Raw Transcript
Hello Everyone, Tom for everystepcalculus.com and everystepphysics.com. I’m going to do partial fraction problem. This person said that it’s a hard partial fraction problem. Let’s see about that. It is hard if you ask me, without my programs. But let’s do it. Index 8 to get to my menu. Scroll down to partial fractions. We’re going to enter the integral. You have to press alpha before you enter anything in these entry lines. We’re going to go alpha and then going to do left parentheses, X squared plus 3 times X plus 1 close off the parentheses divided by, open up the parentheses, X to the fourth power plus 5 times X squared plus 4, close off the parentheses.
I always show you what you’ve entered. Now maybe you’re better than I am but you have to factor this denominator in partial fractions every time. So this it makes it hard for a class in it for a test problem is way too hard for a test probably for homework because I don’t know who would be able to write off from memory to factor that. I say it’s okay. We factor the denominator; here it is right here. And we start doing our partials. The idea is that you are going to deliminate the denominator here by multiplying at times the same thing with the numerator this. This here is the the factored part ion is here and this is the original denominator. And so then we have
the numerator is equal to Ax plus B times x square plus 4 plus Cx plus D times x squared plus 1.You multiply that out using the foil method. Remember the foil method? First outside inside and last. First would be Ax times x squared,outside would be Ax times four, et cetera. I multiply it out here. The calculator uses small letters rather than capitals. No problem. They combine those terms and come up with this. A plus C is x squared, x squared, 4a plus c, et cetera. Now we have to figure out the coefficients for this numerator. And since there’s no X cubed, we have to put one in there, which is 0 times x cubed. And then we have one coefficient here, three coefficient here, and one. So here we have one, three, and one. So then therefore AC is equal to 0, BD equal to one. 4a plus c is equal to three and 4b plus d is equal to one. And we work out, we use subtraction. We’re subtracting like terms. So AC is subtracted. 4a plus c and that equals three. A equals one. Do the same for each one of them. D equals one, C equals a minus one. Notice here we’re replacing A with what we found which is one. Which I do that for you here. When they’re in quotation marks, I replaced it in there. Partial fractions are this. Right here. And I also do the integrals for you. Log of x squared plus one divided by two, et cetera, et cetera. Pretty neat, huh?
everystepcalculus.com. Go to my site and buy my programs.
Thank you
Partial Fractions
Let’s talk about partial fractions and what I’ve found out after programming them. Partial fractions would never occur in real life. Remember the integral is the area under a smooth curve, nothing more from my knowledge, when you graph any original function; it has at least 2 asymptotes. Well that eliminates the smooth curve. Remember an asymptote is a vertical line or sometimes horizontal line where the original function never touches to infinity, so if you choose a range that goes (crosses) over that asymptote there is no computed answer, so no area under a smooth curve. So partial fractions are a Sudoku of math problem, like so many of calculus problems. So let’s get into what I’ve found in programming this area of calculus. The denominator has to be factored to produce partial fractions. Sometimes there are two factors, then three, then two factors with an exponent in between the parenthesis, then even one factor with an exponent outside the parenthesis. The factored denominator is the key as to how to approach the problem.
In my programs, I have a whole separate program to decide if there are two factors or three or any other choices. Most of us are not good at factoring. Maybe simple functions we can, like the difference of squares (x^2-9) = (x-3)(x+3). When it gets a little deeper that this most of us are lost. For instance, what’s the factors of (x^3-x) or (x^4+7x^3+6x^2)? The calculator knows and because of that, so does my programs. But would those appear in your test? In my opinion if they did then most of the class would fail that. No question about it. Let alone complete the partial fractions. So enough about factoring. Remember in algebra when they said you could do anything to one side of an equation as long as you do it to the other side also. They do this in partial fractions. On the left side of the equation in partial fractions they multiply the original function by the denominator, which effectively gets rid of the denominator. Well they do the same thing on the other side (except factored) to get rid of the added or subtracted fractions, and change the A / (x-1) to a product through common denominator. That said, a few trick to remember in partial fractions. If in the factored denominator you have (x^2+6) for the first factor so you want to set that up for partial fractions you would go Ax+B / (x^2+6), if the second factor was (x^2-3) you’d set this up as (Cx+D) / (x^2-3), Do you understand this? Very important and no exceptions!! If the factor turned out to be (x-5)^5(exponent outside of the parenthesis), this is called re-occurring powers, or exponents, and the factors are A/(x-5)^1 + B/(x-5)^2+C/(x-5)^3+E/(x-5)^4+F/(x-5)^5. Now if the power in the denominator is less or equal to the power in the numerator, you have to use short division to find the remainder to the partials.
