17th post

This week in calculus we took two practice AP exams. One was calculator allowed and the other one was not. I found them to be hard but i didn't do to bad on them lol. Anyways, i'll start off by telling some of the things i know pretty well:

Average Value:

This is something i didn't always understand but today i learned how to do it and it seems pretty easy to me now. You start by understand you will have a number in front of the integral and you find it by 1/b-a... then you have to integrate the function and plug in the numbers. Once you have done that you solve.

Another thing i understand well is Rolle's Theorum: If the function is continuous and differentiable when you plug in the two points they give you.. then Rolle's Theorum can be used. After that, you take the derivative of the function and set it equal to zero and solve for x. Your x value will be your answer.

For mean value theorum, you also have to figure out whether the function is continuous and differentiable by plugging in the points given in the original function. This time, instead of setting the derivative equal to zero, you set it equal to the slope of b and a. To find the slope, you plug into f(b)-f(a)/b-a. Then you set the derivative equal to that number and solve for x.

Another thing is finding critical values. To find critical values, you take the derivative of the function and set it equal to zero and solve for x. What ever number or numbers you get for x is your critical values. Absolute extrema is pretty simple as well. If they give you a point in the problem, then you plug those numbers into the original function to get another number. You also solve for your critical values and plug those into the original function as well. Once you get your second numbers, you set each pair into new sets of points. Your highest point will be your absolute max and you smallest point will be your absolute min.

For some things i do not understand:
I am still having trouble with the trapezoidal rule. I am also having problems with trig inverse functions and solving for x with them and setting them up into intervals. If anyone can help me i would appreciate it. Thanks :)

Post #17

Alright everyone, we have Calculus Exam Packets this weekend to start on our exam which is comming up on monday! I say we all get together!..ha
but anyways, this past week we took a practice AP Exam.
The first portion was without a calculator so we need to know and remember our formulas and how to read graphs.
~remember that an odd function is symetric about the x axis and an even function is symetric about the y axis.
~remember how to solve in order to get an equation of a tangent line:
take the derivative and find the slope then since you need a point you'll plug in the number they give you to the original equation
~in order to find vertical asymptotes set the denominator equal to zero.
~know that when you’re looking at the derivative of a graph and you want to find a graph that is the original, if it’s increasing then you know that it will have to have a positive slope.
~Average value is not the same thing as average speed!...remember is one over b-a times the integral of the equation.
Tuesday we were unable to get to school due to flooding, so we were off!
Wednesday when we came back, we took the second portion of the AP Exam which allowed calculators!
~if you need to find values that the slope of a tangent line is equal to a certain number:
Take the derivative and set it equal to that number when you solving with a calculator go to y = and plug in the equation look at the graph and go to calc zero and guess what they are and you’ll have your answer!
~When you want to find the smallest positive value, plug the equation into y = and then adjust window, go to second trace and maximum guess the left and right then guess what the max should be and they’ll tell you the answer!
~remember if you need to find the integral of something then go to fnInt( and plug in the equation}, X, the number, the other number [that’s at the top and bottom of the integral] and that will give you the answer!
~know if you’re looking at the derivative of the graph, in order to find the maximums, look at the zeros. [and maximums must go from positive to negative]
~remember the equation for average rate of change: f[b]-f[a] all over b-a then plug in using the points they give you and solve!
Thursday and Friday we went over all of the things we had wrong on the practice AP Exam! B-Rob said that we should've got at least 10 questions right.


DO NOT FORGET:
We have the packet for our take home portion of our exam to do! Good Luck to everyone!

~ElliE~

fifth reflection

hmm, well i was gonna do my fifth reflection on area of disks, but it is almost the exact same thing as volume, so i decided against it. and i looked through some stuff and decided i am going to do this one on Riemann Sums!!! (particularly the right Riemann Sum because that is my favorite)

The first formula you need to know is x=(b-a)/n [a,b] with n subintervals. You will need to know this because each of the next formulas require that you know what x is.

LRAM- left hand approximation. (this puts the rectangles used to find the area on the left side of the curve) x[f(a)+f(a+x)+...f(b)]
RRAM- right hand approximation. (this puts the rectangles used to find the area on the right side of the curve) x[f(a+x)+...f(b)]
MRAM- approximation from the middle. (this puts the rectangles right on top of the curve, so that the curve goes through the middle of each one) x[f(mid)+f(mid)+...]
Trapezoidal- this does not use squares, instead it uses trapezoids to eliminate most of the empty space inside the curve, and I think this is the most accurate. x/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]


alright, they are all basically worked the same, the only difference is the formulas, so I am just going to work an MRAM problem because that is just a good one i came across.

Calculate the right Riemann Sum for q(x)=-x^2-3x on the interval [-2,1] divided into 4 subintervals

so first, you have to find your delta x which you get from using the formula b-a/n and in this specific example, would be 1+2/4, which is 3/4.

Next you just plug into the formula
3/4[f(-1.25)+f(-.5)+f(.25)+f(1)

so then you plug in those values into the formula they gave you to get:
3/4[-(35/16)-1.25-(11/16)-2]

which is then simplified to -147/32

and that is it. for the other ones you just plug into the other formulas to get the answer, but it is worked basically the same way.

yay, i'm finally done all my reflections

Fourth Reflection

Fourth Reflection!!! volume and area of disks. a disk is a solid object, and to find the volume of it, you just solve for x or y of the given equation. you can tell by the problem which axis its about. so you draw the equation, then you reflect it. the formula is S r^2 dx, and the radius is the equation that is given. and the formula for the area of a disk is the same thing but the radius is not squared.

Example: Find the volume of the solid obtained by rotating about the x-axis the region under the curve from -2 to 1: y=sqroot(-2x^2-10x+48)

so, since the formula needs to you to square the equation, the square root just dissapears. then, you have to integrate the equation, then you do top-bottom, and multiple by pi. so after you integrate it looks like this: -2/3(x)^3-5x^2+48x

then after you plug in everything and do top-bottom it looks like this: -2/3(1)^3-5(1)^2+48(1)-(-2/3(-2)^3-5(-2)^2+48(-2))

then you can simplify it on your own even more, or plug it all into your calculator. i chose to simplify it more by hand first because with that many numbers it is pretty easy to make a mistake in your calculator. So i further simplified it to -(2/3)-5+48+(16/3)+20+96 which equals: 491/3.

so then, you have to multiply that by pi, like i mentioned above, and the final answer is 491(pi)/3.

Actually, I think I am going to do area of disks for the next one because that would make this one way too long.

reflection 3

Alright, my third reflection is on Implicit Derivatives.
Implicit derivatives involve both x's and y's, unlike normal derivatives.
1: So, first you have to take the derivative of whatever they give you as you normally would.
2: Whenever you take the derivative of y, you have to note it with dy/dx.
3: Solve for dy/dx
(if you want to find slope plug in an x and y value)

example: y^3+y^2-5y-x^2=-4
First you just take the derivative, but don't forget to not the derivatives of the y's! So you get: 3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0

Then you have to solve for dy/dx, so you get:
dy/dx(3y^2+2y-5)=2x which then is further solved for to get dy/dx=2x/(3y^2+2y-5)

and that's it for that problem, it's done.

Another example:
Okay, let's say you want to find the slope of 3(x^2+y^2)^2=100xy at the point (3,1)

First you take the derivative, which involves all kinds of product and exponent rule...
6(x^2+y^2)(2x+2y(dy/dx))=100(y+x(dy/dx))

then, you need to foil it n stuff, so you get:
12x^3+12x^2(dy/dx)+12xy^2+12y^2(dy/dx))=100y+100x(dy/dx)

then, as usual, you would have to solve for dy/dx:
dy/dx=(-12^3-12xy^2+100y)/(12x^2+12y-100x)

after you solved for dy/dx, you plug in your x and y value from the point given to get your answer, so I think the final answer would be: 1.84 (if I put it in my calculator right)

that is it for implicit derivatives, and they are really easy to identify, it is the exact same thing as a derivative pretty much, just with x's and y's. Just don't forget to plug in the point that some problems will give you at the end, I have forgotten to do it before.

second reflection

okay, so for my second reflection i am going to go into detail on the second derivative test to find all possible points of inflection and intervals of concavity. remember, points of inflection only happen where there is a change of concavity.

