okay here we're going to identify we've

got eight functions we're going to

determine which of those are exponential

functions so recall an exponential

function is a function of the form a

raised to the power of

X and A has to satisfy a

couple a couple conditions either it has

to be between Z and one or a has to be

strictly greater than

one we don't want a to equal one nor do

we want it to be a negative number okay

we don't want a negative number raised

to a

power so okay so which one of these

maybe all of them maybe none of them can

we write in this form well certainly the

very first function that looks exactly

like an exponential to me we've got a

number raised to a a an exponent a a

variable and our number is certainly

greater than one so G of X works that

would certainly be an

exponential notice we have this next one

H ofx equals 2 raed to the^ of

3x we could write that as

2 cubed raised to the power of X that

still gives us the exact same thing 2

the 3 Power 8 so really this function is

the same thing as 8 raised to the power

of X so that one is also an exponential

function

well P ofx = -4 that's just a constant

that doesn't

work now f ofx = 1 raised to the^ of 3x

+ 4 that sort of right we've got this

number we've got some variables in we

got a variable in the exponent but one

to any power just equals one so that's a

constant so that one doesn't

work m ofx = 50 + x that's a linear

function that doesn't

work Q ofx = 8x - 3 well again 8 to the

power of X we said that's exponential

the minus three is just going to shift

our function down so 8x - 3 that also

looks like an exponential function to me

X2 well that's quadratic that doesn't

work again and N ofx = 5x that's just a

linear a linear function

so it looks like we have 1 2 3

exponential functions

Can You Identify the Exponential Functions?  Which of the Following is an Exponential Function?

Can you identify which of the following functions are exponential functions? Some algebra may be required to make it perfectly clear!  In this video I show why each of the given functions is an exponential function or explain why it is not an exponential function.
These questions would be appropriate for someone who has already seen a quick introduction to exponential functions.
Recall that an exponential function is a function whose base is a number raised to a variable, with the base being larger than 0 and not equal to 1.

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okay in this video I just want to talk

about quadratic inequalities and I just

want to talk about a really common

mistake people make so don't do what I'm

about to do so a second ago or in my

other video we actually solved this

quadratic equation x squared plus 2x

minus 8 and we came up with the

solutions negative infinity to negative

4 and then 2 to infinity a lot of times

what people will do so again what I'm

about to do is very much incorrect so

don't do what I'm about to do but it's

basically I see it

I see it all the time okay so you can

factor x squared plus 2x minus 8 we

factored that a second ago as X plus 4

and then X minus 2 so our inequality has

to be greater than or equal to 0

and now what people will do is they'll

just you know if it was an equation you

set each piece equal to 0 but if it's an

inequality you can't do the following

people still want to do the same thing

they want to take X plus 4 and make that

greater than or equal to 0 and then they

want to take X minus 2 and make that

greater than or equal to 0 so the stuff

down here again was the actual correct

solution

before I lose my focus here and then

they'll subtract four and they say well

X has to be greater than or equal to

negative four and then it says if you

solve this by adding two it says X has

to be greater than or equal to two so

what does that mean does it have okay so

it has to be greater than negative four

or two or is it have to do both of these

in either case I think you're gonna see

that definitely you're not going to get

this part of the the solution that we

found before so this parts okay writing

it but doing this step where you take

each piece and make it greater than or

equal to or less than or equal to or

whatever it is that is very much

incorrect again you have to make this

number line and check all the points so

I just want to point this out so that

hopefully you won't do it alright good

luck

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


so in this question we were given that a

line and the XY plane passes through the

origin and has a slope of 1 over 7 we

want to know which of the phone points

lies on the line is it 0 comma 7 1 comma

7 7 comma 7 or 14 comma 2 again

different ways to do this one way again

I'm not a huge fan of graphing and

general you know it depends on the

question because graphing can be a bit

sloppy but i think this is one where

graphing would work we'll also just talk

about the equation of a line here real

quick as well okay so i know that it

goes through the origin so there's one

point it has a slope of 1 over 7 so

remember the top part that's the change

in Y we sometimes write a little

triangle that triangle Delta means

change so change in Y over the change in

X so it says from the origin we would go

up one unit and then I would go over

seven units to get another point okay

well let's see so you know 0 comma 7 0

comma 7 would be up here on the y-axis

that's definitely not correct 1 comma 7

well let's be careful because this is

the point this is the point 7 comma 1 1

comma 7 would be up here somewhere so

it's not answer choice B 7 comma 7 would

be up here so that's not right oh so by

deduction okay 14 / 2 seems to be

corrected again that makes sense because

from this point if we go up another one

unit will be at two and then we go over

another seven units will be at 14 so

yeah so it looks like 14 over to excuse

me 14 comma 2 it's going to be the

correct answer so answer choice D also

remember that the equation of a line we

can use y equals MX plus B where B

represents the y-intercept and n is the

slope

so we could write the equation of our

line as y equals 1 over 7x plus 0 or y

equals 1 over 7x and you could also

another way to do it so if you didn't

want to graph it you know maybe they

gave you you know some really kind of

weird numbers something not so easy

again you could start substituting these

points into the equation and see which

one result in a true statement so for

example I could ask myself answered

choice a 0 comma 7 work well ok if I put

0 in for X and seven in for y I would

ask myself does 7 equal 1 7th x 0 again

I'm substituting in y equals 7 and 0 for

X well one over 7 x 0 0 does not equal 7

so I would know that answer choice a is

not correct and I could go through your

answer choice B and C that doesn't work

answer choice see that doesn't work

again answer choice D just to show it

works so okay so i'm going to put two in

for y and I'm thinking does that equal

one over 7 x 14 well this is the same

thing as 14 / 7 which gives you two yeah

two equals two so that would be another

way to do it that would be sort of a

more algebraic way to do it without

graphing so again two ways to do it you

know whichever way makes you happy again

if you peeked over at me on the test I

would definitely be doing a rough sketch

because you know the numbers are pretty

small and I could do that pretty quickly

but again you know notice they're trying

to trip you up there giving you a slope

of 1 over 7 they're giving you 1 comma 7

as a solution so make sure you're doing

your change in Y over your change in X

correctly when you graph it

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! SAT Math - Question 12, 2016 Edition and Beyond.
Section 1 - No Calculator allowed!


all right here we're going to do one

more example using the limit comparison

test so here we've got s of 1 over n and

you know the series from one to Infinity

so you know there's not a lot going on

here with this series uh you might first

try to do the test for Divergence and if

you do the test for Divergence well we

can do the limit as in goes to Infinity

so this is what I do if I'm kind of

completely unsure about a series well as

n goes to Infinity you know uh inside

the parenthesis we'll get one over a

large number so that'll approach zero so

it says it approaches sign of zero but

sign of zero is zero so the test for

Divergence would be

inconclusive well I see this series uh

or I see this this this one over n so

maybe we can compare it

to one over n and see what happens and

recall this is the harmonic series again

a Divergent P

series okay so we can do the limit as n

goes to

Infinity S of 1 n / 1

n but again uh as n goes to Infinity we

set sign of 1 over n is going to go to

zero 1 over n is going to go to zero so

we can use lal's rule on

this so we'll have to take the

derivative of the uh numerator so we'll

have to use the chain rule so the

derivative of s is cosine we'll leave

the 1 over n

alone and then we would have to multiply

by the derivative of 1 over n that would

be the chain Rule and I'm going to be

lazy and not even do it because I

recognize in the

denominator well we'll get the

derivative of 1 over n well we would

have to do the exact same thing well

those two terms would simply uh you

those two fact factors would simply

cancel out so I mean you can always

actually compute them but it's going to

be the exact same thing so they're going

to go away well now we're left with the

limit as n goes to Infinity of cosine 1

/ n that's going to give us cosine of 0

cosine of 0 is one again we get a

positive finite number since we're

comparing uh to a Divergent P series

that would tell us that our Series in

question also diverges

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Limit Comparison Test for Series - Another Example 8. In this video, I use the limit comparison test to determine whether or not a given series converges or diverges. I do not delve heavily into the theory in this video (but do in another video if that is what you need to see!)


all right this is going to be an example

of solving a quadratic equation by

factoring and this is kind of one of the

more to me it's one of the it's kind of

the fundamental case simply because the

coefficient on the m s is a one and in

that case what we need is we need two

numbers so we're looking for two numbers

that

multiply to -35

but

add to the value they have to add up to

the coefficient in front of the linear

term the term to the first Power which

is just uh in this case positive two so

we need two numbers that multiply to35

but add up to

positive2 so to get M SED we're going to

use an m and an M an m and an M um and

whatever numbers uh satisfy this

requirement that's what we're going to

put in each we'll put one in each set of

parentheses

so okay to get uh

35 and I guess to get Negative 35 um you

know we could use some combination of 1

and 35 you know one of them will have to

be positive one will have to be negative

but in either case I don't see how

that's possibly going to add up to

pos2 um how about some combination of

five and seven well I guess if we made

the seven positive and the five negative

those would certainly multiply to -35

but add up to positive2 so those are

going to be our candidates we'll have M

minus 5 m + 7 that's how the left side

factors so we'll set that equal to

zero and now once we once we have our

equation factored we just take each

factor set it equal to zero and

solve so the first equation we can

simply add five to both sides sides

we'll get M = 5 and if we subtract

s from the second equation we'll simply

get M -7 and that will be our other

solution so our two solutions here are M

= 5 and M = -7

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving a Quadratic Equation: Factoring a Simple Example.
 In this video, I show an example of solving a basic quadratic equation using factoring.