My programs do all of this for you, as I required my programming to do for me also when I was in class. Hey good luck in your class, I’m always available to help you if you just ask.
Partial Fractions Example 10-Video
Partial Fractions | Example 9-Video
Raw Transcript
Hello Everyone, Tom from everystepcalculus.com and everystepphysics.com. Partial Fraction Decomposition. This time with three Partial Fractions.Let’s get started.Put index 8 in here with the open and closed parentheses, you get to my menu.You can scroll down to start up in the a section but you can scroll down alphabetical to Partial Fractions.Enter the integral by pressing Alpha first. Alpha, parentheses X squared plus 12 times X plus 12,close off the parentheses, divided by, open parentheses X cubed minus 4 times X. I always show you what you’ve entered. How many of you could factor this here? It’s pretty tough, I would think. Do it easy on my program. It comes up with 3 factors. So you write this in your paper. X squared plus 12x plus 12 is the numerator. Divided by the denominator times the denominator. And this is the actual denominator factored so we do the same thing here. And we’re going to use this to multiply times all of this to get rid of all the divisions. Common denominator is the actual words. So we added with x squared plus 12x plus 12 equals A and x times x minus 2 times x plus 2 Bx times x plus 2 Cx times x minus 2 And x equals 2. These quotation marks here is what the calculator does but you would put parentheses wherever you see these. parentheses 2 squared. You’re substituting 2 for every x in these functions. Use parentheses 2 here plus 12, etc. Do that throughout the whole system here. And that turns out to be B equals 5. At x equals zero. Use 0 for every x in all the computations here. And that turns out to be A equals minus 3 and x equals minus 2. Substitute for x in all the equation. C equals minus 1. Partial Fractions are minus 3 over x plus 5 over x minus two plus minus one over x plus two. And we integrate those. One thing about integrating a denominator with no, no exponent. You notice you can’t. This is an exponent of one, theoretically. But if you try to switch it to the numerator, five times X minus 2, it becomes a minus 1. Well in integration, you’re going to add one and then divide by the result. Well if you add one to a minus one, you get zero. In the numerator, anything to the zero is one. So that’s the reason you have to use logs. Whenever you see a no exponent to the denominator, it’s a log integration. So minus three logged minus the absolute X 5 log absolute x minus two minus one log x plus 2 plus C. Pretty neat, huh? everystepcalculus.com
Go to my site and buy my programs.
Thank you for watching.
Concavity-Video
y = x^3-6*x^2-3*x+1
Raw Transcript
Hello Everyone, Tom from everystepcalculus.com and everystepphysics.com. We’re going to do a problem on Concavity. And I’m gonna show you how my program work on that. Index 8 to get to my menu. We scroll down in the c section to Concavity. There is is there, letter D. I always have you start a graph on your paper. You put the x axis and y-axis with these. Keep up with graphing by hand, these functions. And I have a function here of the internet, for example. You have to press Alpha before you enter anything in these entry lines. Press Alpha. And the problem is X cubed minus six times x squared minus three times x plus one. I always show you what you’ve entered, you can check it and see if it’s correct. If it is, we press okay. I know we want Concavity, number 4. I’m going to press the number here, number four. Here’s Concavity. And I give the examples, here’s Concavity down and heres concave up. This is a valley, this is a mountain type situation. See, you take the original function. Take the first derivative here which is this. On your paper and the second derivative which is this. Six x minus twelve. We set 6 x minus twelve to zero to find what the x is at that which gives us critical number, there. And with respect to the second derivative and x equals two. So here we draw a number line here with the with two in the middle. Now we’re going to choose some number above two and some number below two. And and plug it into the second derivative here. I do that for you. I say X equals three. So at the second derivative x equals three, the answer is 6. And you’ll notice the 6 is positive. That means that its going to be concave up. And it’ll look like this on the graph. If we do a number less than 2 which is one. And plug it into the second derivative and we get minus 6. So, this is negative. So this is gonna be down. And I show you this and the next screen. So this one is down when you when you graph this function, it’s going to be down like this is going to be up like. That’s called Concavity. And that’s how you do it . Now on your paper, you must write negative infinity to two and two to positive infinity. And then you’ll be correct in your test or homework. Pretty neat, huh?