Example: f'(x)= 6/(x^(2)+3)

First, you have to take the derivative of that, and you have to use the quotient rule, so the beginning of the problem will look like, [(x^(2)+3)(0)-6(2x)]/(x^(2)+3)^(2), which simplifies to -12x/(x^(2)+3)^2 remember, that was just the first derivative.

Second step is to take the derivative of the first derivative, that would make this step called taking the second derivative.

Once again u need to use the quotient rule, so f''(x)={(x2+3)^2-(12)-[(-12x)2(x^(2)+3)2x} all that over (x^(2)+3)^4 then you get a bunch of stuff, then you simplify, then you cancel, so I am just going to type the end answer of the second derivative. Which is, (3)(6)(x+1)(x-1) all over (x^2+3)^3

The possible points of inflection are found in the numerator of the finished second derivative, in this case, if you look, it would be x=1, and x=-1

so then you set up your points, (-infinity, -1) u (-1, 1) u (1, infinity)

then you plug in. f''(-2)= positive value f''(0)=negative value f''(2)=positive value

then you know that your intervals concave up at (-infinity, -1) u (1,infinity) or x<1,>1

and it is concave down at (-1,1) or -1
and you're points of inflection are x=-1, and x=1

so that is all for the second derivative test, it's not that hard, but some problems have a bunch of quotient and product rule and other stuff, so they can take forever to do, but hopefully we don't get any of those on the exam.

first reflection

okay, so my first reflection is going to be on the first derivative test. the first derivative test is used to find absolute maximums and minimums. To do these, first you take the derivative of the given equation, then you set it equal to zero and solve for x, then set up intervals using the x values, then plug in numbers within the intervals into the derivative to find where your relative maximums and minimums are, then you plug the x values into the original equation to find the absolute max's and min's.

Find the absolute maximums and minimums of the function f(x)=x^2-8x+4

First, you take the derivative of it:

f'(x)= 2x-8

Then you set it equal to zero and solve for x.

2x-8=0 2x=8 x=4

Then you set up the intervals

(-infinity, 4) U (4, infinity)

then you plug in a number found within the interval into the first derivative to find relative max's and min's.

(-infinity, 4)= negative number
(4, infinity)= positive number

so at x=4 there is a minimum

after that, you plug in that x value into the original equation to find the absolute max's and min's. (in this case it is only a min)

x=4 is the absolute minimum (4,-12)

There is a shortcut to finding absolute maximums and minimums, and that is using the second derivative test. all you do for that is take the first derivative, set it equal to zero and solve for x, then take the derivative of the first derivative (hence the name second derivative, then plug in the x value from the first derivative into the second derivative.

And that's all for my first reflection!!!

Academic Detention Blog #3

So once again i am just going to go back to the basics becuse i feel that if i take the time to explain it that it will be a good review for the exam on tuesday which will probably kick my butt.

so average speed and instantanious speed here we go.

lets talk about the differences between the two topics because i often get the mixed up. the difference between average speed and instantaneous speed is that instantaneous speed refers to the speed only during that instant. Average speed, on the other hand, refers to the speed during or over a particulare period of time.

EXAMPLE PROBLEM:

a bag of flower is dropped off of a roof on to the car, what is the average speed during the first two seconds of falling. Given: y=16t^2 to describe the fall.
x=(0,2)- this represents the first two secons.
to find your y's you would plug the first x and then the second x into the given equation.
f(0)=0 and f(2)=16(4)=64
after completin those steps, you plug in to your slope formula.
(y2-y1)/(x2-x1)=(64-0)/(2-0)= 32
the average speed of this particular problem would be 32 m/s.

Optimization- a process that stretches an equation. first you determine all quantities of the equation, then you write a new equation, reduce it and determine the domain of the equation, then from there you determine the max's and min's of the equation. optimization is usefull when dealing with area and volume.

Mean value thereom - somewhat like rolles. says that if a function is continous and differentiable then, f^(c) = f(b)-f(a)/b-a. saying that when its continous and differentiable on the interval then the value c exist on that interval.

Academic Detention Blog #2

I'll just go over the basic rules of the first derivitive test. (yes i know i spell it different on every post, woops)

STEPS:

1. take the deriv.
2. set the deriv. equal to zero.
3. solve for the xmax, mins, horizontal tangents and the critical points, (if you don't know how to do this i'll explain in a second.)
4. set up intervals using the step above.
5. plug in to the first deriv. (which is why it's called the 1st deriv. test)
6. plug in values from above to the original function to find the absolute max and min, but only do this if it asks for it.

EXAMPLE:
x^2-6x+8
2x-6
2x-6=0
x=3 is the critical point.
the intervals are (-infinity, 3)u(3,infinity)
.. after you set up the intervals, plug in numbers within the intervals to see whether or not the intervals are increasing or decreasing and if they are a max or min. so in this case you could plug in 2 and 4.


The second derivative test is also really easy.

STEPS:

1. take the first deriv.
2. take the second deriv.
3. set the 2nd deriv equal to zero.
4. set up intervals
5. pic a number and plug in to the second deriv.

(the second deriv. is used to find whether or not an interval is concave up or concave down and where are the points of inflection.)

Academic Detention Blog #1

So i am not going to say this week in calc because thats a dumb statement for make up blogsss.

thankfully by now the initial derivative rulse have been ingrained into my mind, so i will go over them.

PRODUCT RULE-one of the easiest rules there are
copy the first times the derivative of the second (plus) copy the second times the derivative of the first.

EXAMPLE: 5x
0(x) + 5(1)
=6

Also, some things to remember-
-the derivative of any number alone is 0.
-the derivative of any variable, such a x, is 1.

Now the EXPONENT RULE, also an easy concept, these things are about as easy as claculus gets. all you do is follow this formula: d/dx U^n=NU^(n-1)

an example of this would be:
x^4
4x^3
easyyyy!

another easy rule is the QUOTIENT RULE. this applies when you are taking the derivative of a fraction. all you do is copy the bottom times the derivative of top (minus) copy the top times the derivative of the bottom (all over the bottom squared)

exampleeee: (x^2)/x

((x)(2x)-(1)(x^2))/x^2
2x^2-x^2/x^2
x^2/x^2=1

Now for what i dont understanddd, i really don't have a full grip on optimization still, if i study it right before the exam i'll be fine but i can think of a way to get the concept to stick with me permanently.

Final AD Blog #4

I do not understand substitution and whatever we learned the day before the. I wasn't here and no one has explained it to me. I soooo need help on that!

Well for my last blog, I though I would talk about implicit derivatives. They are pretty simple.

Implicit derivatives deal with x's and y's. Steps:
1. take derivative of both sides
2. every time you take the derivative of y, note it with dy/dx or y^1 (which means y(prime))
3. solve for dy/dx

Simple enough right? Lets try some example problems:

1. y^3+y^2-5y-x^2= -4
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x= 0
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)= 2x
dy/dx(3y^2+2y-5)= 2x
dy/dx= 2x/(3y^2+2y-5)

2. x^2+y^2=9
2x+2y=0
2x+2y(dy/dx)=0
(dy/dx)= 2x/2y= -x/y

Now for second derivative of implicit derivatives:
After you take your first derivative, take the second derivative.
If there is a dy/dx, its derivative will be (d^2y)/(dx^2).
It is pretty much the same thing, really simple.

AD Blog #3

Rolle's Theorem:

1. You have to make sure it is continuous and differentiable on the closed interval [a,b].

2. Plug in your points [a,b] to equation. They have to be equal to continue. If not this theorem cannot be used.

3. Take derivative, solve for x and make sure your values are on the interval.

Example:
f(x)= x^4-2x^2 on [-2,2] find all values c for which f^1(c)=0

1. Is is continuous and differentiable? YES

2. f(-2)=(-2)^4-2(-2)^2=8
f(2)=(2)^4-2(2)^2=8
f(-2)=f(2)=8

3. 4x^3-4x= 0
4x(x^2-1)= 0
x= 0, 1, -1
c= 0, 1, -1

My question is about average speed. What are the steps to solving?