okay in this example I want to evaluate

again a a fractional expression

involving absolute value and here we're

going to have to just substitute in some

values so we want to evaluate the

absolute value of B minus the absolute

value of positive a all divided byga a

and we want to do this when a = 3 and b

= -5 so all right let's just basically

fill everything in and hopefully

simplify it down so we've got the

absolute value of

B in general when I substitute I like to

put things in parentheses I I found it

keeps me uh from making mistakes it's it

ends up being good form especially if

you have lots of things to substitute uh

because you might have to distribute

this negative sign so you'll see that as

you go along um but just in general I

like to put things in parentheses so I'm

simply replacing my B with 5 there's our

minus sign we have the absolute value of

a which is simply

pos3 we're dividing all that by a which

is

-3 and if we keep simplifying this we've

got the negative of5 which inside the

bars inside the absolute value will be

positive five minus the absolute value

of

positive3 again being divided by -3

well the absolute value of postive 5 is

five the absolute value of posi3 is

three so in the numerator we're really

getting the value 5 minus

3 again still over -3 and 5 - 3 that

simply gives us pos2

over3 and that will be our solution

-2/3 um and again you could put the

negative on the on the top or out front

however you want to write it um so all

right just a little bit of substitution

here again you know being careful I

think it's a good rule of thumb to put

things in parentheses um although I

clearly didn't do it on the a part maybe

I I could have um but especially when

you have these negatives because you

have to be again really careful to

distribute things so uh that's the

reason why I'm doing that

there

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Evaluating Expressions Involving Absolute Value - Example 3
A basic problem involving absolute value, a bit of substitution and some simplifying!


okay in this example we've got this

rather you know long formula here m

equal uh R / 1200 multili by 1 + r/ 1200

ra to^ of n / 1 + r over 1200 ra to^ of

nus 1 multiplied P they give you some

context this this has to do with um a

loan payment formula the important part

says which of the following gives the

question is which of the following gives

p in terms of m r and N so again what

that means is it says we want P isolated

we want some equation that says P equals

something which is you know so there's

the answer choices as you can see

they've got P all by itself so the only

thing to me that seems really daunting

in this problem is you know this is kind

of a long formula when you first look at

it you know maybe your eyes kind of

glaze over and you say oh I don't know

so let me show you though that this

really isn't very bad at

all so to Simply solve for p

okay so I'm going to rewrite the

original expression Okay so we've got M

there on the left I'm going to give

myself a little space um before so

here's the original expression there's

our multiplied by P so we've got R over

1200 ultip 1 + r over

1200 raised to the power of n we've got

1 + r over

1200 raised to the^ of n-1 and I can

stick that in Brackets it doesn't change

its value so there's the original

expression that I have right there well

all I want to do is just multiply so

that I can get rid of some of this stuff

well I've got a r over 1200 in the

numerator if I put one in the

denominator when I multiply I would be

able to cancel cancel the same thing to

get rid of this 1+ R over 1200 raised to

the power of n I would need a 1 + r over

1200 raised to the power of n and to get

rid of the

denominator I would need 1 + r/

1200 raised to the^ of

nus1 and notice now everything on the

right side is going to cancel except for

p the the numerator and denominator

would cancel and again now I've gotten

rid of each of these factors that were

originally in the numerator as well well

again if I do it on the right I've got

to do it on the left

so there's 1+ r R 1200 raised to the^ of

n

-1 and then R over

1200 ultip 1 + r over

1200 raised to the power of n and again

you can think about this as being M over

one if you want to again everything's

going to cancel

out on the right side except for

p so there's our P equals and again

there's nothing really to do on the left

you just you know I think the answer

Choice had it just written like this so

I'm just going to rewrite it one more

time the exact same thing that we just

had raised to the power of n and then

they had the M sticking off to the side

and again that just means you're

multiplying by M over one so that's it

so I believe that was answer Choice B

would be the correct answer so again to

make a long story short all that you

have to do for this one is you just take

you know this original okay forget about

the P for a second take this original

fraction and you just multiply uh by the

flip of it on the left and that gives

you your solution

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! SAT Math - Question 7, 2016 Edition and Beyond


all right in this video we're going to

do a couple examples related to finding

the difference quotient of a function

and the idea is if you have some

function f ofx the difference quotient

is what's known as the quantity f ofx

plus h minus F ofx over H um if it's the

first time you've ever seen the

difference quotient it feels like kind

of an arbitrary you know like where the

heck is this coming from sort of thing

it actually has a lot of applications um

it has to do with sort of average rates

of change um slopes of tangent lines and

those actually tie into uh surprisingly

maybe maybe not surprisingly it actually

ties into some important ideas and

calculus so it's one of these things it

looks arbitrary but uh who whoever's

throwing these sorts of problems at you

knows knows that uh that there's a a

reason for it for sure so um okay in

part A we've got this function f ofx is

2x - 5 we're not going to do all the

difference quotient we're just going to

kind of find the numerator f ofx plus h

minus F ofx well we've already got our

our F ofx to get f ofx plus h what do we

do we just well everywhere there's an X

we replace it with X+ H so x + H - 5 so

it says this difference um f of x + H -

F ofx we would get 2 * the quantity x +

H - 5 there's my f ofx plus h and then

it says subtract away the original

function which is f ofx I'm going to put

that in parentheses 2x - 5 to make sure

I distribute the negative and now we've

got it I'm going to simplify but this is

our F ofx +

H minus our F

ofx so let's see um notice if we if we

multiply this out we would get 2 * X and

A minus 2 * X those would cancel it

looks like we would get a 2 * H left in

there um and notice we have a -5 minus

A5 well that's -5 + 5 so that would also

cancel out so I claim that the

difference quotient here is just the

quantity uh 2 H and that's

it okay so suppose we've got um suppose

f ofx is 3x + 2 and we actually want to

find this difference quotient um f ofx

plus h minus F ofx over

H so I'm going to squeeze it all in here

okay so again if we find f ofx plus h

that just says replace the X with

x + H we've got our + 2 and then it says

subtract away the original function f

ofx which is 3x + 2 and then it says

divide all of that by

H so if we distribute out the numerator

again notice we would get a 3X plus 3 H

in the first set of

parentheses um if I take the 3x minus

the 3x that would cancel so I would be

left with 3 H notice we have a plus two

and we would get a -2 when we distribute

so those would also cancel out we've

still got the H in the denominator but

now we can simply cancel out our H's and

it says the difference quotient of the

function 3x + 2 is just the value the

number three

❖ The Difference Quotient - Example 1 ❖

The Difference Quotient - Formula & Examples

Description:
In this video, we explore the concept of the difference quotient, sometimes referred to as the subtraction fraction (but only by me). You'll learn the formula for the difference quotient and see how it's applied through several examples with different functions. Understanding the difference quotient is a fundamental step in calculus, as it serves as the basis for the derivative and provides insight into the rate of change of a function.

This video will guide you through:

The formula for the difference quotient:  ( f ( x + h ) − f( x )) / h.
Applying the difference quotient to various functions.
Simplifying expressions to reveal the structure of the difference quotient.
Gaining foundational knowledge for future topics in calculus, including derivatives and instantaneous rate of change.
If you find this video helpful, please subscribe, comment, like, and share! I appreciate it if you tell your friends, teachers, or parents about my videos :)

For more support and exclusive content, visit my Patreon:
https://www.patreon.com/patrickjmt?ty=c

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okay in this example we're going to plot

three more points we're going to plot

the point negative 3 comma negative 1 4

comma 0 and 0 comma negative 4 so define

net the point negative 3 comma negative

1 well from the origin I'm going to move

3 units to the left so 1 2 3 units to

the left so now I'm sitting at the x

coordinate of negative 3 and then the y

coordinate is negative 1 so we'll go

down one unit so that will be the point

negative 3 comma negative 1 the next

point is going to be 4 comma 0 so from

the origin will go to the x coordinate

of 4 or 4 units to the right and then

for the Y value it says don't go

anywhere don't go up or down so the

point 4 comma 0 that's going to be

sitting on the x axis and last but not

least we have 0 comma negative 4 so I'll

move 0 units away from the origin

horizontally and then we'll go down 1 2

3 4 units so there's the point 0 comma

negative 4 that's going to be sitting on

the y axis

GED 'Most Missed' Math Practice Problems , Code A.5.a,  #2

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


all right in this video I want to do an

example using deor theorem to raise a

complex number to a power and it's

actually extremely easy um if you have a

complex number Z in polar form so R *

the quantity cosine theta plus I sin

Theta it says if you take your uh

complex number and raise it to a power

we just take R and raise it to that

power of n and then our angles Theta we

simply multiply that by n so

actually uh not too bad at all a lot of

times just you know uh the time

consuming part is going to be just

putting it in polar form in the first

place so let's see if we can squeeze it

all in here so here we're going to find

the value of 2 * cosine of 60° plus I

sin of 60° all of that to the second

power okay so um all we have to do in

this case so here to the second power

that's going to be our value of

n so here's our R here's our Theta and

our Theta this will be our value of n so

all we have to do is just plug

everything into our formula it says we

would get well 2 to the second power and

then we would have cosine of 2 * Theta

which is

60° plus I * s of 2 * 60°

have enough parentheses here I think one

more um and now you can always clean it

up a little bit so 2^2 is 4 this will be

cosine of

120° plus I of

120° and you know you can always keep

going you can evaluate cosine of 120 s

of 120 um simplify this a little further

or we can always just leave it in the

polar form uh which is what I'm going to

do here so again very simple uh you know

kind of a one more formula to remember

but it's easy to use uh just raise your

your R to the power multiply your thetas

by that power and you get your solution

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! DeMoivre's Theorem: Raising a Complex Number to a Power, Ex 1.  In this video I give DeMoivre's theorem and use it to raise a complex number to a power.  Note that our number must be in polar form!


