0

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! X-Intercepts and Y-Intercepts of a Functions and Finding Them! Example 2


okay here we're going to find the

equations for the hyperbolas asymptotes

and we're going to write these in

point-slope form so we have the

hyperbola X plus 16 quantity squared

over 100 minus y the quantity Y minus 9

squared over 400 equals 1 so this is to

me just kind of writing things in a

proper form and just knowing some some

corresponding formulas so kind of the

formulas that we're going to use we're

going to write our at the hyperbola in

the form X minus H quantity squared over

a squared minus y minus K squared over B

squared equals 1 and the idea is once we

write it in this form this standard form

it says the asymptotes it says the

asymptotes are going to be of the form Y

minus K equals positive and negative B

over a times X minus H okay so again

what I need to do is I need to figure

out a value for K I need to figure out a

value for B I need to figure out a value

for a and I need to figure out a value

for H all right well let's do that okay

so I can rewrite this so it says we're

going to write it as X minus something

so X minus well we want it to be

equivalent to X plus 16 so we can write

it as X minus negative 16

well squared I want to write the

denominator as some number squared well

I want that number to equal 100 well

what number squared is going to be 100

and you could say well hey could be

negative 10 but we always let our a and

B values be positive so we're going to

use positive 10 squared and then we've

got well we've got Y minus 9 we can

leave that part alone this is all over 4

100 again I want to write that as a

number squared a positive number being

squared well we can write 400 as 20

squared so now I've got all my values

okay it says this is going to be our

value for H H is going to equal negative

16 it says 10 is going to be our a value

it says positive 9 again we kind of

forget about the the negative so we

could write this as Y minus 9 okay so

that's going to be our K value and then

20 is going to be our B value so I'm

just going to fill that in and we'll

have our asymptotes so it says our

asymptotes will be Y minus K which is 9

equals positive or negative B over a so

again we said B had value 20 a was equal

to 10 then we do X minus the H value so

the H value is negative 16 and of course

we could simplify this down just a

little bit but now we've done everything

so it's just kind of again just writing

things in the proper form and just

knowing formulas so 20 over 10 is going

to be 2x minus negative 16 will be X

plus 16 and now we have we have our

asymptotes

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, Hyperbola , Shifted : Find Asymptotes.  Here we find the point-slope form of the asymptotes of a hyperbola.


all right here we're gonna look at some

examples of rewriting fractions as

decimals and if we have to we'll round

off all right basically to write a

fraction as a decimal you just do long

division that's kind of the whole the

whole trick okay so we've got seven over

eight seven divided by eight so again we

can write that as well

7 divided by 8 and I'm gonna go ahead

and write seven I'm gonna add in some

some decimal places so I'm gonna write

this as seven point zero zero zero tens

hundreds thousands maybe one more place

in case we have to round okay so usually

if somebody doesn't say round to a

certain place I'll just put some number

of zeros down and if I need more I'll

add more so okay so eight will go into

seven zero times

so zero times eight zero again we can

just subtract 7 - 0 7 and we'll drop

down our zero so now I think eight will

go into seventy how many you know how

many times without going over let's see

I guess 8 times 8 is 64 I think that's

gonna work and again we just subtract 70

minus 64 will leave us with six we can

drop down our zero so now let's see

eight will go into 60 I guess

seven times seven times eight will be 56

again we can subtract 60 minus 56 will

leave us with four drop down our next

zero well eight will go into 40 exactly

five times and now five times eight

would be 40 we could subtract we would

get zero but then we would just simply

drop down you know our neck Siro and we

would say well eight goes into 0 zero

times and we would just now start

getting zeros forever and ever and ever

or

since there's no remainder that tells us

that we have a fraction that we can

write using a decimal representation

where the decimal terminates so in this

case again we just keep the decimal

place exactly lined up so it says we can

write seven over eight as the decimal

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Writing a Fraction as a Decimal, Example 1