AD Blog #2

I'm always confused with first and second derivative test. So I decided to relearn and hopefully explaining it will help me too.

First Derivative Test:

Example:
Find the critical pts of f(x)=(x^2-4)^(2/3)

(2/3)(x^2-4)^(-1/3)
4x/3(x^2-4)^(1/3) =0
2x=0
4x=0
x= 0, +2, -2 <---critical points

Now set up intervals and plug in numbers in between them. You are plugging into the FIRST DERIVATIVE. You don't really have to know the number, but you need to know if it will be positive (+ve) or negative (-ve) in order to know if it is decreasing or increasing.

(-infinity,-2) (-2,0) (0,2) (2,infinity)
f^1(-3)=4(-3)/3(9-4)^1/3 = -ve/+ve = DECREASING
f^1(-1)= -ve/-ve = INCREASING
f^1(1)= +ve/-ve = DECREASING
f^1(3)= +ve/+ve = INCREASING

Mins and Maxs?
x= -2, 2 <--mins
x= 0 <--max

Second Derivative Test:

You take the derivative and the second derivative. The values you get are NOT critical values; they are more like points of interest. Now when you plug in this time you must plug into the SECOND DERIVATIVE. This helps tell about concave up (+ve) and concave down (-ve).

6/x^2+3
1st deriv= -12x/(x^2+3)^2
2nd deriv= (36(x+1)(x-1))/(x^2+3)^3
x= +1, -1 <--not critical values

(-infinity,-1) (-1,1) (1,infinity)
f^(double prime)(-2)= +ve = concave up
f^(double prime)(0)= -ve = concave down
f^(double prime)(2)= +ve = concave up

pts of inflection x= -1, 1

My question is about the shortcut..I still don't understand how to do it and why we do that.

Academic Detention Blog #1

I always forget how to find the equation of a tangent line, but it is so simple. So, I thought I would discuss the steps.

Finding an equation of a tangent line:

Steps:
1. if no y, plug in x to original problem
2. take derivative
3. plug in x (gives your slope)
4. write in point slope form

Example:
y=sin2x at x= pi/2

1. y=sin2(pi/2)=sin(pi)= 0 (pi/2, 0)
2. y^1=cos(2x)(2) *(y^1 is y prime)
2cos(2x)
3. 2cos(2(pi/2)=2cospi= -2
4. y-0=-2(x-(pi/2))

If they ask you for the normal line, it is the same steps except your slope would be perpendicular.


My question is about instantaneous speed. Do I just plug into the formula
lim f(x+h)-f(x)/h ?
h-->0

Make up #11

Area between curves

In Calculus we recently learned how to find the area and volume between curves. The formula for it is:

Int (Top Equation)-(Bottom Equation) on the interval [a,b]

You can find a and b by setting the equations equal to each other and solving because a and b are their intersections.

Two rules needed to know: If the area is on the y then a and b need to be y values and solved for x. If the area is on the x then a and b need to be x values and solved for y.

Example:

Find the area of the rigion bounded by f(x)=2-x^2 and g(x)=x.

2-x^2=x 2-x^2-x=0 (-x-1)(x+2) x=-1 x=2

int (Top)-(Bottom) dx [-1,2]

To find the top and bottom equation just graph them on your graphing calculator. You'll see that 2-x^2 is on top with x on the bottom.

int (2-x^2)-(x) dx [-1,2]

int (2-x^2-x) = 2x-(1/3)x^3-(1/2)x^2

Solve like an ordinary definite integral.

2(2)-(1/3)((2)^3)-(1/2)((2)^2)-[2(-1)-(1/3)((-1)^3)-(1/2)((-1)^2)]=(3/2)

Volume is little different because there are two ways to find the volume of a region, depending on the region itself. The two methods are discs and washers.

Discs: (π)int [R(x)]² dx [a,b]

Example:

(π)int √(sinx)² dx [0,π]

(π)int sinx dx [0,π]

(π)(-cosx) [0,π] -cos(π)-(-cos(0))

π(1+1)=2π

Washers: (π)int (Top equation)²-(Bottom equation)² dx [a,b]

Example:

√(x) and (x²)

(π)int (√(x))² - ((x²))² [0,1]

((1/2)x^2) - ((1/5)x^5) 1/2(1)-(1/5)(1) - [1/2(0) -(1/5)(0)]= 3/10

(π)(3/10)= (3π)/10

Finally done all eleven blogs!

Make up #10

Indefinite Integration

Integration, the basis of all Calculus classes, is probably the hardest thing ever to learn. Integration, in lame man's terms, is just the opposite of differentiation. That means that instead of going from x to 1, you from x to (1/2)x^2.

First of all, there is a new sign: its an S-shaped symbol that just stands for opposite of derivative, or integral.

Next, there are two types of integrals: Definite and Indefinite.

Indefinite Integrals are just doing the opposite of a derivative.

int x dx (1/2)x^2 +c

int 5 dx 5x +c

int cosx dx sinx +c

See the pattern there. ALWAYS ADD A +c AT THE END OF ALL INDEFINITE INTEGRALS!!!

Definite integrals just involve one more step after this, plugging in an x value or values

int x dx [2,4] (1/2)x^2 (1/2)(4^2)=8 (1/2)(2^2)=2

int x dx [2,4]=6

You have to subtract f(a) from f(b). This gives you the area under the curve of the function on the interval. Integration problems are very very common on the AP exam and all Calculus students will know everything about basic integration before they can pass the exam or even the class.

Substitution is the only trick that integration needs. Substitution takes the place of the chain rule, multiplication rule, and even the quotient rule. The steps to substitute are as follows:

1. Find derivative inside integral.
2. Substitute u for the non-derivative then differentiate u
3. plug back in with original.

Example:

int (x^2+1)(2x) dx

you may think this is impossible but its not using substitution

1. Find derivative inside integral. u=(x^2+1) du=(2x)

2. Integrate u. (1/2)u^2 +c

3. Plug back in (1/2)(x^2+1)^2 +c

You can simplify from there but I prefer not to since I'm lazy.

Make-up #9

Riemann's Sums

Riemann's Sumvs help approximate areas under curves using rectangles or trapezoids. There are four different formulas each with different properties.

LRAM-left-hand approximation

▲x[f(a)+f(a+▲x)+...+f(b-▲x)]

http://upload.wikimedia.org/wikipedia/commons/c/c9/LeftRiemann2.svg

RRAM-right-hand approximation

▲x[f(a+▲x)+f(a+2▲x)+...+f(b)]

http://upload.wikimedia.org/wikipedia/commons/4/45/RightRiemann2.svg

MRAM-middle approximation

▲x[f(mid)+f(mid)+...+f(mid)]

http://upload.wikimedia.org/wikipedia/commons/d/d2/MidRiemann2.svg

Trapezoidal-trapezoid approximation (most accurate)

(▲x)/2[f(a)+2f(a+▲x)+2f(a+2▲x)+...+f(b)]

http://upload.wikimedia.org/wikipedia/commons/7/76/TrapRiemann2.svg

I bet your all wondering what ▲x is. For these types of problems your always given an interval [a,b].▲x is just (b-a)/n. N? yes n is the number of subintervals within your intervals. The question will usually tell you how many subintervals you have.

Say your giving a problem tell you to find the LRAM of the interval [0,4] with four subintervals. Since your ▲x=(b-a)/n you would subtract 0 from 4 and then divide that number by 4 giving you ▲x=1.

So now that you know that, you can solve for LRAM.

▲x[f(a)+f(a+▲x)+f(a+2▲x)+f(a+3▲x)]=LRAM.