okay in this video we're going to talk

about vector addition and scalar

multiplication and all it really says is

if you have two vectors if you want to

add or subtract them

you you either add or subtract their

corresponding components it says if you

multiply a vector by a real number it

says you just multiply each component by

that real number so nothing worse than

that

let's look at a couple examples here so

here we've got three vectors v has

components zero two u has components

negative one four

w has components five comma negative

three and we wanna find the components

of a couple different vectors so let's

do part a here first we'll do vector

w minus vector u

well the components for vector w we said

that was five comma negative three

and then we're going to subtract away

the components of vector u which are

negative one and four

all we do is we take five minus

well negative one so i'm just

subtracting uh the first components and

then we'll take negative three

uh minus four

and well after we simplify we'll get

five minus negative one that's five plus

one so we'll get our first component to

be six

negative three minus four will give us

negative seven

so those would be the components of the

vector

w minus u

okay so

part b kind of combining both ideas so

it says take 3 times vector w

which is five has components of five and

negative three

plus two times vector v which are has

components zero and two

minus four times vector u which has

components negative one and 4.

well if we distribute the 3 we'll get 15

and negative 9 for our first

the components of our first vector here

after multiplying

i'll distribute you can fill you know

it's basically feels like distributing

so 2 times 0 will be just still 0. 2

times 2 will give us 4.

if you want to you can imagine

distributing the negative 4 so

you could always think about this as

being plus negative four

so if we distribute that negative four

negative four times negative one will be

positive four

negative four times four will be

negative sixteen

and now it's addition um you could do

these pairwise you could do them two at

a time if you want to but the basic idea

is if you have a bunch of addition

you just add

all of the respective first components

so 15 plus 0 plus 4

and then our next component would just

be negative 9 plus 4

plus well a negative 16.

and let's see i think we can clean this

up without too much trouble 15 plus 4 is

going to give us 19.

negative 9 plus 4 will be negative 5.

negative 5 plus negative 16 will be

negative 21.

and now we have the components of this

new vector

3w plus 2v minus 4u

408 Vector addition and scalar multiplication Patrickjmt


all right in this video I want to do

another example of using sum and

difference identities to find a value of

a trig function so suppose we want to

evaluate s of

195° so sometimes kind of the tricky

part is just finding out you know um how

to write this as a sum of uh angles that

we already know well we know 45° we know

how to evaluate s and cosine at 45°

90° 135° I'm just basically taking

multiples of

45° 180° Etc um let's see let's maybe

stick 30° up there we can do 30 60 we

know 90 120 um

Etc so uh what I'm going to try to do um

you know you could do the same thing

with 60° 60° 120° we know 180° again

just multiples of

60° I'm either looking for sort of two

numbers in here that either add up to

one 195 or whose difference is

195 well there's definitely uh more than

one way to do this but certainly it

looks like we could use uh 60° and

135° so s of

195° is the same thing well that's going

to be 60°

plus

35° and now we just have to use our

identity um and recall our sum identity

for sign here it says we just take sign

of the first one

60° we take cosine of the second one

135° since there's a A plus in between

we keep that and then it says we use

cosine of 60° and then we'll use S of

135° so I'm just using my uh very top

identity and now hopefully these are uh

values that we know so s of 60° that's

the same as s of Pi 3 um that's going to

going to give us < tk3

over2 cosine of

135° that's going to be in the second

quadrant um that's going to give us <

tk2 over

2 cosine of 60° um that's going to give

us

12 s of

135° that's going to be positive < tk2

over 2 and now we can just simplify this

down if we multiply we'll get negative 3

* two we can write that Asun 6 uh 1 * Ro

< tk2 is simply < tk2 uh we would have

our common denominator of 2 * 2 which is

four so to me that looks like a nice

simplified solution for S of

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 for Sine and Cosine, More Examples #1.  More examples of using the sum and difference identities to find value other trig values. For more free math videos, visit http://PatrickJMT.com


so in this example we have our friend

Patrick who's building a rectangular

patio and that's going to be centered on

one side of his yard and the rest of the

yard is grass okay so I've kind of got

it a little uh you know kind of marked

here the outside part's going to be

grass and if the measurements in the

diagram are in feet we want to know

what's the square footage of the grass

portion of the yard so we want to know

you know how much basically how much uh

uh area is the grass covering now

there's a couple different ways you

could do this you could start trying to

you know say okay well this is nine and

you could start you know saying okay

this is centered this is 15 you could

start trying to chop this up into a

bunch of little rectangles but that

would definitely be Overkill um that

would be much more uh uh be a lot a lot

more trouble than what you need to do so

the idea that we're going to use in this

one is well we're going to figure out

the

area of the entire uh rectangle

and then we're going to

subtract we're going to subtract away

the area of the

patio and if we take the entire

rectangle and subtract away the area of

the patio that's just going to leave us

with

the with the area that the grass is

covering so that's the idea we're going

to figure out the area of the entire

rectangle we'll subtract away the area

of the that the patio covers and that's

going to give us the area of the

grass all right so really we're just

finding areas of rectangles here and

just doing a little bit of arithmetic

so the area of the of the large

rectangle well that's going to be 30

multiplied by 18 so let's do that here

so um 30 multiplied by 18 well 8 * 0 is

0 8 * 3 is 24

and then I'm going to drop down some

people put a zero I always uh we always

did little X's so 1 * 0 is 0 1 * 3 is 3

and now we just add those up so we've

got 0 4 + 0 is going to be 4 and then 2

+ 3 is going to be 5 so the entire

rectangle has an area of 540 square ft

let's figure out the area of the

patio well again that's is going to be

the length time the width so 9 *

15 well 9 multiplied by 5 that's 45 so

we'll drop down the five and carry the

four 9 * 1 is 9 if we take 9 + 4 that's

going to give us 13 so the area of the

patio is 135 squ ft so now we just need

to do the subtraction to get the area of

the grassy part so we'll take 540 we'll

subtract away 13 5 let's set this up

um

vertically all right 0 mult or excuse me

zero subtract five can't do that so

we'll take one from the four we'll carry

that so that'll give us 10 10 - 5 is

going to give us 5 3 - 3 that's just

going to leave us with 0 and 5 - 1

that's going to leave us with four so it

says that the area of the grassy part

that's going to be

405 Square ft

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


all right um in this video I'm going to

just do some uh math calculation tricks

nothing maybe that mind-blowing I picked

up a book the other day and I don't know

I thought some of the tricks were kind

of interesting so this is just going to

deal with squaring numbers close to 100

so I'm going to do 106 squar 94 squ 112

squar and 88 squar

um and the idea to do these uh maybe

let's look at 106 squar

First notice all of these numbers again

are close to 100 and that's basically

the trick um on all of them that's the

that's the important part if you think

about

106 notice that 106 is actually six

bigger than 100 what we do is we take

the number six and we add it to 106 that

gives us 112 so that's how we get our

first three digits we think well it's

six away so just add them and then to

get the last two digits all you do is

take this number you square it and you

just tack it on okay so 106 squared

would be

11,23 same idea for 94 squared the only

thing that's different now is instead of

being um six units above 100 it's six

units below so we do the same thing we

add these numbers so -6 and 94 that's

going to give us 88

but then we do the same thing we Square

the number so -6 squared would be the

number 36 and we get our value

8836 um maybe two more that are just a

little bit

trickier um we're going to do the same

thing here if we look at the number

112 well 112 is 12 bigger than 100 so if

we add 112 + 12 that'll give us um 124

but notice now if we Square the number

12 12 squar is

144 we're going to get too many digits

if we tack on the

144 what you do is you carry the one to

the 100 spot drop your 44 um at the end

and then you just add them together so 1

two 5 44 that's going to be 112 squared

and you do the same thing for 88 that's

going to be 12 units

um smaller than the number 100 so if we

take -12 and 88 that's going to give us

76 um and again if we do the same thing

if you square -12 you'll get 1 44 we've

got to carry the one we'll get

77 uh 44 so

7,744 is our final answer so I don't

know I thought it was kind of neat um

don't get rid of your calculator just

yet but I'll probably post some other

videos on some neat little math tricks

if uh if people find these interesting

so all right um I hope this was mildly

entertaining and um feel free to post

comments or questions

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Fun Math Trick! Squaring Numbers Near 100.
A neat way to square numbers near 100 without having to multiply it out the long way!
For more free math videos, visit http://PatrickJMT.com


okay here we're going to find the

vertices of the hyperbola y - 20 2 / 4 -

x + 2 2 / 36 = 1 so the first thing I'm

going to do is pick out the center so I

see x + 2^2 but instead of positive2 we

use the opposite sign or -2 and then we

have y - 20 2ar instead of -20 we'll use

positive

20 so I know that my Center again you

don't have to sketch this you could just

use some formulas but I always like to

think about it in terms of a graph it

helps me remember so there I've

got -2 comma 20 I'll put a little X

there that's where it's centered at

since it's of the form y^2 - x^2 = 1 I

know it's going to open

upwards I know my hyperbola is going to

open upwards and

downwards so I'm looking for the

vertices well to get the

vertices what we do is for our hyperbola

we look at the the the first term when

we have it in standard form Okay so

we've got y - 202 / 4 if we take the

square of four we'll get two and that's

going to be our a

value and that a value tells us from the

center how far how many units you move

to get to the vertices

so in this case since it opens upwards

and downwards I know from the center

that I'm going to move up and down to

get my vertices and again I just take

the square root of that first term the

square root of the pos4 which is two and

that's going to give me my

vertices okay so since the x coordinate

doesn't

change that'll just remain -2 it was

originally centered at -2a 20 so if we

move up two units we'll be at -2 comma

22 and if we move down we'll be at -2

comma 18 and those will be our vertices

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Vertices of a Hyperbola - Easy, Basic Example
In this video I show how to find the vertices of a hyperbola by working an example.