everystepcalculus.com
Thank you
Chain Rule Example Solved-Video
Raw Transcript
Hello everyone. This is Tom from everystepcalculus.com.
I’m going to demonstrate the Chain Rule right now.
Let’s do it. Index eight to get to my menu for calculus.
And we’re going to
scroll down to the C section. They’re all alphabetical.
Here’s chain rule here.
And I gave you a choice of what kind
problems you’re gonna do. Sine, Cosine, Tangent, etc.
Or given Y, etc.
Here’s the, in parentheses, with an exponent.
And here’s the square root of something. I’m going to do number 2 here.
Press alpha before you
put anything in these entry lines here.
Alpha first and the left parentheses which it gives you up here for example.
And we’re going to 5 times x squared
plus 7 to the fifth power
I always show you what you entered. You can change it if you want.
We’re going to differentiate that.
I say it’s okay. Here’s the formula here for chain rule.
and I go through the steps for you. It’s kinda like U-Subsitution
in a way for differentiation. U-Subsitution is really for integrals.
So after we put the terms into the into the formula. Mark everything down in your paper you see it.
Here’s the derivative of g of x
which is 10 times X
and keep writing and here’s the answer.
Solving Definite and Indefinite Integrals – Q&A
Dear Tom,
I would like to know if your program also solves integrals consisting only of symbols like x, a or b in both definite and indefinite matters?
Example:
or
Would solving does definite integrals be possible using your programs showing the solution process step by step?
Kind regards,
Ben
Answer
The first integral, sounds like your professor didn’t teach you, so you will want my programs to teach you or me to teach you. A common endeavor in my experience. When you see the da that’s “with respect to” in other words, you are integrating that integral with respect to “a”. In my programs you’d change all the “a’s” to x”s.
Now could you solve the integral if it looked like:
x x
∫ (1/x)dx or ∫ ( 1/[(c+b)-(k+x)] ) dx
o o
Now these would never appear on a test unless your professor demonstrated them on the black board, so they are either homework or experimenting off the internet. Both integrals are definite integrals because they show a range to integrate over. When you solve the integrals you have the area under that functions curve. That’s all integrals do is solve area under curves. You can graph 1/x on your calculator and you can see what the first integrals function looks like. Now, anytime you see a function without an exponent in the denominator, it is a natural log answer. The reason is is that you can’t integrate term with a division sign.
In this case the x has an exponent of 1 ( x^1), you always have to move the denominator to the numerator to eliminate the division sign before you can integrate. If you move the x^1 to the numerator to eliminate the division sign it becomes x^(-1) so when you do the integration process which is always add 1 to the exponent and divide the term by that answer, However when you add 1 to a -1 you get zero, and any term to the zero exponent = 1, so that’s why the natural log comes in. In the first integral above = ln(x), then you can enter the ranges for x that you are given, do the top range first and subtract the bottom range to get the answer (area under the curve of the function). The second integral above you’d break it into two integrals ∫[1/(c+b)] dx which equals x/(c+b) and then the second term 1/[(k+x)]dx which equals ln(k+x).
U-Substitution using Log Rule on TI-89
U-Substitution using Log Rule on TI-89
Raw Transcript
Hello everyone I’m Tom from every step calculus dot com.
Im gonna do a U-substitution problem
that requires the log rule. And uh,
let’s get into it. index8() is my,
you have to put that in the end line here to get to my menu.
I’m already at U-substitution.
You scroll down with the cursor to get there.
And wait for it to load here for a second.