1[(0)+(1)+(2)+(3)]=6

Make-up #8

Related Rates FTW!!!

x and y are both differentiable functions of t and are related by the equation y=2x^3-x+4. When x=2 dx/dt=-1 Find dy/dt.

Some people are freaking out in their heads by now but not to fear, related rates problems are similar to optimization and implicit derivatives.

Steps for Related Rates Problems:

1. Identify all variables and equations.
2. Make a sketch and label. *VERY IMPORTANT*
3. Write an equation that involves your variables. *You can only have one unknown so a secondary equation may be given* (Usually this step is needed for word problems that ask you to find the rate at which a side of a triangle moves or the volume of a sphere changes.)
4. Take the derivative with respect to time, which is t.
5. Plug in all variables and solve.

Sure there are a lot of steps involved but they flow from one to the next so simply its like there is only 1 step: solve.

Step 1. Identify all variable and equations.

y=2x^3-x+4 x=2 dx/dt=-1 and dy/dt=?

Step 2. Make a sketch and label.

Um...since I am doing this on a blog I will not sketch and label the problem.

Step 3. Write an equation involving all variables.

Already given to us (y=2x^3-x+4)

Step 4. Differentiate with respect to "t".

dy/dt=6x^2(dx/dt)-1(dx/dt)

Step 5. Plug in a solve.

dy/dt=6(2^2)(-1)-1(-1)
dy/dt=-23

TA-DA!!! Is this not simple?

Makeup #7

Implicit Derivatives

The only difference between implicit derivatives and regular derivatives is that implicit derivatives include dy or y', the actual derivative of y.

y=x+2 y'=1

In an implicit derivative, you are always asked to solve for y'.

Example:

x^2+2y=0

1. Take derivative of both sides first.

2x+2y'=0

2. Then solve for y'.

y'=(-2x)/2

Some examples include:

4x+13y^2=4 y'=(-4/26y)

cos(x)=y y'=-sin(x)

y^3+y^2-5y-x^2=4 y'=2x/((3y+5)(y-1))



Some Derivative Formulas just for refreshing:

d/dx c=0 (c is a number)
d/dx cu=cu' (c is a number)
d/dx cx=c (c is a number)
d/dx u+v=u'+v' (also works the same for subtraction)
d/dx uv=uv'+vu'
d/dx u/v=(vu'-uv')/v^2
d/dx sinx=cosx(x')
d/dx cosx=-sinx(x')
d/dx tanx=sec^2x(x')
d/dx secx=secxtanx(x')
d/dx cscx=-cscxtanx(x')
d/dx cotx=-csc^2x(x')
d/dx lnu= 1/u(u')
d/dx e^u=e^u(u')

Make-up #6

Optimization

Optimization can be used for anything from finding the maximum amount of fencing to make a pen to finding the least amount of volume for a cylindrical cone. This concept is used commonly throughout the world and needs to be mastered for college level mathematics.

Steps in order to optimize anything:

1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.

2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.

3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.

4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers.

Example:

I have an enclosure against a barn for my pigs that has an area of 180,000 ft². What dimensions of my pen would maximize my perimeter?

1. My two equations are P=2l+w (my barn takes up 1 side) and A=lw.

2. I will solve for w because i have one using my secondary equation.

180,000=lw 180,000/l=w

3. Plug in to the primary then differentiate.

2l+(180,000/l)=P 2+(180,000/l²)=0 180,000=2l² 90,000=l² l=300m

4. Plug back into the secondary equation.

(300)w=180,000 w=(180,000/300)=600m

l=300m
w=600m

Academic Detention Reflection #3

The third blog of this detention I will start by explaining the area between curves. The formula is bSa top equation-bottom equation
1. draw the picture of the graphs
2. subtract the two equations from each other
3. put the like terms together and integrate the result

Next I will talk about trig inverse integration. The trig inverse integration formulas are: (sr=square root)
1. S du/sr(a^2-u^2)=-1/sr(u)arcsin u/a +C
2. S du/a^2+u^2=1/du(a)arctan u/a +C
3. S du/u sr(u^2-a^2)=1/du(a)arcsec lul/a +C

Another thing I will talk about is substitution. Substitution takes the position of derivative rules when integrating. The steps for substitution are:
1. Find a derivative of something else inside of the integral.
2. Set u equal to the non derivative found in the integral
3. Then take the derivative of u
4. Substitute back in
5. Solve

Example problem:
S (2x^2+5) (4x) dx u=2x^2+5 du=4x dx
S u du
1/4 u^2+C
1/4 (2x^2+5)^2+C

Also I am going to talk about taking implicit derivatives. The steps for taking implicit derivatives are:
1. Take the derivative of both sides like you would normally do
2. Everytime the derivative of y is taken it needs to be notated with either y ' or dy/dx
3. Solve for dy/dx or y ' as if you are solving for x.

For what I do not know is something like Find the area of the region enclosed by y=-4x^2+41x+94 and y=x-2 between x=1 and x=7.

Make-up #5

Rolle's Theorem

To understand Rolle's Theorem, you must first understand the Extreme Value Theorem.  The Extreme Value Theorem states that a continuous function on the closed interval [a,b] must have both a maximum and a minimum.  They can be on the endpoint a and b though.

So Rolle's Theorem gives the conditions that guarantee the existence of an extrema in the interior of a closed interval.

Rolle's Theorem-If a function is continuous on the closed interval [a,b] and differentiable on the open interval (a,b) and f(a)=f(b), then there is at least one point, designated "c", where the derivative of f(c)=0.

Example:  Find all points "c" where the derivative of f(c)=0  f(x)=x^4-2x^2   [-2,2]

1.  I first found that f(a) equals f(b).

(-2)^4-(-2)^2=8
(2)^4-2(2)^2=8

2.  Once finding that f(a)=f(b), I set the derivative equal to 0 and solved for x.

4x^3-4x=0
4x(x^2-1)=0
4x=0  x^2-1=0
x=-1,0,1

There are three point of extrema within the [-2,2] of the function x^4-2x^2.  One at x=-1  another at x=0 and the last at x=1.


Mean Value Theorem

The Mean Value Theorem states that If the function f(x) is continuous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number "c" in (a,b) such that f'(c)=(f(b)-f(a))/(b-a).

This is essentially the same thing as average speed, which finds the slope of the equation.


Intermediate Value Theorem

The Intermediate Value Theorem states that if a function f(x) is continuous on [a,b] and K is any number between f(a) and f(b), then there is at least one number "c" where f(c)=K.

Make-up #4

First Derivative Test

The First Derivative Test is used for relative and absolute maximums and minimums.  The Advanced Placement test is loaded with both multiple-choice and short answer questions that involve the First Derivative Test.

Example:

Find the absolute maximums and minimums of the equation (1/3)x^3-2x^2+3x.

1. Take the derivative of the equation.

x^2-4x+3

2.  Set it equal to zero then solve for x.

x^2-4x+3=0   (x-3)(x-1)  x=3 x=1

3.  Set up intervals.

(-infinity,1) (1,3) (3, infinity)

4.  Plug in a number found within all intervals into your first derivative (relative maximums and minimums).

(-infinity, 1)=positive number      at x=1 there is a maximum
(1,3)=negative number               at x=3 there is a minimum
(3,infinity)=positive number

5.  Plug in x values to original equation to find absolute maximums and minimums.

x=1 is the absolute maximum (1,2/3)
x=3 is the absolute minimum (3,0)


Second Derivative Test

The Second Derivative Test is useful for finding concavity or absolute maximums and minimums.

Example:

Find the absolute maximum of the equation y=x^3-3x

1.  Find first derivative.

x^3-3x
3x^2-3

2.  Set equal to 0 then solve for x.

3x^2-3=0  3x^2=3  x^2=1  x=1 x=-1

3. plug into second derivative.

3x^2
6x    6(1)=6   6(-1)=-6   x=6 there is a absolute maximum.