okay here we're going to identify the

quadrant or axis that contains the point

8 comma 6 so the first coordinate of

eight that tells us how far we travel

along the x axis so we'll travel let's

see so uh 1 2 3 4 five six 7even 8 so

we've moved over eight units on the

xaxis and 1 2 3 4 five six uh units on

the y AIS and we'll put our DOT right

there our point right there so our point

in this case is going to be in quadrant

one so just to remind you remember this

is quadrant

one quadrant 2 quadrant 3 and Quadrant 4

so so our Point here will belong to

quadrant one

How to Plot a Point in the X-Y Plane

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! How to Plot a Point in the X-Y Plane.
Just a quick discussion for those beginning algebra on plotting the point (8,6).


all right here we've got a question

related to vector functions and velocity

acceleration and position so here we've

got a car and it travels with a velocity

vector given by V of T equals T squared

times e to the T plus 1 T's measured

measured in seconds and the vector

components are measured in feet if we

know the initial position of the car is

R of 0 equals 1 comma 3 we want to find

the position of the car say one second

later well to do this if we're given

velocity to get back to our position

vector R of T all we're going to have to

do is we're going to integrate our

velocity vector with respect to T okay

so we can write this as the integral

we're just going to integrate each part

so the integral of T squared DT and then

we'll have to integrate e to the T plus

one DT well let's see if we integrate

the first part the antiderivative of T

squared will be T cubed over 3 we have

to add our arbitrary constant here see

and then when we integrate the second

part well we'll just get e to the T plus

T again we have to introduce a new

constant let's call it D since we've

already used C all right so but we are

given an initial condition we're given

that R of 0 equals 1 the vector with

components 1 and 3 so now what we can do

is simply plug that in ok so so it says

if T equals 0 well the first component

we would get 0 cubed over 3 plus C that

would have to equal the first component

of 1 and likewise if we plug in T equals

0 well we would get e to the 0 plus 0

plus D that would have to equal the

second component which is going to be 3

well 0 cubed over 3 is just 0 so we'll

get C equals 1 e to the 0 well that's 1

plus D equals three if we subtract we'll

get D equals two all right so really we

know that we can describe our position

vector R of T as T cubed over 3 plus C

which again is just going to be 1 and

then the second component will be e to

the T plus T plus D again which we said

had a value of 2 all right so we wanted

to figure out we wanted to find the

position of the car after 1 second so

really what we want to find is we just

want to find R of 1 well all we have to

do is plug that in we'll get 1 cubed

over 3 plus 1 comma e to the first plus

1 plus 2 well 1 third plus 1 is going to

be 4/3 and then we would have E Plus 1

plus 2 which would give us a positive 3

and now we've got the position of the

car one second later

Vector Functions: Position, Velocity, Acceleration Word Problem, 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 !! Vector Functions: Position, Velocity, Acceleration Word Problem, Ex .  In this video, we are given the velocity vector of a car as well as its initial displacement.  We have to find the position of the car after 1 second.


all right in this video I'm going to do

one more example of multiplying complex

numbers and I'm going to finish off the

example that I didn't finish a second

ago which was to multiply I * the

quantity 6 + 4 I by the quantity 2 - 7

I all right so again since you're

multiplying we can either multiply the I

by the first uh the first factor or we

can multip the I by the second Factor

you definitely don't distribute because

there's not a plus or a minus in there

and I think what I'm going to do first

is I'm actually going to do the

multiplication between the 6 + 4 I and

the 2 - 7 I again a couple different

ways that you could simplify this stuff

down for sure but this is the way that

I'm going to do it so again we we still

have to distribute as in the previous

example um so 6 * 2 is going to give us

12 we'll say we'll get 6 * -7 I that'll

leave us with a - 42 I on the inside of

when we take 4 I * pos2 that'll give us

positive 8 I and then I think pos4 * -7

is going to give us

-28 then we have I * I which is I

2ar okay so I'm going to continue

simplifying inside the brackets

so the I is just hanging out front we

have 12 let's see we've got -42 I + 8 I

I believe that's going to give us a -34

I we still have our minus 28 but we can

replace our I squared again with

-1 so really on the inside we have 12 we

have a -28 * a -1 so that's really 12 +

28 12 + 28 is going to give us POS

40 - 34 I so I think I've done about all

the simplification I can inside the

brackets so now I'm going to distribute

the I to both

terms so uh what we have I * 40 which is

simply 40 I and then our -34 we have an

i * an i which is I

squared so again we can use this trick

one more time that i^ 2 is equivalent to

1 so really we have 40

I + 34 and again typically you'll write

uh the convention is to write the real

part

first and the imaginary part which is

positive 40i we'll write that second and

now we have multiplied out this original

expression you could have certainly

multiplied the I you know for example to

the six and the 4 I I and get 6 I + 4 i^

2 simplify that down and then distribute

all that to the 2 - 7 I collect

everything and again at the end of the

day you'll end up with this expression

this complex number 34 + 40i is your

answer

Complex Numbers:  Multiplying - 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 !! Complex Numbers:  Multiplying - Ex 2.  In this video, I do an example of multiplying three complex numbers and simplifying!


alright so another example here for

finding the equation of a parabola in

this case with a focus of five six and a

vertex in one six so I'm going to make a

little sketch here so there's one

positive six there's the vertex it's got

to focus over at two three four five two

three four five comma six so again

that's not a point on the graph but I'm

just going to put a little X or a little

mark there since the focus sits inside

the parabola I know that the parabola

would have to be opening to the right so

to me this tells me that this is uh

the form the equation of this parabola

is going to be I'm going to take X minus

the x-coordinate of the vertex which in

this case will be one equals y minus the

y coordinate of the vertex which is

going to be six squared but in front of

this I'm going to put my value for P and

again all P represents as P is the

distance from my vertex to my focus well

since we went from the x coordinate of 1

to the x coordinate of five in this case

my value for P would be P equals four we

would have to move over four units so it

says well the equation of this parabola

then would just be X minus one equals

four times four times y minus 6 squared

so again X minus one equals 16 times y

minus 6 squared or we could write this

as x equals 16 times y minus 6 squared

plus one and now we've got the equation

of our parabola

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Conic Sections: Parabolas, Part 5  (Focus and Directrix).  In this video, I find the equation for a parabola given the vertex and given the  focus and/or directrix.


all right so one more example here of uh

doing a one-sided limit and this one I'm

not going to quite graph all the way so

we've got H ofx = the < TK of x^2 - 25

and we're doing uh X is approaching five

from the right so again we're looking at

the limit as X approaches 5 from the

right the square < TK of x^2 minus

25 well again intuitively um I'm going

to take numbers a little bit bigger than

five so maybe you know

5.01 so I'm getting closer and closer

and closer to the number uh to the

number five so I'm bringing it in closer

and closer and closer okay so again

we're we're approaching from the

right well if I take numbers a little

bit bigger than five and square them um

as they get closer and closer to five

the number that I'm squaring is going to

get closer and closer to 25

and if you subtract 25 well it's going

to get closer and closer to zero so

basically the moral of the story here is

um this function would be defined for

numbers a little bit bigger than five

and we can just plug it in uh we would

get 5 squar -

25 which would be the square < TK of 25

- 25 which would be the square < TK of0

which is just zero and that's all we've

got to do is just plug it in notice uh

you'd have to be a little

careful if somebody said what's the

limit as X approaches 5 from the left

um here I would say this does not exist

it doesn't really make sense to ask this

question um because notice if you take

numbers a little bit smaller than five

well if you square them um it's going to

be a little bit smaller than 25 so that

when I subtract 25 away I would

basically end up getting a negative

number

underneath my square root and we don't

take square roots of negative numbers uh

we're going to restrict ourselves to

real numbers and this would give us an

imaginary number so it wouldn't really

make sense to plug in five from the left

again algebraically if you plug it in

youd just get zero and you'd say hey

it's zero so um these are things you

have to again kind of be careful about

um you have to really think when can I

just plug it in and when can I not so

again here if I take something bigger

than five it's going to be bigger than

25 after I Square it well if I subtract

25 I'm basically going to get something

close to zero but it's going to be a

little bit bigger than zero but again if

you take the square root of that number

it's going to get arbitrarily close to

zero

One Sided Limits, Example 3

In this video we look at finding a one sided limit given a function involving a square root.  In order to find the limit we must choose the appropriate 'piece' of the function depending on whether our limit approaches the given value from the left or from the right.
This is a typical limit type of question in calculus and relates to continuity.