all right here we're going to look at uh

some mixed numbers and what we're going

to do is we're going to write the number

31 / 4 as a mixed number so to do this

um I'm just going to write it as long

division we've got 31 / 4 and to do this

I think well okay so let's see four

certainly won't go into three uh four

will go into 31 I guess seven times so 7

* 4 is

28 and then we subtract uh 31 - 28 is

going to be 3 so this is our

remainder all we have to do is we take

whatever numbers on top which is

seven and then we take our remainder

which is three and we divide it by the

same original divisor so it says 31

over4 says we can write that as 7 and

34s so that's the way to write it as a

mixed number

I'm going to go back sort of the

opposite direction suppose we started

with the uh you know the right side and

we wanted to write it just as a a

fraction a single fraction with no whole

number out front that's what I'm going

to do next so I'm going to convert 6 and

23 to what we call an improper

fraction the way that we do

this is pretty easy what we do is we

take the bottom number and we multiply

so we say 6 * 3 and then we add it to

the top number so we're going to

multiply and then add okay so let's see

6 * 3 plus two is what we're doing so

that's going to give us 18 + 2 which is

20 and what we do is when we rewrite

this number this number that we have

here is what goes in the numerator of

the fraction and again we keep the same

divisor okay so when I look at this I

say well 3 * 6 is 18 + 2 is 20 I get 20

over3 and now we have our mixed number

written as an improper fraction

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Improper Fractions and Mixed Numbers : Switching Between Them.  Here we take an improper fraction and write it as a mixed number and then take a mixed number and write it as an improper fraction.


okay so one more example here of

integrating a trig function so here

we're gonna integrate cosine x times

sine of cosecant X all over sine squared

so you know kind of when I look at a

problem like this again the only thing

you know the only way I can integrate

sort of sine is just to get it to be

sine of a single variable

you know the sine of cosecant X I'm not

really sure what to do with it

so in a sense I think well the only way

to get rid of it would be to use au

substitution so I'm gonna let u equal

cosecant X and we'll see what happens

here so recall the derivative of

cosecant X is negative cosecant x times

cotangent X well I don't see that in the

problem immediately but recall you know

cosecant again that's just 1 over sine X

cotangent X that's cosine X over sine X

and well I guess let's not forget our DX

on here if you multiply this will get

negative cosine X over sine squared X DX

okay that is in the problem we have

cosine X over sine squared x times DX we

don't want the negative so we can just

multiply both sides by negative one and

now it looks like to me are you

substitutions gonna work out rather

nicely

so we would have the integral of sine of

U so we replacing the cosecant X with

you and again the cosine x over sine

squared DX all of that's just gonna

equal negative 1 D U and now we can

integrate so the antiderivative of sine

is negative cosine but what the extra

negative will just get positive cosine

of U plus C and now all we have to do is

just sort of free substitute so again u

was equal to cosecant X so it says our

antiderivative will be cosine of

cosecant X plus C

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Integrating Trig Functions , More Examples #3.  Here we integrate a trig function using a u-substitution.