Academic Detention Reflection #2

The second blog of this detention I will start by explaining linearization. If the word approximate is used then you should know to use linearization. The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation

Example problem:(sr=square root) Use differentiability to approximate sr(4.5)
f(x)=sr(x) sr(4)+(1/2 sr(4) )(.5)=1.125
sr(x)+(1/2 sr(x) ) dx error=.005
dx=.5
x=4

The next topic I will talk about is integration. Integration finds the area under a curve. The Riemann sum approximates the area using the rectangles or trapezoids. The Riemanns Sums are:
LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
delta x=b-a/number of subintervals

Example problem: Find the area of f(x)x-3 on the interval [0,2] with 4 subintervals.
delta x=2-0/4=1/2
LRAM=1/2[f(0)+f(1/2)+f(1)+f(3/2)]
1/2[-3+-5/2+-2+-3/2]
1/2[-9]= -9/2
RRAM=1/2[f(1/2)+f(1)+(3/2)+f(2)]
1/2[-5/2+-2+-3/2+-1]
1/2[-7]= -7/2
MRAM=1/2[f(1/4)+f(3/4)+f(5/4)+f(7/4)]
1/2[-11/4+-9/4+-7/4+-5/4]
1/2[-8]= 4
Trapezoidal=1/4[f(0)+2f(1/2)+2f(1)+2f(7/2)+f(2)]
1/4[-3+-10/2+-4+-6/2+-1]
1/4[-16]= 4

For things I do not know I forgot how to do average speed and seem to have lost my notes on this so if someone can remind me on how to do these problems I would greatly appreciate this.

Academic Detention Reflection #1

The first blog of this detention I am going to talk about related rates. The steps for related rates are:
1. Identify all of the variables and equations
2. Identify the things that you are looking for
3. Sketch a graph and then label that graph
4. Create and write an equation using all of the variables
5. Take the derivative of this equation with respect to time
6. Substitute everything back in
7. Solve the equation

Example problem: The variables x and y are differentiable functions of t and are related by the equation y= 2x^3-x+4when x=2 dx/dt=1. Find dy/dt when x=2.
dy/dt=? dy/dt=6x^2 dx/dt-1 dx/dt
x=2 dx/dt=1 dy/dt=6(2)^2(-1)-(-1)
y=2x^3-x+4 dy/dt=-23

Next the angle of elevation is the topic. The angle of elevation is pretty much the same thing as rate of change except the difference is you look for an angle of a triange instead of a rate.

Example problem: A balloon rises at a rate of 5 meters/second from a point on the ground 50 meters from the observer. Find the rate of change of the angle of elevation of the balloon to the observer when the balloon is 50 meters from the ground.
tan(theta)=4/x
tan(theta)=5/50
sec^2(theta)=1/10 d(theta)/dt=1/10 dy/dt
d(theta)/dt=cos^2(theta)=1/10(4)
1/2(1/10)(4)= 1/5 radians/second

For something I do not know: I still do not fully understand how to do optimization problems. I can start them and get the starting equations to use I just do not remember how to finish them.

Make-up Reflection #3

Graph Interpretation


Every AP exam will include at least one graph on it's short answer section that requires the test-taker to interpret certain requirements from it in order to receive credit for that question.

Interpreting a graph is very easy but in order to do it properly one must understand certain properties.

1.  If the original graph is increasing, the slope is positive.
2.  If the original graph is decreasing, the slope is negative.
3.  An interval with a positive slope on the first derivative means that there is a downward concavity on that interval in the second derivative.
4.  An interval with a negative slope on the first derivative means that there is an upward concavity on that interval in the second derivative.
5.  Upward concavity (bowl-shaped) is positive.
6.  Downward concavity (umbrella-shaped) is negative.
7.  There is a horizontal tangent where the slope=0 on the original graph.
8.  Wherever there is a horizontal slope there is either a maximum or a minimum value.
9.  A x intercept on the first derivative is either a maximum of a minimum on the original graph.

Make-up Reflection #2

Tangent Lines

Tangent lines can be used for many purposes. Finding a tangent line requires only a little knowledge of calculus but a substantial amount of algebra.

Example:

Find the line tangent to the graph y=2x²+4x+6 at x=1.

1. Identify the equation and point of tangency. If not given a y value, plug the x value into the original equation.

y=2x²+4x+6 y=2(1)<sup>2

2. Differentiate you equation.

dy/dx=4x+4


3. Plug in x value then solve for dy/dx.


dy/dx=4(1)+4=8




Your dy/dx value is your slope from here on you can create your equation of the tangent line. There are three forms that your equation can be presented in: point-slope, slope-intercept, or standard form. For these purposes, I am using point-slope.


(y-12)=8(x-1) is your final answer.


Another Example:

Find the line tangent to the graph y=x^3 that is parallel to the perpendicular graph of y=-(1/3)x.

1. Find the slope of the perpendicular line of y=-(1/3)x

-1/3 =m Perpendicular m=3

2. Set first derivative =3.

3x^2=3

3. Solve for x.

3x^2=3 x^2=1 x=1 x=-1

4. Plug x back into the original equation to find y values.

X^3 (1)^3 (-1)^3 y=1 y=-1

Final equations are y-1=3(x-1) and y+1=3(x+1)</sup>

Make up Reflection #1

Average Speed

First of all, remember that a slope is the y value, or dy, of a derivative.

Example:

A ball is flung from a little child. It's path is projected as y=4.9t²m in "t" seconds. What is the average speed of the ball from 0 to 3 seconds?

1. Set up equations and intervals: (f(b)-f(a))/(b-a) 4.9t^2 [0,3]


2. Plug in a and b values for t: f(b)=4.9(3)²=44.1 f(a)=4.9(0)²=0

3. Plug into main equation and solve: (44.1-0)/(3-0)=14.7m/s

Average speed is used for many different things, from finding the speed at which a cannonball was launched out of a cannon from how fast a cheetah runs in a straight line trying to catch it's prey. The concept behind average speed is a fairly simple concept that many people understand right away. You're basically finding the slope of the equation using calculus and algebra. If I ask someone what the average speed of a ball from [3,4] if it's path was graphed as y=x.

y=(4) y=(3) (4-3)/(4-3)=1

so yea the slope is also 1. wow a big breakthrough in the world of mathematics.

Academic detention #3

OK explanations for my 3rd academic detention.

Integration

riemann summs...

Lram-left hand approximation
deltax[f(a)+f(a+deltax)+.....f(b+deltax)]

Rram-right hand approximation
deltax[f(a+delta x)+....f(b)]

Mram
deltax[f(mid)+f(mid)+....]

trapezoid
deltax/2[f(a)+2f(a+deltax)+2f(a+2deltax).....f(b)]


Another integration process is washers and disk

Disk are used with solid objects and the formula is

=(pie) b/s/a[R(x)]^2dx

and washers are for objects with holes and the formula is

=(pie)b/s/a top^2-bottom^2 dx

Ex Find the volume of the solid formed by revolving the region bounded by the graphs of y=squareroot of x and y=x^2

after graphing in your graphing calculater you find that you need to use washers

so you get =(pie)S(squareroot of x)^2-(x^2)^2 dx

x=1 so (pie)[(1/2)-(1/5)]-0

3(pie)/10 is your awnser

and for another big problem i have in calulus is optamization

Academic Detention #2

For my second Academic Detention blog ill explain two more things

1st: Related RAtes

1. Identify all variables and equations
2. identify what you are looking for
3. make a sketch and label
4.write an equation involving your variables
5. take the derivative with respect to time
6. substitute in derivative and solve

Ex: The variable x and y are differentiable functions of t and are related by the equation y=2x^3-x+4 when x=2 (dy/dt)=-1 Find (dy/dt) when x=2

(dy/dt)=6x^2(dy/dt)-(dy/dt)
=6(2)^2(-1)-(-1)

(dy/dt)=-23

Another problem i have and still have is Angle of elevation i don't know where my problem is i think its just starting the problem thats really hard for me so if someone can help with that thanks.

Academic Detention#1

Since i got an academic detention for not doing my comments on blogs this is my first of three new blogs to make up for the lost work.
I'll just start from the beggining since there is no one specific week i can talk about.