#math #algebra #khanacademy #khan #geometry #education #montessori #homeschooling #functions #notation #organicchemistry #organicchemistrytutor #calculus #derivative #powerrule #limits  #limit #graph #graphing #conjugate #onesided #continuity


okay here we're going to look at a

question related to proportions and all

we want to figure out is which of the

following does not form a proportion

with the value 3 over five so we have

four different choices here all you

basically have to do in this case is uh

basically think what fractions would be

equivalent to the fraction 3 /

5 well there's a couple different ways

to really think about this notice 3

fifths is in reduced form

so another thing we could do is just

reduce these other fractions and see if

you get 3 over five at the end so uh 6

over 10 well if we multiply the

numerator by 1/2 and the denominator

also by 1/2 we would get 3 over five so

certainly uh part A does check out the

same thing with part C 30 over 50 well

we could just cancel out the zeros and

be left with 3 over five so that one's

okay okay let's see uh are there any

common factors between uh 24 and

35 and I think there's not this is

actually uh already reduced well 24 over

35 doesn't reduce to 3 over

five so uh answer Choice B would be our

our our value that doesn't form a

proportion with 3 over5 and just to be

complete notice 12 over 20 um if we

multiply the numerator by 1/4 and the

denominator also by 1/4 equivalently

you're dividing top and bottom by four

we would get 3 over five so again D does

work

Equivalent Proportions / Fractions

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Equivalent Proportions / Fractions.
 Here I do a quick example of finding an equivalent proportion to a given proportion.


okay so here we're gonna do another

example involving implicit

differentiation so suppose we've got the

curve given by x over X minus y equals y

squared minus one and we want to find

dy/dx so this one same ideas before it's

just going to be a little more tedious I

think really because of this fraction

more than anything so alright we're

going to have to take the derivative

with respect to X of both sides well on

the left side we'll have to use the

quotient rule so the quotient rule says

you get whatever's in the denominator

and then we take the derivative of

what's in the numerator so the

derivative of 1/x will just be one minus

then we leave the stuff in the numerator

alone and then we take the derivative of

the denominator

so the derivative of One X is just 1 and

the derivative of negative 1 Y will be

negative 1 dy over DX and then that's

all over the denominator has squared

okay on the right side we'll have to

again take the derivative using implicit

differentiation so we'll just get 2 y

times dy over DX the derivative of

negative 1 is just 0 so all right now

what's probably the easiest way to solve

for dy DX well I think I'm gonna do a

couple steps at once in the numerator we

would be left with just X minus y and

then we would have minus so if we

distribute the negative x we would have

negative x times 1 or negative x and

then it looks like to me we would have a

positive x times dy over DX well I guess

I won't do so many steps at once let's

slow it down a little bit here all right

well okay I guess we can do a couple

things here now I am going to do a

couple things to get rid of this

denominator I'm just going to multiply

both sides by X minus y squared and by x

minus y squared so those will cancel on

the left notice we have an X my

sex so that would actually those would

just cancel out so really we would be

left with negative y plus x times dy

over DX on the right side I'm gonna pull

kind of all the variables stuff out

front so we have 2y times X minus y

squared times dy over DX getting pretty

close now I want to put all the dy DX's

on one side so maybe I'll just pull all

those dy DX's over to the right side so

I'll leave the original stuff over here

just like it was but then I'll subtract

the x times dy over DX from both sides

and then we'll have negative Y leftover

on the left and now we're almost there

all we have to do on the right side is

just factor out that dy DX so if we

factor out the dy over DX and the

brackets or parentheses here brackets we

would be left with the two Y times X

minus y squared and then minus X on the

right side we're left with negative Y so

to solve for dy DX we just divide so our

derivative will be negative y over all

of this stuff to Y times X minus y

squared minus X and now we have our

derivative

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Implicit Differentiation - More Examples #4
Yet another implicit differentiation example!


all right in this video i'm going to do

another example of predicting the types

of solutions to a quadratic equation

using the discriminant

so again the discriminant is just the

part under the square root the b squared

minus 4ac

so in this case we've got the problem 2x

squared plus six x

minus six equals negative eight

uh again before you start plugging

things in you know the quadratic formula

or to the discriminant you always need

one side of the equation to be equal to

zero

so the first thing i'm going to have to

do in this case is simply add 8 to both

sides

so then we're going to have 2x squared

plus

6x negative 6 plus 8 is going to give us

positive 2

equals 0 and again the positive 2 that's

going to be our a value

the positive 6 that's going to be our b

value

and the positive 2 that's going to be

our c value

so when we fill in the b squared minus

the discriminant we're going to end up

getting

so we'll have b which is 6 squared

minus 4 times a which is 2

times c which is also 2.

so for the discriminant 6 squared is

going to be 36

4 times 2 is 8 8 times 2 is

16. so we're going to have 36 minus 16

which will give us a positive 20.

so since the discriminant is a positive

number what that tells us in this case

is that we have two real solutions

i should say two real you can even

clarify distinct

solutions

and therefore we have zero complex

solutions

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Using the Discriminant to Predict the Types of Solutions to a Quadratic Equation - Example 2


okay so this is going to be question

number one for the 2016 version of the S

and again no calculators in this part so

this one's not hopefully too terribly

bad it says if the quantity X - 1 all

divid 3 equal K and K = 3 we want to

know what's the value of x is it 2 4 9

or 10 so this is just a basic just a

basic linear

equation so we have x - 1 over 3 equals

well they give us that k equals 3 so we

can substitute that in and now it's just

going to be a matter of solving for x so

I've got x -1 / 3 we can simply multiply

both sides by 3 on the left side the

threes will cancel out and we'll be left

with x

-1 on the right side we'll three

multiplied by 3 is going to give us 9

and now to solve for x we can just add

one to both sides

and that'll leave us with the value of x

= 10 which was answer Choice d

SAT Math - Question 1, 2016 Edition

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! SAT Math - Question 1, 2016 Edition and Beyond
Section 1 - No calculator allowed
This video is part of a playlist https://www.youtube.com/playlist?list=PLANMHOrJaFxPm2HZBsDfs1_ie3cmyremr


okay in this video I just want to

quickly look at inner products and

orthogonal functions so uh the

definition of an inner product uh we say

the inner product of two functions f ofx

and g ofx on the interval A to B is a

number denoted by we just put them in

parentheses so parentheses F comma G

close the parenthesis and to find the

inner product all we do is we multiply

those two functions f ofx * G ofx and

then we calculate that definite integral

from A to B uh integrating with respect

to X so all we do to find the inner

product just multiply them integrate

over the desired

interval if this inner product ends up

equaling zero so when we calculate the

definite integral if we get zero we say

that those functions f and g are

orthogonal and functions being

orthogonal uh that ends up being

something something important and we'll

see that in some some following videos

so very quick simple little

example show that the functions 1 + x

and x - x^2 we're going to Simply show

that those are orthogonal on the

interval neg -2 to

two well all we have to do in this case

we just integrate from -2 to two all

we're going to do is just multiply those

functions

together calculate the definite integral

and we better get zero if they are in

fact

orthogonal okay well this shouldn't be

too terrible it's just a matter of

multiplying it out so if we distribute 1

* X is X we'll have 1 *

x^2 we'll have positive x and x which

will be positive x^2 and then it looks

like we'll havex the

3 okay so we can certainly simplify that

our Nega x^2 and positive x^2 are going

to cancel so now we're we're just

calculating from -2 to 2 of x - x

3r DX well this isn't too bad at all

when we integrate x we'll have x^2 over

2 when we integrate x the 3r we'll have

X to 4 over

4 and now we simply have to evaluate

that from -2 to 2 well both of these are

even functions so they're clearly we're

going to get the same value from -2 to 2

so maybe it's obvious that we get zero

but let's I guess let's go ahead and

evaluate we'll have 2^2 which is 4 4 / 2

is 2 uh 2 4th is 16 16 / 4 is going to

be 4 so that's our uh upper limit of

integration and again since we're

squaring and raising to the fourth power

when we substitute in -2 we're going to

get the exact same

thing okay so we have -2 minus -2 -2 +

2 well that certainly equals zero so yes

in fact those two functions are

orthogonal what I'm going to do in my

next video is actually look at an

orthogonal set of functions and in that

case we'll justify that that entire set

of functions um are orthogonal and uh

we'll also give a I'll give a exact

definition of what we mean when we talk

about orthogonal sets of functions

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Inner Product and Orthogonal Functions , Quick Example.
In this video, I give the definition of the inner product of two functions and what it means for those functions to be orthogonal.  I work a quick example showing that two functions are orthogonal.