And we’re gonna enter our function. You have to press alpha before you enter anything in my
entry lines in my programs.
Alpha 8 times
X divided by
parentheses
X squared plus
:00
I will show you what you’ve entered. The reason you need to do a log rule is
notice in the denominator your have x squared plus one.
This is an exponent of 1. If you we’re to
transfer this up to the numerator the exponent of
one would become a -1. And in integration you always add,
the first thing you do is always add to the numerator.
And one plus minus 1 is zero so the answer will always come out to be
-1. That’s the reason anytime you see this without, without an exponent other than
one. The one is of course hidden. Um,
you know it’s a log problem.
So we’re going to press there.
Say its okay. Here’s the original
function. And all these you have to rewrite them. Notice there is a constant error
of 8. You have to bring that outside the integral, that’s very important. And we’re going to also
take the x
here and put it over by the dx.
And we’re going to put one in the numerator,
for X squared plus 1 in the denominator.
So U is x squared plus 1 the derivative of that is 2x.
Another trick to all these, took me a long time to figure out.
Was d you need to have the divisor of 2
You have to move that over as a divisor here, and leaving
this here. You’ll notice that X DX is the same as x dx.
That means it’s a U-substitution problem if it wasn’t it’s not a
U-substitution problem. I ask you that though just
to make sure you know what you’re doing. I say yes.
So we have i8 outside the integral and one over you
du number 2. Now we have a constant here at one half.
So we have to bring that outside the integral, which we do over here.
8 and 1/2 that equals 4.
And then we have four times log of U. Anything over one of U is
log of u, plus c.
And the answer is, after we substitute the u back in.
4 times log of x squared
plus 1 plus c. Pretty neat huh?
Every step calculus dot com. Go to my site buy my programs, pass your calculus
class and
have these programs for the rest of your life your kids or grandkids or
your sister, whoever might take calculus in the future,
you have these forever, that’s what’s good about them. And also subscribe to me on my
you know, so that you can see the future
movies or the other blog.
Partial Fraction Decomposition on TI-89
Partial Fraction Decomposition on TI-89
Raw Transcript
This is a video from EveryStepCalculus.com demonstrating how my programs work on a TI-89 Titanium calculator and other calculators in the TI system for physics and calculus problems.
Okay, Partial Fraction Decomposition. Let’s get started, to get to my menu you have to press 2nd alpha to put the i_n_d_e_x in here you have to press alpha to put the 9 (). Press Enter and you’re into my menu. Scroll down to wherever you want all the way up and down.
But I’m at Partial Fractions right now which we want to do and we’re going to enter our function. You always have to Press alpha before you enter anything in a these entry lines in my programs. So Alpha we’re gonna do a quantity of 4 times X plus 9, divided by the quantity X minus 1, times the quantity of x plus 1, times the quantity of X plus 4.
I always show you what you’ve entered so you can change in case you made a mistake and we’re into the problem. You factored the denominator. It’s already factored here but in case it wasn’t, I would. And then you multiply, you multiply by the factor of the denominator. You get these. They’re all exactly right. Mark it down on your paper exactly like you see it. And if x = 1, the idea is that you cancel certain factors, certain variables. And here’s when you put -1 the function here is where they all set.
Mark that on your paper, turns out to be 5=A*(0)+B*(-6)+C*(0) 5=B*(-6) 5/-6. Mark all that on your paper. X equals one which will negate the second term. You mark these all these… put all these in replace or um x in your function there. A couple of them become zeros so that you can solve so a equals 13 tenths. And then you have enough variables, enough answers to solve the other c. Here’s the answer.
Pretty neat, huh? EveryStepCalculus.com. Go to my site, buy my programs and pass calculus.
More Partial Fraction test questions solved on your TI-89 calculator
Partial Fraction Test Question #1
U Substitution e^(x) Solved on TI-89
U Substitution e^(x) | Every Step Calculus
Raw Transcript
Hello Tom from every step
Calculus dot com. Here’s a problem on u-substitution.
L,et’s get started index8() to go to my menu.
u-substitution is the subject.
Press alpha before you enterr anything in these entry lines here.