Average Speed:
An anvil falls off the roof onto Ryan's head. What is its average speed during the first two seconds if y=16t^2 describes the fall.

0: y1=16(0)=0
2:y2=16(2)^2=64 [0,2]

y2-y1 64-0
-------- =m so... m=-------------- =32
x2-x1 2-0

M=32

and for another explination

Instantaneous speed:

lim f(x+h)-f(x)
---------------
h->0 h

Find the instantaneous speed at t=2

y=16t^2 lim 16(2+h)^2-16(2)^2
f(t)=16t^2 ------------------------ =64
h->0 h



I seemed to have lost my notes on Point of Inflection and i don't remember how to do them at all so can someone post the notes and an example for me thanks.

reflection 2

for my academic detention again :( ha

LRAM-left hand approximation

remember to start with the interval all the way to the left and add delta x until you reach your b interval term and substitute that into the original function. add all of those and multiply by delta x:

Dx [f(x+Dx) + … + f(b-Dx)]

RRAM-right hand approximation

same as LRAM, except you start from the right.

Dx[f(b) + f(b+Dx)…f(a+Dx)

TRAM-trapezoidal approximation

remember that delta x is over two and that you should multiply all of the numbers, except your functions of the intervals, by 2.

Dx[f(a) + 2f(a+Dx) + 2f(a+2Dx) +…f(b)]

MRAM-midpoint approximation

Dx[f(mid) +…]

You find the numbers in the midpoints of by adding delta x from a to b.

TRIG INVERSE FUNCTION FORMULAS:

The trig inverse integration formulas are: sqrt=square root
1. S du/sqrt(a^2-u^2)=-1/sqrt(u)arcsin u/a +C
2. S du/a^2+u^2=1/du(a)arctan u/a +C
3. S du/u sqrt(u^2-a^2)=1/du(a)arcsec lul/a +C

reflection 1

for my academic detention :(

LN INTEGRATION:

natural log integration actually isn't as hard as it seems. msot of the time, the top of the fraction is the derivative of the bottom of your fraction, just like trina said, ex: 4/4x.. but sometimes you have to substitue in something.. like you were given 1/4x, you would need to substitute in a 4 because the derivative of 4x is 4, not 1. so your u=4x du=4dx, and you would simply but a 1/4 in front of the integration symbol. this is ln integration.

DISKS AND WASHERS:

disks are solid objects, washers are objects with holes in them (they will have two equations). you mostly just need to remember the equations for these two, which are not hard at all! volume is squared, area is not! for both disks and washers, same equation, just not squared when it asks for area.
if it asks for volume, always put a pi in front of the integral. if it asks for area, you don't put pi.

formula for volume of a disk
pi S r(x)^2dx
formula for area of a disk
S r(x)dx

r=radius. radius is your given equation. it's that easy. so pretty much plug in the equation into your formula, integrate it, and solve.

formula for volume of a washer
pi S top^2 - bottom^2 dx
formula for area of a washer
S top-bottom dx

you are given two equations.. plug both into your calculator and see which equation is on top and which is on bottom on the graph that is given.

Help with selected topics

Here are a few sites to help with selected topics.

Trig inverse integration

http://www.intmath.com/Methods-integration/6_Integration-inverse-trigonometric-forms.php

http://www.ltcconline.net/greenl/Courses/106/explogtrig/invtrigint.htm

Knowing where a graph is increasing decreasing etc

http://tutorial.math.lamar.edu/Classes/CalcI/ShapeofGraphPtII.aspx

http://www.franklinroadacademy.com/teachers/zab/APCalculus/APCalculusInaNutshell/Text/FirstandSecondDerivatives2004.pdf

Blog!

1 BLOG!
In an attempt to further explain Reimann Sums, I shall first give the formula for each Middle RAM, Left RAM, Right RAM, Trapezoidal RAM then explain each in layman’s terms, all while attempting an example problem pertaining to all of the aforementioned sums.

*One must remember that these sums are all ways of estimating the area under a curve (hence the fact that this is under integration). Particular ones work better at estimating certain problems. However, as a general rule, trapezoidal is the most accurate when, and only when, the delta x is not zero. (I have found this out by working numerous problems in one of the MANY packets given to us)

LRAM-left hand approximation

When doing this, a trick is to start with the furthest left interval (hence the LEFT Reimann sum) and add delta x until right before you reach your b interval term and substitute those numbers into the original function. Then sum all of those values and multiply by delta x. Simply stated:

Dx [f(x+Dx) + … + f(b-Dx)]

RRAM-right hand approximation

This procedure is the same as the left except, you guessed it, you start with the farthest right interval or b!

*note, this formula may not be the same as in your notes, but since addition is commutative, you can start with whatever you like.

Dx[f(b) + f(b+Dx)…f(a+Dx)

MRAM-middle hand approximation

The trick to finding the middle sum is having the capability of finding all of the midpoints in the given interval.

Dx[f(mid) +…]

You are going to find the numbers in which you find the midpoints of by adding delta x from a to b. (will explain later)

TRAM-trapezoidal approximation

For this particular one, remember that delta x is over two and that you should multiply all of the numbers besides the functions of your intervals by 2.

Dx[f(a) + 2f(a+Dx) + 2f(a+2Dx) +…f(b)]

Stay tuned for part 2 of this blog in 2 BLOG!

2 BLOG!

These are my examples for the above explanation of the four sums.

Given f(x) = x^2 - 3 with 3 subintervals on the interval [1,4], find the LRAM, MRAM, RRAM, and TRAM.

Beginning, one must first find delta x. Delta x is found by subtracting b minus a and dividing by n (subintervals).

So in this case:

Dx = 4-1/3 = 1

LRAM:

1[f(1) + f(2) + f(3)]

1[-2 + 1 + 6]

5

RRAM:

1[f(2) + f(3) + f(4)]

1[1 + 6 + 13]

20

TRAM:

½ [ f(1) + 2f(2) + 2f(3) + f(4)]

½[ -2 + 2+ 12 + 13]

½ [-1] = -½

MRAM:

1 2 3 4

3/2 5/2 7/2

1[f(3/2 + f(5/2) + f(7/2)]

So after fully breaking down practically step for step the process of finding MRAM, LRAM, RRAM, and TRAM, do you believe you have it? You must keep your formulas straight. And try not to stress over it. But on another note, you also may be asked certain unnoticeable questions either on the wiki or on the AP exam.


D

Ash's 16th Post

So, what I learned this week:
~Malerie can sing
~Malerie can have the patience of a saint
~The Saints are cheating
~My street has some kind of invisible force field that does not allow rain or snow in
~Louisiana's weather can fluctuate from snow to sweating heat in 12 hours
~There's not enough time in the day to do Calculus, understand it, re-do what you didn't understand, work for your 6 other classes, eat, shower, and live at the same time...end of story
~I'm going to need mental help after this class to A) tell me if psychologists REALLLLLY use tons of Calc B) see if I can get some lost sanity back
~It's extremely convenient to have a best friend in Advanced Math that you can get graphing rules from
~Apparently, I've been doing integration wrong THE ENTIRE TIME I'VE BEEN DOING IT
~The correct way of integration is so much easier than what I was doing

So
Let me explain SIMPLY how to do integration (btw, I think having a little sister helped Malerie explain to me what I was doing wrong..)

Integrate: 4x^2 + 3x + 4 dx

Step 1. add one to the exponents
___x^3+___x^2+___x

Step 2. divide the coeff by the new exponent
4/3x^3+3x^2+4x

TA DA!!

What I was doing was..no I'm not going to even bother trying to explain what goes on in this black abyss that I call a head and confuse people =]

Now for what I don't get:

How to identify fraction integration!
There's trig identities, ln integration, etc!
If you can make up, delete, and flat out ignore numbers, how do you know what to do?