okay here we're going to identify the

conic that's given by the polar equation

R = 5 / 10 - 15 * sin Theta to do that

to determine what type of of conic

section it is we'll first find the

eccentricity and using the excentricity

we can find the directrix so this is

almost in the form

Ral e * P / 1 - e * sin thet so to put

our equation in that form we'll we'll do

is we'll Factor the 10 out of the

denominator well 10 * -15 over 10 would

give us

-15 and we can reduce 5 over 10 will be2

we can just put that in the numerator

then we have 1 minus well 15 over 10

will be 3 * sin Theta so already we get

that our eccentricity here our

eccentricity is going to equal three

Hales and since that's greater than one

we know that we have a

hyperbola and we want the numerator to

be in the form e * P so e * P we want

that to equal the numerator which is 12

well again we know that e is equal to

three

Hales so to solve for p we can just

simply multiply both sides by

[Music]

2/3 so we'll have that P = 1/3 and and

for this type of polar equation we know

that the directrix will be P units below

and parallel to the x-axis so there's

our x axis we go P units down which

would be 1/3 units down it's going to be

parallel to the x-axis that's going to

be the directrix D well since we're 1/3

units down and we have a horizontal line

the

directrix would have an equation Y =

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Polar Form of a Conic: Finding the Type, the Directrix and Eccentricity.
Here we have the polar form of a conic.  I determine what type of conic section it is and then find the directrix and eccentricity.


all right in this video we're going to

do an example of using some sum and

difference identities to simplify an

expression and uh here are kind of the

uh the six identities that we'll make

use of again each one you know has a

positive or A negative giving you a

different identity in there so the first

one here uh we're just going to simplify

s of 20° time cosine of 40° plus cosine

of 20° Time s of 40° well it's just

picking out the identity that it looks

like well there's signs and cosin mixed

up so that has to do with the first

one um if there's a positive sign in

between again for our sign identity it

says if there's a positive sign uh we

keep a positive in between so it says

this is simply going to equal s of well

20° plus

40° and that simply gives us well s of

60° and let's see s of 60° again on the

unit circle that's the same thing as pi

over

3 and sign at that uh at that angle uh

the unit circle s of 60° is going to

give us simply the value < TK 3 over 2

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 1.  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

look at some examples related to the

product rule and the product rule again

says if you have a function times

another function and we want to take the

derivative what we do is we take the

derivative of one of the functions

multiply it by the original of the other

then we stick a plus sign in there and

then the function we took the derivative

of the first time we leave it alone and

then we take the derivative of the other

function um notice a couple things here

uh you know multiplication and addition

are commutative so we can you know we

could have done um you know FG prime

plus frime G um it doesn't really matter

the order that you do these things as

long as you remember to take the

derivative of one part leave one alone

plus and then just kind of switch them

out so people uh you know sometimes are

worried about that a little bit so here

we're going to find the derivative of uh

a couple different functions and we'll

do these uh maybe uh one at a time here

so so uh in part A we have f ofx = x - 2

uh * x + 3 now normally what I would do

before if I were going to take the

derivative of this function I would

actually probably multiply it out first

um combine my like terms and then take

the derivative but just to illustrate

this product

rule so again the the product rule says

uh take the derivative of one of these

two factors so I'm going to take the

derivative of the X x - 2 well the

derivative of 1x is just one the

derivative of a constant is just zero so

I'm not even going to write it in there

and then we'll have the x + 3 so we're

leaving that part

alone and then we'll stick our plus sign

in between there and now we'll leave the

x - 2 alone and then we'll take the

derivative of x + 3 well the derivative

again derivative of 1 x is just one the

derivative of plus three is just 0 so

that would be our derivative and I'm

going to go ahead and combine these we

would get x + 3 plus

X+ oh almost did it wrong so we would

get x + 3 + x - 2 and when we combine

those we have x + x which is 2x pos3 - 2

will give us pos1 and that's our

derivative notice if you had distributed

this out at the very beginning um if you

had distributed out we would have x * X

which is x^2 we would have a posi

3x -2X - 6 so we would have x^2 + x - 6

um if we did the multiplication uh first

now if we just use our actual product

rule U well the derivative of x^2 is

going to be 2x the 1st the derivative of

POS 1X is just going to be POS 1 uh the

derivative of -6 is just zero so again

just illustrating hey you do get the

same thing um but again just in this

video uh just trying to illustrate the

use of the product rule here a little

bit so all right I think I'm going to uh

do the other ones in separate videos and

we'll also talk about maybe trying to

simplify these down and clean them up a

little bit because sometimes that can be

pretty tedious

Basic Product Rule 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 !! Please consider supporting PJMT on Patreon! https://www.patreon.com/patrickjmt?ty=c
Basic Product Rule Example #1


okay in this example we're going to

graph the hyperbola x^2 over 2 = 1 + y^2

over 2 and what I'm going to do is sort

of write this in a you know a more

standard form so the first thing I would

do is you know I'm just going to

multiply both sides by two uh just to

get rid of those denominators to me it

just kind of is something to uh kind of

clutter up the problem and just uh just

maybe confuse us a little bit so uh the

twos will cancel out leave us with X2 on

the left same thing on the right the

twos will cancel out we'll have 1 + y^2

and now if we subtract y^2 from both

sides we'll get x^2 - y^2 = 1 and we

could even write this as x^2 over 1 -

y^2 over 1 =

1 and well you know really we could even

write one as one squared and again one

is one squared so now we have it in this

form you know x^2 over a 2 - y^2 over b^

2al one and to me you can already start

reading off you know a lot of uh a lot

of information about this so for one

thing this is going to be centered this

is going to

be centered at the

origin okay since it's of the form X2

over a number minus y^2 over a number

that tells me that the parabola or

excuse me the hyperbola the hyperbola is

going to open you know to the left and

to the right okay so um so we know that

much of it as well um the value

underneath x s since it's 1 squ that

tells me that a equals 1 and again a

equals 1 that's simply the distance from

wherever it's centered out to one of our

vertices so that tells me that the

vertices are in fact going to be at 1

comma 0 and also at 1 comma

0 okay so uh we've got that much and I

guess the last little bit that we could

use here maybe to help sketch our graph

we know that the ASM tootes are going to

be of the form y = positive negative A

over B * X well again a is equal to one

it's the number that's being squared

underneath X and B is also going to be

equal to one again it's the you know the

number that's being squared underneath

the uh the Y squ

term so our asmp tootes will be of the

form positive negative 1 over

1 x which will just be you know positive

and negative 1X so yal positive X will

be one of our ASM

tootes so there's y = postive x and then

y = x will be one of our other asmt

so of course we should make the graph

kind of uh you know start to approach

those ASM tootes but that would be a

very you know a very rough sketch now of

the of the hyperbola okay it's just uh

kind of getting it you know in that

certain recognizable form again that

tells me if it opens left and right or

up and down um once I figure out my A

and B I can get my ASM tootes I can get

my vertices and again I think that's

enough to make

a good decent rough sketch

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 : Graphing a Hyperbola.
In this example, we are given the equation of a hyperbola; I show how to do a rough graph.


okay here we're going to look at finding

the volume of a cone so we're going to

find the volume of a right cone that has

a circular base with a radius of five

and a height of 15 and the units are in

centimet we're going to approximate Pi

by using

3.14 so the volume is 13 multipli the

base times the

height

well the base in this case is a circle

so recall to find the area of a circle

we use pi times the radius which in this

case is five squared and then we

multiply that by the height which is 15

okay so now I'm going to approximate

here I'm going to replace Pi with

3.14 well 5 SAR will be

25 and then we simply multiply that by

15 so let's see here um 1/3 of 15 so 1/3

of 15 would be five still multiplying by

3.14 *

25 5 * 25 would be

125 and then we still have to multiply

that by

3.14 so I'm going to use a calculator

here so

one so

25 multiplied

3.14 I'm getting this to be equal to

cm

Volume of a Cone

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Volume of a Cone.  Here we look at an example of finding the volume of a cone.


all right here we're going to do one

more limit problem um here we're going

to look at the limit as n goes to

Infinity of n raised to the negative n

to the thir

power the first thing I'm going to do is

just rewrite

this so n raised to negative n the 3

okay we can always put that over one

well we can make negative exponents

positive by sticking that exponent in

the denominator so now we have 1 over n

raised to positive n the 3

and you know if you think about it again

we're letting n Go to Infinity so

certainly for n values greater than or

equal to one I claim that well n raised

to n the 3 power that's always going to

be whoops I almost got my inequality

backwards n raised to n the 3 power

that's always going to be greater than

or equal to n the 1 I mean imagine if

you plug in one you'll have one to the 1

and on the left and one to the first on

the right if you plug in two for example

though you're going to have 2 raised to

the power of 2 Cub so 2 to the 8 is

certainly greater than or equal to two

again if you plug in three you know 3

raised to the 3 cubed or 3 raised to the

27 power certainly greater than or equal

to three so I think this is uh pretty

clear I don't think many people would

argue with

that well that tells us that if we take

one over this

number that's going to be less than or

equal to 1/ n again you know our

fraction now the fraction on the left

has a bigger denominator well if a

fraction has a bigger denominator and

the same uh numerator that fraction is

going to be smaller well uh again since

we're using positive numbers n is some

positive number likewise 1 over n ra to

n 3r that's always going to be greater

than or equal to

zero so now we can bring in our squeeze

theorem and we can simply look at the

limit as n goes to Infinity of the left

side which is zero well that limit

certainly is zero likewise if we look at

the limit as n goes to Infinity of the

right side well the limit as n goes to

Infinity of 1 over n that's also

definitely zero so what this now tells

us is that by the squeeze

theorem so by The Squeeze theorem the

limit as n goes to Infinity of 1 and

raed to n 3 must also equal

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Squeeze / Absolute Value Theorem for Sequence : Ex 3
Here I look at an example where I use the squeeze and or absolute value theorem to find the limit of a sequence.