Diamond key to get to the
e(x) problems this is 7 times
X. Divided by.
Anytime you using the denominator you always have to have parentheses
around the whole function.
And then we’re going to go
to e(x) again. 7
times X plus
8.
I always show you what you’ve entered, you can change it if you want, I say it’s okay.
You have to rewrite this. Notice the idea is that the seven ex/dx is the same as
here. Otherwise it’s not a u-substitution problem.
And then I check whether that’s that’s true.
So they’re equal so we say yes.
And here’s the answer 7 times log e to the x
plus c. Pretty neat huh? Every Step Calculus dot com, go to my site buy my programs
pass calculus, also subscribe to
my channel so you can see more movies
or blogs.
3 Calculus Questions Solved by TI-89
I’m really trying to figure out how to use your program with a couple different problem examples:
“Find the following derivatives of f(x). Do not leave any negative exponents”
1. F(x)= 4^x-ln(2x^2-x)
What program can I use to solve this?
another is f(x)=2csc(x)+5arccos(x)
This is fundamental to calculus and you must start to learn it or know it.
Notice the minus sign between the terms (if a plus no difference)
This means that the derivatives of the terms are separate or individual
You differentiate each by itself and one at a time
There could be 15 more terms each separated by a minus or plus and you’d do each one separate.
So: f(x) = 4^x f’(x) = x*4^(x-1) Notice when differentiating you take the exponent
Multiply it times the front of the term and take one away from the exponent.
You must do this process in your sleep and fast or you’ll have trouble passing calculus
Another Example
f(x) = 15x^7
f ‘(x) = derivative
= 7*15*x^(7-1)
= 105x^6
Integration is the exact opposite
You add one to the exponent first
Then divide by that new exponent
∫[105*x^6] dx
= 6+1 = 7
So: 105*x^7 / 7
= 15x^6
Problem 1) the trig problem
Reload the attached program
First Delete the same named program from you calculator
Then reload the attached changing to “Archives” first
For the problem go to:
“trig” in my menu
Scroll to csc(x)
Find the derivative and add the 2 in the front of the answer
For arcos(x)
Scroll to cos-1 in the menu of trig
And find that derivative adding the 5 in front
“Answer the following questions about the given function: f(x)=x^3-6x^2+9
A) Find f'(x)
B) Find f”(x)
These again are separated by minus and plus signs so each is differentiated separately
f(x) = x^3 – 6*x^2 + 9
f ’(x) = 3x^2 – 12x + 0
f ‘’ (x) = 6x – 12
Notice I took the exponent and multiplied it by the front of the term
And then took one away from the exponent in each case.
Do these in your sleep!!!!!
Use log differentiation to find the derivative of the following function with respect to variable x.
(x^3+2)^3sqrt(2x^2+3)
Notice the “times” sign so you’d be thinking “product rule” or logs to differentiate
First it must be expanded:
I have expanded log problems to another base
But haven’t programmed to expand natural logs, but am in the process.
Should be done soon
(x^3+2)^3 * √(2*x^2+3
ln((x^3+2)^3) + ln(√(2*x^2+3) (always you change a square root to ^(1/2)
= ln( (x^3+2)^3 ) + ln( (2*x^2+3)^(1/2) )
= 3*ln( x^3+2 ) + (1/2)*ln( 2*x^2+3 )
Now you can choose in my menu
“log Problems”
Choose “ln(x)”
Choose “differentiate”
And add the above ln tems one at a time
Put the problem back together when finished with the individual solutions.
Related Rates with TI-89: Triangle Heated
The area of an equilateral plate, being heated, is increasing at a rate of 150 mm²/min. At what rate is the length of a side changing when the sides are 250 mm long?