Also, THOSE CLICKER QUIZZES!!
Can I just say "AHHHHH!!" Just when I thought I might have a shot in this class v.v How can you identify what to do, how to do it, and work it in such a short amount of time??
Math and I just don't click. (Haha)
How can I fix that?

mhers 16th post

so this week we learned all kinds of new stuff, and we practiced using the little clicker things to take quizzes... so, here are some things we learned:

washers are like disks, but they have holes in them. washers has two equations, volume of washers is S top^2-bottom^2 dx, and area of washers is S top-bottom. to do washers, all you have to do is first draw the graphs of the two equations, then you use whichever formula applies, either for volume or area, then you set them equal and solve for x to find the bounds. then you plug in the bounds and then integrate.


a disk is a solid object, and to find the volume of it, you just solve for x or y of the given equation. you can tell by the problem which axis its about. so you draw the equation, then you reflect it. the formula is S r^2 dx, and the radius is the equation that is given. and the formula for the area of a disk is the same thing but the radius is not squared.

another thing we learned was trig inverse integration, and the three formulas for it are:
S du/(sqrta^2-u^2)=arcsin u/a +c
S du/(a^2+u^2)= du(1/a) arctan u/a+c
s du/(usqrtu^2-a^2)= 1/a arcsec lul^e/a+c

things I don't understand: to be honest I hardly understand any of the stuff I just mentioned, I have almost no idea how to do anything we learned in like the past three weeks... so everything from like basic integration until now I do not get. And I don't know what to do about it..

post 16

Alrighty, so this week in calc we learned disks and washers... volume and area. went over that the rest of the week, got some packets to work on. all that good stuff. we also learned how to use the clickers which is SOOOOO cool :-) haha, anyways.

disk-solid object without holes
washer-object with a hole

volume of a disk. given equation, solve for x/y. it tells you in the problem which axis it is about. draw your picture and reflect it. formula for volume of a disk is S r^2 dx. your radius is the equation that's given. area is the same except it's just r not r^2.
so the steps all together for finding volume of a disk: solve equation for variable. draw your picture. plug equation into formula. take integral. tada :)
steps for area are exact same except plugging into area formula.

washers are given two equations. graph them and find out which is on top and which is on bottom. volume of washers formula is S top^2 - bottom^2 dx. area is the same formula except neither top nor bottom is squared. same steps. given equation, solve for variable, plug into formula, take integral. tada :) area, just plug into area equation. (obviously)

what im still confused on is LRAM, RRAM, MRAM. never really caught onto it :/

post sixteen

This week was one more week closer to mid terms. This week we learned several things.

Some of the things that we learned were intergration of washers and disk.

The formula for the volume of disks is S (top)^2 - (bottom)^2 dx

The formula for the area of washers is S (top) - (bottom)

The steps are:
1. Draw the graphs of the equations
2. Subtract top graph's equation by the bottom graph's equation(in disks each equation would be squared)
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. Integrate

These steps are fairly simple but you must remember to draw the graphs.

One thing that i am not catching on to is intergration of trig functions. Like i know the formulas but cant work a problem.

Posting...#16

WEll this weekend is bad i basically played in the wet freezing cold rain for some reason that is unkown to mankind all exsitiance of life and the results were oustanding Friday i felt bad Saturday i felt worst Sunday i felt horrible and i don't think im going to school tomorrow(monday). I BELIEVE I HAVE THE FLU AHHHHHHHHHHHHH!!!!. This week we basically just messed with those clickers all week and me i grabbed the wrong number because i did not pay attention to the numbers and i still don't know my number so i might grab yours next be afraid very afraid.

This week in Calc we were still on integration we learned washers and disk
The formula for the volume of disks is (top)^2-(bottom)^2 dx
The formula for washers is S(top)-(bottom)

The steps are:
1.Draw the graphs of the equations
2. subtract the top graph's equation by the bottom graph's equation
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. integrate

I understand washers and disk which is great but theirs still a bunch of things in integration i don't understand

1. I still have trouble wiht Trapazoid
2. I don't really know when to integrate and start working on the problem
3 EVERYTHING WE LEARNED

Those are my 3 biggest problems so can anyone help me wiht 1 and 2 but i thinkg that 3 is to broad of a topic for anyone to help me with.

AS always thanks.

Post Number Sixteen

The last few weeks in Calculus have got me so crazy it's unbelievable. I feel like i'm not retaining any information I'm taught. But here we go..

This week we learned about trig inverse integration and volume by disks and washers.

In class I never caught on to trig inverse integration. I know that it's supposedly easy if you recognize the three formulas, but i'm just not getting it. The three formulas are:
1) 1/du arcsin u/a + C
2) 1/dua arctan u/a + C
3) 1/dua arcsec absolute value of u/a + C

Now on to volume by disks.
First off, what is a disk? It's a solid object! I actually understood this in class :)

The formula is pi S[r(x)]^2dx

First you have to sketch the graph. Just plug into your y equals in your calculator and sketch! Depending on what axis it tells you to do the problem on is how you reflect the graph. Once you sketch it, shade the graph and then start working the problem!

Ex: Find the volume of the solid formed by revolving the region bounded by the graph of f(x) = square root of x and on y=0 x=3 about the x-axis.

First you would sketch the graph, which looks like a curved line coming out of (0,0) heading to the right to x=3. You then reflect that about the x axis and shade, which alternatively gives you a shape that looks like half of a football. Now to start the problem.
Pi S(0 - 3) (square root of x)^2 dx
Pi S(0 - 3) x dx
Pi[1/2x^2] 0 to 3
=9pi/2

Volume by washers is the same as volume by disks except it has a hole in it. That means it is bounded by two graphs. The formula for volume by washers is Pi S top^2 - bottom^2 dx.
This also i understood in class.

One thing i still don't understand is MRAM and TRAM, so if anyone can help me with that please do.

The packet we got in class friday i understood and i was actually getting them right, but of course at home i get lost and am not becoming successful with this packet. It's like once i get out of the room, all my knowledge is lost.

Also, we got to try out turning point this week. I got all of the answers right the first day, but it all went downhill from there. Hopefully this will help me to remember everything we've learned before.

Can't wait for another overwhelming week of calculus, sikeeeee.

post 16

And we are getting close to exams now that another week has come and gone. One week to be exact before exams, so because of this we learned a few more things before we start to review.
We learned about the area of washers and volume of disks, the area between curves, and trig inverse integration.
The area between curves formula is bSa top equation-bottom equation
1. draw the picture of the graphs
2. subtract the two equations from each other
3. put the like terms together and integrate the result

The formula for the volume of disks is S (top)^2 - (bottom)^2 dx
The formula for the area of washers is S (top) - (bottom)
The steps are:
1. Draw the graphs of the equations
2. Subtract top graph's equation by the bottom graph's equation(in disks each equation would be squared)
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. Integrate

The trig inverse integration formulas are: sr=square root
1. S du/sr(a^2-u^2)=-1/sr(u)arcsin u/a +C
2. S du/a^2+u^2=1/du(a)arctan u/a +C
3. S du/u sr(u^2-a^2)=1/du(a)arcsec lul/a +C

For what I need help with is remembering all of this. I can handle these when the formulas and stuff are in front of me but without it sometimes I forget what I am doing so if there are any tricks on remembering these formulas i would appreciate if you would let me know.

Post #16

This week in calculus i actually understood what we were doing and got most of the clicker work right too! It was a good week and i needed that boost of confidence before i completely turn the other direction when we are preparing for the exam. Which i'm a little, well, really nervous for. So the thing i am so anxious to tell you about is volume and area of discs and washers..they are so easy!

So, i'll say this now: AREA IS THE SAME FORMULA, JUST DON'T SQUARE IT!

Now, lets do a volume problem for discs. What's a disc, well i'll answer that question. It's a SOLID object WITHOUT holes. You will have one equation solved for x or y and it will ask you to find the volume of a solid about either axis and give you certain bounds. You simply draw your picture, plug into the formula, take the integral, and then plug in your bounds.