okay here we're going to solve the

system of equations -2x + y + z = 5 3 y

+ z = 13 and Z equals 4 so this one's

basically uh almost almost done for us

and the idea is eventually what we're

going to try to do is we're going to

make more complicated systems of

equations eventually resemble something

like

this well we're already told that Z

equals 4 So eventually our solution

in this

case it will be a point it will be an

ordered triple so we'll have an x a y

and a z value we already know that the Z

value is four well to get the Y value we

can just substitute that back into the

second equation so we have 3 y + z = 13

but again Z equal

four well if we subtract four from both

sides we'll get we'll have 13 minus 4 is

9 and then if we divide both sides by

three we'll have simply that y equal 3

well now we know the Y value and the Z

value so we can solve for the x value as

well we'll have -2x + Y which is

3 plus Z which is

4 that equals

-5 well 3 + 4 that's 7

if we subtract s from both sides we'll

have -5 - 7 which will be -12 and then

if we divide both sides by -2 we'll have

that xal 6 so our solution to our system

of equations is when the value of x is 6

Y is 3 and Z is four

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Solving a System of 3 Equations, 3 Unknowns Using Substitution - Basic Example.  In this example, most of the work is already done for us.  All we have to do is substitute in our known value to find the other values.  In general, we would like to take a system of equations and algebraically manipulate it so that we create a system as shown at the beginning of the video and proceed as in this example.


okay in this video we're gonna start

looking at solving some multi-step

linear equations just where we're gonna

have to combine some like terms maybe

like in Part C get rid of some

parentheses by distributing so just uh

just problems that'll take a few more

operations to to to get to the solution

so Part A here we have 21 equals 3w

minus 12 plus 5w the first thing I'm

gonna do is I see that we have like

terms on the right side of my equation

the 21 I'll just drop that down but I

see a term involving W and another term

involving W so we can simplify those so

again I just kind of almost pretend like

the negative 12s not there well 3w plus

5w would be 8w well I do need to

remember that the negative 12 is there

so I just drop that down and again I'm

trying to get the W by itself so what

I'm gonna do is I'm gonna add 12 to both

sides

well 21 + 12 is gonna we'll get a 33

we're left with 8w on the right side and

again to solve for W since we're

multiplying by 8 I'm gonna divide both

sides by 8 so on the left side we're

left with 1 w 33 divided by 8 you know

that doesn't reduce as a fraction you

can always turn it into a decimal or a

mixed number but I'm gonna leave it

right there that's typically how I would

I think that's how you would typically

probably see the solution in any event

so doesn't have to be a nice whole

number obviously that happens but again

nothing too crazy just a matter of

combining your like terms getting the

term involving W by itself and then just

getting rid of the coefficient by

dividing

Solving Multi - Step Linear Equations

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 we're going to

do an example of actually just trying to

find an identity uh for this expression

Cent x * tangent xus cosine squ and uh

this one looks maybe a little tricky at

first I don't know it's definitely uh

turns out to be not too bad and again

the idea I'm going to use usually if I'm

not sure what to do um I don't see any

identities that are readily apparent um

I just turn everything into SS and

cosiness so Cent X that's cosine X over

sin

x tangent of X is just sin x over cine X

and then we still have our minus cosine

2 x left

over well we can simply just start

canceling out factors we've got a cosine

over cosine we've got s over s again you

can think about there's being uh ones on

the top and bottom again that wouldn't

change their value so we're really left

with 1 over one or 1 - cosine s X but

again recall our identity that sin^2 X

plus cosine 2

x equals 1 whoops uh so if we simply

subtract the cosine squ from both sides

we'll get that sin^2 xal 1 - cosine 2 x

so that's simply what I'm going to plug

in uh it says 1 - sin^2 is sin^2 X and

that would be a nice you know sort of

much more uh reduced certainly less

cluttered expression than what we

started with

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Proving an Identity - Other Examples, Example 2


all right in this video we're going to

do an integration by parts problem and

again recall that the integration by

parts formula says if you integrate udv

that equals UV minus the integral of vdu

and we'll talk about exactly what that

means so the idea is you're starting

with the left side of this integral

we're going to label something to be U

and we're going to label something to be

DV so in this problem I see two things I

see an X and then I see the e to the X

and we'll attach the DX to one of those

so I have to pick one of those to be U I

have to pick one of those to be

DV and as a general rule of thumb you'll

pick U to be the thing that gets more

simple when you take the derivative of

it so so if I let U equal x the

derivative of that will be one if I let

U be e to the X the derivative of that

is just e to the X which really doesn't

get any better so I'm going to let U

equal x and then from that I'll get that

du equal 1

DX okay I'm going to pick DV to be well

the rest of the problem e to the x

DX and to get back to PL old V well

since I have the derivative of V to get

back to V I would have to find the

anti-derivative but the anti-derivative

of e to the x is just e to the X and on

these problems you don't have to worry

about the plus C on this

part okay so I've now picked

U I've now picked DV in my problem so

the idea is I've used the left side of

this

equation I've rewrote this as udv and

now I'm going to rewrite it using the

right side of the equation so it says

you take whatever U is in this case X

you multiply that by V which in this

case is e to the X and then it says from

that we subtract away the integral of V

okay that's e to the

X time du and again du in this case is 1

DX so the idea in general is hopefully

you've went you've basically turned one

integration problem into a new

integration problem and

notice part of the problem has no

integral in it at all so you're done

with that part of it hopefully the new

thing that you have left that you still

have to integrate has somehow gotten

better than what you originally L

started with if you're getting something

more complicated chances are you picked

the wrong U and the wrong

DV but in this case I can certainly

integrate e to the X so my x e to the X

from the first part just drops right

down minus and now I find the

anti-derivative of e to the X which is

just e to the X now that I've completed

all the anti-derivatives I tag on my

plus C and that's it I'm done I found

the

anti-derivative

❖ Integration by Parts - Integrate: xe^x ❖

📚 Master Integration by Parts with a Classic Example! 📚

In this video, we delve into the powerful technique of Integration by Parts in calculus. Join me as I demonstrate how to effectively integrate the function xe^x, a classic example that showcases the method's utility.

What You’ll Learn:

Understanding Integration by Parts: A brief overview of the formula and its applications in calculus.
Step-by-Step Example: Follow along as I walk you through the integration process for xe^x, breaking it down into manageable steps.

Tips and Tricks: Discover useful strategies for selecting  u and dv to simplify your integration tasks.

Why Watch This Video?

Ideal for Students: Perfect for high school and college students studying calculus.
Clear Explanations: Easy-to-follow instructions that demystify a key integration technique.
Enhance Your Calculus Skills: Build confidence in your ability to tackle a variety of integration problems.

📈 Don’t Forget to:

LIKE this video if you find it helpful!
SHARE with classmates or friends looking to improve their calculus skills!
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all right in this video we're going to

look at an example of finding the

interval of convergence so here we're

going to find the interval of

convergence for e to the x squared the

series representation for e to the x

squared which is N equals 0 to infinity

X to the 2n over N factorial what we're

going to do to find this interval of

convergence

is we're just going to use the ratio

test so the ratio test against as we

look at the limit as n goes to infinity

of a sub n plus 1 over a sub n so in

this case we've got the limit as n goes

to infinity so everywhere there's an N

I'm going to replace it with n plus 1 so

we'll have X raised to the second times

n plus 1 over n plus 1 factorial and

again normally you well you know we

divide by the original but again we can

just multiply by the reciprocal so we'll

multiply by n factorial over X to the 2n

so let's see I'm going to just clean

this up a little bit this would be X to

the 2n plus 2 I'm going to put the X to

the 2n underneath that then we've got n

factorial over n plus 1 factorial well

if we simplify we have the limit as n

goes to infinity X to the 2n plus 2 over

X to the 2n we're just subtracting

exponents that will leave us with x

squared recall you can just expand out

in factorial over n plus 1 factorial

that's going to leave us with 1 over n

plus 1 and in this case you can treat

the X like a constant so we can pull it

out technically you need to keep things

in absolute value here you're squaring

it so it really doesn't matter but just

as a force of habit I would say when you

bring constants out keep them an

absolute value so we've got the absolute

value of x squared times the limit as n

goes to infinity of 1 over n plus 1

well this is going to leave us with x

squared times well let's see the limit

as n goes to infinity of 1 over n plus 1

that's going to be 1 over a large number

which is 0

well and recall what we want so we want

you know we want our ratio to be less

than 1 remember for the ratio test for

the series to converge the ratio has to

be less than 1 so I'm kind of thinking

you know what values could we plug in

for X and multiply it by 0 and get

something less than 1 well any number

that you plug in and multiply it by 0

you'll get something less than 1 so what

that tells us is it says the interval of

convergence it says the interval of

convergence is just going to be all real

numbers so everything from negative

infinity up to positive infinity

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Finding Interval of Convergence for a Given Power Series Representation. In this video, we find the interval of convergence for a given power series.