Raw Transcript
Hello everyone. Tom from everystepcalculus.com and everystepphysics.com I’m going to do a test problem on an equilateral triangle with regards to related rates. Let’s see the problem is the area of an equilateral triangle plate being impeded is increasing at a rate 150 millimeters squared per minute. At what rate is the length of a side changing when the sides are 250 millimeters long? So, let’s do this. You have to put in index 8 open and close parentheses to get to my menu. And we’re going to scroll down to average rate of change. You could scroll to related rates. Let’s do that. They are both the same. And scroll down to the R section here. Here’s Related Rates there. And we’re going to hit number 7 for Equilateral Triangles. I’ll show you the formula here. Write this stuff on your paper. And of course you differentiate both sides with respect to time. What’s given here, they’re giving you the rate of change of the area so we’re going to go and enter the change. I’m going to press alpha to put anything in here. Alpha 150. And it gives millimeters. Choose number 2. And it gives minutes. Choose number 2. And then the side is Alpha 250. Now we’ll show you what you’ve entered so you can change it if you want. Say it’s okay. Write all this down.
Exactly as you see it. Here’s an answer. .69282 millimeters per minute. Pretty neat, huh.
everystepcalculus.com. Go to my site, buy my programs, and subscribe to me so that can see other videos
Related Rates Test: Ladder Against Wall Video
Related Rates Test: Ladder Against Wall Video
Raw Transcript
Hello. Tom from everystepcalculus.com
Related Rates. These are word problems. Very difficult
even though seemingly simple but
let’s get started. Index eight is calculus one in my program to get to my menu.
That comes with my instructions of course and we’re going to scroll down to
from the menu on a scroll down to related rates. Could be on a test
homework or something and we find it all alphabetical
and in this case
we’re gonna do a ladder against the wall which is the usual
calculus problem related race
problem. It’s used in the Kahn Academy
for their example and also in this example here
from number nineteen. So when I scroll down to that
and we’re going to press enter.
I show you what it looks like. Here’s the ladder
and here of course is the y axis and the x axis. All related rates
are right triangle basically. And they
I ask for them at ladder length. You have to press alpha before you enter anything in
the my entry lines in my programs
and they give it as five meters.
And then you have to decide which is changing.
Is the ladder moving down the y axis
or is the bottom moving? Do they give you the bottom moving right or left?
So, in this case is the y is
the ladder is moving down to give you that change.
So we’re going to choose number two here.
Press Alpha again and they give that to you
as one meters per second so we’re going to choose one. Press enter.
And then they give you
certain distance on the x axis so we’re going to
press and that you need to use three. So we have these
parameters given. I ask you if that’s correct.
You can change it if you want. I say it’s okay.
The first thing with these ladder problems is that you have to find out the other side so
of course x squared plus y squared equals z squared and the
one that we don’t know is the y. We know the x is three and we know the
ladder is 5 so we need to figure that out.
So we do that automatically for you here.
It’s 4 units. I say units because they may be centimeters it, it might give you millimeters, it might give you something else.
You decide yourself and put those in for units.
In this case are getting in meters.
And then they the use implicit differentiation to
to differentiate these
terms here and so we have to do the derivative
of x squared, the derivative of y squared with respect to T
and T is time. The x axis is actually time also.
and z squared. We know z squared is 5 so you write all
this on your paper.
and it turns out to be a -1.33 units per second.
Pretty neat. everystepcalculus.com Go to my site, buy my programs and subscribe
to me also so you can see future
movies and blog.
Definite Integral Test Question Video
Definite Integral Test Question Video Example
Raw Transcript
Hello everyone. Tom from come everystep calculus.com and
everystepphysics.com
I’m gonna do it Definite Integral for you.
this came up on my Facebook and I want to answer it here in a video.
Let’s get started. Index 8 to go to my menu.
We’re going to scroll down to Definite Integral. It’s all alphabetical so it
should be in D’s here
and here’s Definite Integral.
And we’re going to enter the function. You have to press Alpha before you enter anything
into these entry lines here.
parentheses one
plus cosine
of two times x,
and close of the main parentheses,
divided by two. I always show you what you’ve entered. You can change it if you want.
I say it’s okay. Now we’re going to enter
the range. At a range of, the lower limit was
was alpha 0
and the upper limit was
alpa pi
divided by two. Again I show you, this is the problem that you’re asking
and was asked.
Now we do the derivative of that which is sin of 2x plus 2x
divided by four and
and pi over two is the upper limit and the lower limit is zero.
We substitute the upper limit for all of x’s.
Here were substituting pi over two for x, pi over two for x.