all right in this video we're going to

look at determining the limit of a few

different sequences so in part A we have

uh the sequence associated with one over

n cubed so what we'll do is again we

just take the limit as n goes to

Infinity of uh one over n cubed well as

n gets really really really big the

denominator is going to get really

really really

big we're going to have one over a very

very big number so one over a large

number gets arbitrarily close close to

zero so the limit of the following uh

would just simply equal

zero um Part

B same thing we'll just take the limit

as n goes to

Infinity so recall the shortcut um if

the degree of the numerator is equal to

the degree of the denominator we just do

the ratio of the

coefficients um so I can just simply

look at this and say well this is 1 n^ 2

and 3 N2 the degree of the top equals

the degree of the bottom we'll get 1/3

it's a very useful little shortcut to

remember when you do sequences in series

but again kind of to show it with a

little bit more work we can divide

everything by n^ 2 so n^2 over n^2 is 1

2 N over n 2 will be 2 over n then we'll

have 4 over n 2 3 n^ 2 over n^2 is 3

then 1 over n^2 and again the idea is

now it's more clear because as in goes

to Infinity this term will go to zero

the 4 over n^2 will go to zero the 1

over n^2 will get arbitrarily close to

zero so again we're just left with

1/3 as the limit uh for the second

sequence last but not least here we have

n the 3 * e the

N if we do a limit as n goes to

Infinity

notice we'll get uh you know as n goes

to Infinity n the 3 will go to Infinity

but we'll get e to the negative infinity

and if you take e and raise it to a uh

quote unquote big negative number that

gets close to zero so this is going to

be an indeterminant product so you may

have to use lal's rule on these

problems I'm going to leave the N cubed

in the numerator I'll put the e to the

negative n in the denominator but then

that would uh be written as just e to

the N so now if we use lal's

rule well we'll get 3 n^2 over e to the

N so the derivative of the top and the

derivative of the bottom this is still

Infinity over infinity an indeterminate

form we can use lal's rule

again we'll get 6n over e to the N again

as n goes to

Infinity both top and bottom will go to

Infinity

so we can use lal's rule one more

time then we'll be left with six in the

numerator we'll still have e to the n in

the denominator but now this is going to

be six over a really really big number

which is going to get arbitrarily close

to zero so we would say this last

sequence uh converges to the number

zero

❖ Finding the Limit of a Sequence, 3 more examples ❖

Finding the Limit of a Sequence, 3 more examples, #1.  Just another example of finding the limit of a sequence by taking a limit as n approaches infinity.
One often will encounter sequences in a Calculus course prior to dealing with Series.  Make sure you can find limits of sequences as it will be an important skill to have when working with series.

#maths #sequence #series #limits #converge #convergent #diverge #diverges #divergent #pseries #exponential #rationalfunctions


okay in this video we're going to look

at an example related to the slope of a

tangent line and we were given the

function um f ofx = x Cub - 2x and

already given that the derivative is

3x^2 - 2 we want to answer a couple

questions what's the slope of the

tangent line um at the point 2 comma 4

and also what would actually be the

equation of that tangent

line well again to get the slope um at

the point uh in this case uh at at xal 2

and yal 4 our derivative you know the

derivative formula in a sense it says

all you need to know uh to figure out

the slope of the tangent line is the x

coordinate so in this case the slope is

going to be F Prime of 2 we're plugging

in the x coordinate of 2 into our

derivative formula again that's what the

derivative will do it tells you the

instantaneous rate of change or um in

terms of a graph it tells you the slope

of the tangent line so we'll just get 3

* 2^ 2 -

2 well 2^ 2 is

4 3 * 4 is going to be

12 12 - 2 is 10 and again that's going

to be the slope of the tangent line at

that point well to get the equation of

the line remember uh we can just use

point slope formula here so y - y1 equal

the slope x - X1 recall this a point

slope formula for a

line well now all we have to do is just

fill in our information um the point was

given to us again we're given the point

2 comma 4 so if we fill that in we'll

just take Yus the y coordinate which is

well

four equals the slope of the tangent

line that's what we kind of had to work

to figure out and then we'll take x

minus the x coordinate which is a

positive2 and we've now got the equation

of our line if you want to we can always

put this in slope intercept form so if

we distribute the 10 we'll get 10 x 10

and -2 would be

-20 we can add four to both sides and

then we'll simply be left with y = 10 x

16 um as the equation of our tangent

line so again both of these are

equivalent

um sometimes again people uh like to see

things in slope intercept form so again

that's it okay so just need to remember

the equation of a line um point slope

formula again you're given the point for

free the only thing you really have to

figure out is the slope but to get the

slope you just plug in the uh you know

plug the point into your derivative

formula and that gives you the slope of

the tangent line

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 Equation of a Tangent Line Using a Derivative.  In this video, we find the slope and the equation of a tangent line simply by using the derivative!


alright in this video I want to talk about
finding limits graphically

and so here we've got our function f(x) we've got a little

removable discontinuity at -4 we have a
jump discontinuity here

at 5 so part A we just wanna ask yourself
does the limit as X approaches -4

exist for this function: well again as X
approaches -4

I'm thinking about what's happening to
the y values do they approach

some specific single y value so
notice that we take

X values  a little bit smaller than -4

I think about on the graph what are the
y values getting closer to

well the y values are getting closer and
closer to the

the value positive 2 likewise as I come
from the right

ok take numbers a little bit bigger than
-4

if I think about what's happening to
the Y values on the graph

again those are getting closer and
closer to 2

so I would say yes and we can write

the limit as X approaches -4

of our function we would say that that
equals

positive 2 again the y values are
getting closer and closer to 2

so again notice the actual the y
value

at -4 is something totally different

but again it doesn't matter what
actually happens at that point

we just think as you get close to -4 on
the left

and on the right are the y values
they're getting close to some

specific number yes and in this case
that answer

that number is 2. Same question

does the limit as X approaches 5 exist
for f(x)?