all right in this video i want to do an

example of a word problem involving

consecutive integers

and here we want to find two consecutive

integers whose product is nine less than

the square of the larger

okay so again i'm gonna try to turn this

this restriction into an equation that

we can do some math on

so again my two consecutive integers

i'll call one of our integers

x and to get to an integer right after

that we would just add one to it

so those are certainly consecutive

integers again assuming

x is an integer okay now it says we

their product is nine less than the

square of the larger

so okay their product is simply going to

be

x times x plus 1.

again make sure you put things in

parentheses because we're going to

distribute things out

so that's our product it says it's 9

less than the square of the larger so

this nine less i usually kind of do that

at the end so

whose product is okay

nine less than the square of the larger

well x plus one would be the bigger of

these two numbers

it's square would simply be x plus one

squared and now i think okay this nine

in here i'm either going to put a minus

nine or a plus nine on the left side to

make everything

work out okay so it says this product is

actually

nine smaller than what we have on the

right side

so to make it equal i would have to

add 9 to the left side and now

everything is balanced out

okay so again we've got a little bit of

algebra to do here now so

i'm going to distribute the x to my

first set of parentheses so we would get

x

times x which is x squared x

times positive 1 is positive 1x we have

our

plus 9 left over now on the right side

be careful

because again x plus 1 squared is really

the same thing as

x plus 1 times x plus 1. so we're going

to distribute that out don't you know

you don't want to put

x squared plus 1 down here that would be

incorrect so

but if we multiply it out x times x

would be well

x squared x times positive 1 would be

positive 1x

1 times x again would be a positive 1x

and then positive 1 times positive 1

will give us a positive 1.

so now we're left with x squared plus x

plus 9 on the left not doing anything

there

on the right i can collect my like terms

we've got x squared

we've got a plus x and a plus x which is

positive 2x

plus 1 and again i'm trying to solve for

x

so the first thing i'm going to do is

i'm kind of recognizing

that i could subtract x squared from

both sides

and that would just cancel out my x

squared

then i would be left with x plus 9.

equals two x plus one

i think i'm going to subtract x from

both sides

so on the left side i still have

positive nine on the right side i have

two x minus

x which is 1x

plus 1. and again now to simply get x by

itself we'll subtract 1

subtract 1. so it says the smaller of

the two integers is going to be equal to

and again if we add one to get to the

other one

it says the other integer

it says the other integer would have to

be the number nine

so it says our two consecutive integers

that

that fulfill this requirement are the

integers 8

and 9.

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! Word Problem: Finding Consecutive Numbers That Satisfy a Given Requirement - Ex 3


in this video I'm going to do a very

simple definite

integral and a definite integral is

different than an indefinite integral

first off notice we have these numbers

in the top and bottom of our integral

sign these numbers are aptly named the

bottom one is called the lower limit of

integration lower

limit of integration

and the top one as you might be able to

guess is called the upper

limit of integration so just some

terminology

there and basically what you do in these

problems is you find the

anti-derivative normally it without the

numbers you're finding the most General

anti-derivative a definite integral is

going to turn into a number so long as

this function is continuous over this

interval but basically you just find an

anti-derivative just like normal so the

anti-derivative of x^2 is X cubed /

3 the anti-derivative of x to the 1st

will get

x^2 over

two the anti-derivative of -2 will be

and in these problems we'll leave off

the plus C and we'll talk about why in

just a

second now we make a little bar over

here four and three are upper and lower

limits of integration and this is just

reminding us now to plug them in so the

first thing you do is everywhere you

have a variable you take your upper

limit of

integration and you plug it in so I plug

it in here here and here all right well

let me do

that so if I substitute that in I'm

going to get 4

cubed over 3 + 4^

2/ 2 - 2 *

2 and then what we do is we subtract

off when we plug in the lower limit and

just like we plugged the upper limit in

we're going to plug the lower limit in

everywhere as well it's a good idea to

use parentheses because you're going to

have to distribute the negative to every

term if you don't simplify first so I'll

get 3

cubed divided by 3 so now I'm plugging

threes in + 3^

2ar /

2 - 2

* 3 and I realize I just made a mistake

here I I plugged a two in here I should

be plugging a four in here so sorry

about that in case you're wondering what

the heck's going on absolutely right so

I've got 4 cubed 4^ 2ar 2 * 4 and then

I'm going to subtract off plugging a

three in

everywhere and we could simplify this

down you could definitely simplify

everything thing down in the parentheses

in the first part simplify everything

down in the parentheses in the second

part take that number minus the new

number

and that'll be your solution so this is

the basic idea with the definite

integral this number is not something

completely just out of the blue it does

have a geometric interpretation and

we'll talk about the geometric

interpretation of a definite integral in

another video

Basic Definite Integrals

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! This video has a couple of examples of calculating relatively simple definite integrals.

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all right this video is going to be the

first video um in kind of a series of

videos here talking about the lass

transform and the llas transform is used

to solve certain types of uh

differential equations so the basic idea

in this case we're not going to do

anything in this case um in this video

the basic idea is is as follows so what

we're going to do is we're simply going

to start with some you know a type of

differential equation what we're going

to do is we'll then use this notion of

what it's called the Lao transform okay

so the Lelo transform is just what's

known as an operator um it transforms

one function into a different type of

function okay so what we're going to end

up getting is a nice U algebraic

equation uh after we do our lass

transform what we'll then do is we solve

this algebraic equation and what that

solution is going to do it's going to

help us recover um information about our

original um about solution to the

original differential equation and once

we solve this algebraic equation uh we

use What's called the inverse Lao

transform and that's going to somehow

get us back to the solution to the uh

original differential equation so uh

obviously a lot of little uh things

involved in the process here so um I

don't know hopefully uh we can talk

about them all and try to make some

sense out of it so uh the next video

what I'm actually going to do is give

the definition of the Lelo transform and

actually talk about calculating the llao

transform for a function

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! The Laplace Transform - The Basic Idea of How We Use It.
This video is the first of (what will be) many videos about the Laplace transform.  In this video, I discuss the basic idea of how we will use the Laplace transform.  There are no computations or anything heavy in this video! Just an outline of 'where we are going'.


OK, so this will be the last problem from the first section of the SAT practice test.

This is the last problem where you are not allowed to use a calculator.

OK, here we are given that a equals five multiplied by the square root of 2

and two times a equals the square root of two times x.

We want to know what's the value of x.

So, this one is just a matter of doing a substitution.

So, again we got two times a equals the square root of two times x.

Well we were told that a equals five times the square root of two.

So I'm going to replace the a with five times the square root of two.

and I am going to simplify the left side so two times five is ten

multiplied by the square root of two, we have the square root of two x on the right side

I can break that up as square root of two multiplied by the square root of x

so again I am trying to solve for x

so the first thing I am going to do is now I am going to divide both sides by the square root of two

So, the square root of two will cancel and I am left with ten equals the square root of x

and now I can simply square both sides

so the square root of x squared is just going to leave me with  x

10 squared is 10 times 10 which will be one hundred and that will be our solution in this case.

so not to bad, again, just a little bit of a substitution and

you know, just solving this basic equation that involves a square root.

Thanks to all of you who support me on Patreon. You da real mvps!   $1 per month helps!! :)  https://www.patreon.com/patrickjmt !! SAT Math - Question 20, 2016 Edition and Beyond.
Section 1 - No Calculator allowed!


all right in this video we're going to

look at another example involving trig

functions so here we're going to

integrate tangent cubed x * cosecant SAR

x minus one and you know to get started

on this one the first thing I would

think you know if I distribute it out is

that going to make life any easier um

and you know maybe we could there's you

know kind of different ways to get

started I guess but the main thing I

think about is I see this cosecant squar

and I remember there's an identity that

involves cosecant squ one of them recall

is that 1 +^ 2 X = cose 2

X well if we just subtract the one over

we'll getent 2 xal

cose cose 2 X - 1 so I'm just going to

replace the cose 2 X - 1 with Cent s so

really we would be left with Tangent

cubed times ENT

squared

but if you want to kind of break this

apart you know tangent is sin cubed over

cosine

cubed um Cent squ that would be cosine 2

X over sin^2 X and well we've got sin

cubed over sin squar that would leave us

with a sin x in the

numerator we would have cosine squar

over cosine cubed um that would give us

cosine X DX and this is the same thing

as the integral of well tangent of X so

haha um and we already saw how to do

that in the previous example so the

anti-derivative of tangent of X we get

the natural logarithm of see X plus C

and again we found our anti-derivative

so really uh you know just being a

little sneaky this original function is

just a you know an equivalent way of

writing the function tangent X

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 #2


okay here we're going to look at an

example of writing a fraction in uh the

simplest form or most reduced

form and being able to go from a

percentage to a fraction to a decimal is

certainly useful uh to be able to switch

interchangeably between those three so

to write a percent as a fraction we take

whatever the uh the number is in this

case it's 32 and we just simply put that

over 100 and that's always what we do

and at that point it's just a matter of

reducing it so let's see what number

goes into both 32 and 100 well it's even

so I know two does at least let's see 32

divided two would be 16 100 divided by

two would be 50 but those are both still

even so let's see uh 16 divided two

would be 8 50 ided two would be 25 so we

could have just divided top and bottom

by four at the very beginning but I

usually just always just start dividing

by two if I see they're even 8 over 25

well let's see one 2 four and eight go

into eight none of those numbers other

than one go into 25 so this would now be

our fraction in simplest form

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 Percent as a Fraction in Reduced Form.  Here we write 32% as a fraction in reduced form.