We come up with pi over four
or approximately .785. Substitute the lower limit
for x. Notice we substitute for x 0
and 0. Here’s 0 and the answer is 0.
So the Definite Integral is this
and we take the upper limit and minus the lower limit.
Pi over four minus 0 answers Pi over four square units or
.785 square units. Remember and integral is
an area under a curve.
So go to my site. Buy my programs if you want to
get help with calculus and subscribe to me also
if you wanna see future movies
or whatever, Thank You.
Calculus Q & A
Can the Calculus App do the Following?
F(x)=6x^-5
F(x)= 6x^4-4x^3+5x^2
Answer:
f(x) = 6x^5 would be “derivative/algebra” for differentiation
Partial Fraction Decomposition Video
Partial Fraction Decomposition on the TI-89
Raw Transcript
Hello everyone, Tom from everystepcalculus.com, going to do a partial fraction decomposition right now, letís get started.
Index 7 if you bought both programs and loaded them into the calculator with the instructions and youíre going to scroll to partial fractions, thatís what we want to do with the problem and weíre going to add our function, you have to press alpha before you add anything in this entry lines here. Alpha four divided by parenthesis parenthesis three times X minus one close up parenthesis times X and it will show you what youíve entered and notice this is already factored, this is actually three X squared minus X but itís already been factored here in this problem and we say itís ok, it allows you to change it in case you made a mistake, so here is the factoring of it, you always the denominator, if the denominator canít be factored itís not a partial fraction *** your problem, then we go through the calculations. A is over X and B is over 3X-1 and we multiply times the factoring because we want to eliminate this from the other side, when we multiply something thatís been divided we can eliminate it and it also makes these into linear variables such as this and then we try to make one zero so we could solve for the other one which we do here and B equals 12, Iíll go through it quick, now we have X equal to zero and so A equals -4 and hereís the answer, the partial fraction here is -4/X + 12 and when you do the X integration of it, it becomes a log problem and I do that for you too, -4 log of log of absolute value of X + 12 log of 3X, pretty neat, everystepcalculus.com, go to my site, buy my programs, pass your calculus class and also subscribe to new videos if you want at my channel.
U Substitution Video: Sine on the TI-89
U Substitution Sine on the TI-89: Raw Transcript
Hello everyone, Tom from everystepcalculus.com and everystepphysics.com, Iím going to do a U substitution problem that a customer sent me regarding sine and letís get started.
Index 7 to get to both menus, both programs are installed and weíre going to go to U substitution in the menu, we are going to enter our function, we have to pres alpha before we enter anything into this entry lines, alpha square root of X times sine of X to the three halves minus one. I always show you what you have entered, here we have square root of X times sine of X to the three halves minus one, notice the parenthesis, you have to use good parenthesis in math, learn that, itís important and I say itís OK, Iíll give you a chance to change it and when weíre evaluating this we rewrite it because this over here, the square root of X over here matches what weíre going to do in the next screen. U we chose is X to the three halves minus 1, the derivative of that is three halves square root of X and we always take the *** from this side and put it over to di divide by three halves, notice when youíre dividing by three halves youíre going to invert that to two thirds, always remember that important step in algebra and math and here is the same, we wrote it over here so we know itís a U substitution problem, if that wasnít the same and you canít make it to be the same then thatís not a U substitution problem and so here is sine of U is equal to sin of this right here which we know and ** with the integral of the sine of U we have du divided by three halves, here we convert it, two third times the integral of U, two third times the derivative of the integral of the sign of U is minus cosine of U plus C, answer is minus two third cosine of this as we substitute back in for U, pretty neat, everystepcalculus.com or everystepphysics.com, go to my site, buy my programs and pass your calculus class.
Related Rates Video on TI-89: Square Area Increasing
Related Rates Video on TI-89: Square Area Increasing
Raw Transcript
Hello everyone Tom from every step calculus dot com.
Related rates problem.
Right off of Yahoo on the Internet.
Where they ask the questions and
a lot of them try to answer it. Let’s get started, index8() to go to my menu.
And
going to, all the instructions come with my programs of course how to do all this
stuff. We’re going to,
go down to related rates.
Scroll down related rates because that’s the problem
we’re doing.