well if I come in from the left okay so
here the y values are getting closer and

closer to some

you know whatever it is let's pretend
it's up here at +4 just to give

it a specific value

so here as we approach from the left the
y values are getting closer and closer

to 4

but then when we come from the right
well

if I look at x values a little bit
bigger than 5

well now the y values are getting closer
and closer to something

different let's maybe pretend that
that's +1

well since when you come from the left

the y values on one side get close to
4 and as when you come from the right

the y values get closer and closer to
something different

in this case +1 if the limit
as you approach from the left and from

the right is not the same

we say no that limit does not exist in
kind of another way I always thought

for a limit to exist for example at -4
notice if I can't put my fingers on the

graph and bring them closer and closer
and closer to -4

they're gonna basically touch ok
they're gonna get closer and closer and

closer

and that y value where they would sort
of be touching would be the y value

of 2

and a limit exists if I do the same
thing at x = 5 if

I put my finger you know on the graph
and try to you know

move them closer you know take x values
closer and closer to 5 I think well

are my fingers gonna be touching are they
getting closer and closer together?

no cuz one's up here one's down here if
that happens we say the limit simply

does not exist

Finding Limits from a Graph

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding Limits from a Graph.  In this video, I show how to find limits using a graph.


okay so one more example here uh about a

little introduction to Polar coordinates

so suppose we want to figure out what

are the polar coordinates for the point

with a rectangular coordinates of 3

comma 3 so hey we just move three units

to the right three units up that's our

good old regular uh rectangular

coordinates well we'll have to figure

out two things we'll have to figure out

the

angle uh that's made by the uh the posi

xais and that point and we'll also have

to figure out our r

value um both of these are pretty

straightforward to figure out

though because really all we're going to

do is just use a

little little trigonometry so we've

moved three units to the right three

units up to get our r value we can just

use Pythagorean theorem so 3^ 2 + 3^ 2

would equal R 2 so that's what 9+ + 9

which will be 18 = R 2 uh so normally

when we take the square root we get

positive and

negative but I'm going to use the

positive square root in this

case and if you want to you can even

simplify the square root of 18 I guess

that's the sare of two times the square

< TK of 9 uh the square of N9 is just

three so we could write this as 3 < tk2

as our value for R so again we write R

and then we write

so our R would be 3un

2 um to figure out this anglea well we

can just use tangent so tangent of

theta would be the opposite over the

adjacent so 3 over 3 so that's just

going to give us tangent of theta equal

1 well we're sitting here in a quadrant

one this is definitely going to be Theta

= pi 4 or

45° so uh one way to describe this point

would be to use the polar coordinates 3

< tk2 comma pi over

4 uh again notice there's uh lots of

different ways that we could describe

this point as well so uh notice we could

have even used 3 < tk2 comma say how

about -7 pi over 4 because -7 Pi

over4 um

we would go around that would put us at

the exact same angle okay -7 Pi over4

would put us at the same angle and again

we would just go at a distance of 3 <

tk2 so this would also be an equivalent

solution um so you can always add or

subtract multiples of 2 pi from your uh

your value your your your angle we could

have even used uh say a different one

suppose we use the angle negative

uh how about the angle um let's

see -3 pi over 4 so normally positive 3

Pi over4 would put us you know up in the

second quadrant negative 3 Pi over4

would put us down in the third quadrant

um if we use a positive radius though it

would keep us in the third quadrant well

to reflect things back we would just

take the negative of that so we could

use <

tk33 < tk2 comma -3i over4 as another

way to describe that uh that point using

polar coordinates so maybe more than you

wanted to see again I think little

things that are definitely uh good to

think about though but certainly this

would be the the answer choice that I

would go with

Intro to Polar Coordinates, 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 !! Intro to Polar Coordinates.  Just a few quick questions related to the basics of polar coordinates.