Here we are here, and we
these are all alphabetical everything is you wanna get to
this problem involving a square the area of a
square. So we’re gonna go down, that choice on the menu.
And
related rates always deals with a formula and then
changes that to derivatives with respect to time.
And so you write all this down area equals s squared.
I give you this little hint, s equals a side and we’re gonna find
da/dt the how the area is changing
with regard to time. And so
now we’re gonna do the calculation here’s the a equals s squared where there’s a
formula and we do the derivative
left side derivative of the right side with respect to T
time. And that turns into
the derivative of area with respect to time and then we have
derivative of s squared which is 2s.
And then ds/dt. This problem they give you area so we’re gonna choose that.
Otherwise they might give you side.
You try to make all those choices for you.
You have to press alpha before you enter anything in these entry lines. I’m gonna press alpha
give you 64
as the area.
I always show you in case you made a mistake I say it’s okay.
and of course the square root of the area is the side.
The formula is inside so we have to change it to that, so that’s 8
And then they give you centimeters.
They might give you any these. And the side is increasing at alpha,
8 centimeters per second
Again I show you its
8 centimeters for the side and 8 centimeters per second for the
change in the side which is ds/dt with regard to time.
Ok, so here’s the
answer you do the problem,you insert variables
Here’s the formula what we found before you insert the variables and you come up with this
answer.
Pretty neat huh? every step calculus dot com You can go to my site buy my program
pass your calculus test and maybe subscribe to me also
on the site so that
you can see my future movies and blog.
Related Rates Video: Square Area Heated
Related Rates Video: Square Area Heated
Raw Transcript
Hello everyone, Tom from everystepcalculus.com, we are going to do related rates versus a square and the area how it increases the feet, letís get started.
Index 8 to go to my menu and since its related rates average rate of change is the same as related rates, Iím going to press 2 here and wait for it to load here for a second and weíre going to scroll down because when you have an arrow like this it shows you thereís more in the menu and itís all alphabetical, so we want to get square area, this is problem number 7 book and so we want to do that, we want to show you how itís done. Related rates always deals with the calculus formula of theyíre dealing with and then you differentiate that on both sides with respect to time and weíre asked to find the change in area which is the derivative of A versus change in time and here is the actual formula for area and you write this all in your paper, this is exactly what you should write and theyíve given us the side in this problem number 7, so you have to put in, first you have to press alpha before you enter anything into these entry lines here. Alpha 12 is given, centimeters and the side is increasing, so they give you alpha .08 and thatís per minute instead of seconds, I gave that choice in my program, itís pretty neat there. So you have 12 centimeters per side and you have ds/dt which is the change of the side with respect to time .08 cm/mn, Iíd always give you a chance to change your in case you made a mistake, Ok and here is the answer. Pretty simple but remember you can do these problems for the rest of your life if you get these programs, if you get my programs versus if you just memorize it now and forget it later, so everystepcalculuc.com, go to my site, buy my program, pass your calculus class and also subscribe to me so you can see other videos and blog.
Related Rates: Given dx/dt find dy/dt
Related Rates Test Question: Given dx/dt find dy/dt on the TI-89
Raw Transcript
Hello Tom from every step calculus dot com.
We’re going to do another related rates problem with a function this time.
Uh, put index8() into here to get Calc one menu.
And we’re going to scroll scroll up
get to the bottom of the alphabet here. We need to get to related rates
which is what we’re doing. And we’re going to choose number three because
we’re going to do a function.
Gonna enter the function we have to press alpha before you enter anything into my
entry lines here. And y
in this problem.
Is alpha three times
x squared.
I always show you what you’ve entered so you can change it.
If you made a mistake. And they’re asking in this problem
for dy/dt. We’re gonna choose number one here.
And we’re gonna enter what the problem is, which is.
dx/dt is 2. And
X equals three, alpha 3.
I say it’s okay.
And here’s the
answer ,36 units.
Pretty neat huh? Every step calculus dot com. Go to my site buy my programs and subscribe
to me
for future movies.
Also if you have a related rates problem that you, that’s
not in my programs if you want me to do you can always contact me
and I’ll set it up for you.
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