So, volume of a disc formula:
S (radius)^2 dx
*ps, your radius is the equation that is given :)

Now, what's the area formula?
S (radius) dx
*it's the same thing, just not squared :)

Now, lets talk about washers.

Washers are an object with a hole. You are given two equations and then MUST graph them to find out which one is on top. Then, you plug in the formula, take the integral, and then plug in bounds.

Formula for Volume of Washers:
S (top)^2 - (bottom)^2 dx

Formula for Area of Washers:
S (top) - (bottom)

This has got to be my favorite thing so far in calculus because every problem is the same! I love it!

Now, my question is the LRAM, RRAM, and whatever other nonsense was with that..it was kinda crazy..and i still don't understand it. Can anyone help?

WEEK 16!

okay so this week was my first week back in two weeks, ehhh i was lost
thankfully, i got extremely excited friday bc i finally had an idea of what i was doing with those packets! and guess what... i acutally finished a packet all by myself!

This week we learned about inverse integration and volume by disks and washers.
i think that the volume by disks and washers are the probably the two concepts i have caught on to the quickest so far this year.

THE DIFFERENCE BETWEEN A DISK AND A WASHER: a disk is a solid object, circle..
a washer has a hole/holes in it, kind of like the gameee!

FORMULA FOR DISKS: pi S[r(x)]^2dx, this formula is almost exactly the same as when you are asked to find the area of a disk, however, while finding the area you do not square the radius (the given equation).

FORMULA FOR WASHERS: pi S[(top)^2-(bottom)^2]dx

EXAMPLE: find the volume of the solid obtained by rotating about the x-axis the region under the curve from 0 to 4. y=sq root of -x^2+6x+27
4
pi S (sq root of -x^2+6x+27)^2dx
0
pi[-1/3X^3+3X^2+27X]= pi [0-(404/3)]= 404/3 pi

it is not necessary for you to graph the equation when you are doing things involved with disks, but when doing washers you MUST graph both of the equations to see which one's the top equation and which one is the bottom equation.

Some things i still don't understand are the trig invers stuff and i know that when i see a fraction and it asks me to integrate it i DON'T USE QUOTIENT RULE, that's stuck in my headdddd but i dont know what to do if i dont do that, i forgot. Help would really be appreciated, thanks guys. ohh and the formula for the trapezoidal rule, i lostt it.

Post #16

Calculus Week Number 16

So this week was a lot of different things. I'll start my post off by saying that I completely did not retain ANY of the information about the trig inverse functions and how to do all of those...I guess the only way I'll remember is by looking at my notes but I forgot my notebook for the weekend so this is all coming from my head...


Okay so the things I do remember

Volume by Discs

Basically you are given an equation, an interval, and what it is to be rotated about...either a line or one of axis's. Your basic steps are to first draw the graph. You need to know what you're looking at. So draw all the graphs as best as you can and determine in a somewhat 3D looking-way what the object or volume you will be finding is. Most of the time, with Volume by disks, you'll wind up with a sphere looking thing. Anyway, so a hint to remembering when to use this is when you have an equation and a line... if it was an equation with an equation you'd use volume by disks because you'd have a hole somewhere. Anyway, the way you do this is you take your equation, square it, and take the definite integral from a to b times by pi.

For a problem where you have to rotate the solid formed by two equations, you basically do it by washers. What you do is you do the top equation squares minus the bottom equation squared. You then take the definite integral from a to b and times that by pi to get your volume.

Both of those are really really simple...it's just a matter of getting it right. Sometimes you have to shift it because you always want it to be rotated about an axis so...what you do is just minus whatever you need from both equations, no big deal.

Anyway, I don't think anyone can really help me with the trig inverse stuff but if you think you can, have at it! :-P

post 16

we learned are between curves and we also learned disks and washers

To find area between curves:
The formula you use is b(int)a (top eq.) - (bottom eq.).
If a and b is not already given to you, then you much set the equations equal to each other and solve.
You find which equation is top/bottom by graphing both and simply looking to see which one is on top.
If the area is on the y-axis, then the a and b values need to be set as y-values, and the equations must be solved for x.

First of all: Disks are solid objects with one equation bounded by the equation and usually an axis. Washers are objects bounded by two equations.

Formula for Disk:

(Pi) b(int)a [R(x)]^2.

Formula for Washers:

(Pi) b(int)a (top eq.)^2 - (bottom eq.)^2 dx.


Washers:

volume by washers is basically the same concept except that it is not just one formula squared, it is pie times the top equation squared minus the bottom equation squared. You figure out which formula is the top and bottom by graphing the equation by hand or on your calculator, once you figure that out, you then square the functions and integrate them and solve. Again, the same rules apply to the bounds.

limit rule is something good to hold onto

limits:

Rules for Limits:…
1. if the degree of top equals the degree of bottom, the answer is the top coefficient over bottom coefficient
2. if top degree is bigger than bottom degree, the answer is positive or negative infinity
2. if top degree is less than bottom degree, the answer is 0

some things im not too great on is related rates and and all the mram rram and lram is confusing to me somehow lol

Weeks 14 and 16

Week 14:

In week 14, we learned many things. We learned substitution, area between curves, e integration, and integration with natural logs.

Substitution needs to be done when there is a product, quotient, or chain rule in an itegral. We need to use substitution because product, quotient, and chain rules cannot be used in integration. Substitution is basically forcing a problem by any means to work. In substitution problems, there will be a derivative and a non-derivative usually multiplied or divided by each other. U will be set equal to the non-derivative and du will be set equal to the derivative. In most of the simple probles, the derivative is exact in the problem, but if it isn't, you would need to put the reciprocal of whatever is needed on the inside on the outside of the integral. The steps for substitution are as follows:

1. Find a derivative inside the integral

2. Set u= to the non-derivative

3. Take the derivative of u

4. Substitute u back in

If there is an extra variable in the derivative, it cannot be taken out by using the reciprocal on the outside. You would have to set u=0 and solve for the variable. You would then put this in the place of the variable and solve the same way you would solve any other substitution problem thereafter.

There is also substitution with e and natural log. Many people see e in their problems and automatically think the problem will be hard, but e integration is the exact opposite. In e integration, u will always be e's exponent. Du will be the derivative of u, which will most likely be the other term in the problem. If it is not, you would fix this just as in regular substitution either with putting the reciprocal outisde of the integral, or solving u for the variable.

Natural log integration is also very simple. Natural logs are found in fractional integrals. U will be the bottom, with du being the top. Once this is found, you plug u into: ln u + c



Week 16:

Last week, we learned trig inverse integration, how to find volume by using disks, and by using washers. The term disk is used when there is no hole in the volume, and the term washer is used when referring to a volume with a hole.

For trig inverse integration there are four different formulas:

sr = square root

1. S du/sr(a^2-u^2) = 1/du(arcsin(u/a)+c)

2. S du/a^2+u^2 = 1/dua(arctan(u/a)+c)

3. S du/u(su(u^2-a^2)) = 1/dua(arcsec(u/a)+c)

Most people get confused with this, as I do. The biggest problem people make is they don't square u and a after they put them into the equation.

Volume by disks is used to find the volume of a solid object

The formula for this is: pi S a-b [R(x)]^2 dx

For example, if you are given a problem that asks you to find the volume of a solid by rotating the region bounded by f(x)=2-x^2 on the x axis from 0 to 2

You would first draw the graph. This graph would be a curve and it would be rotated about the x axis, making it a solid, circular-shaped figure. You then just plug it into the formula and solve.

For volume by washers:

I personally think washers is easier. Maybe it is because I know how to do area under a curve, and this is what I relate it to. The formula for this is:

pi S a-b top^2-bottom^2 dx
*you will always have two equations.

For washers, you also graph the equations, but this time to see which curve is on top. The curve on top will be the first equation in your problem. The one on the bottom will be the one you'll be subtracting by. This is why I relate it to area under a curve.

For the things I do not understand:
This week was pretty fast paste, and it seemed like once we went over one thing, we were moving straight on to the next, and I was left completley lost. I understand the formulas and how to plug into them for trig inverse and disks, but I do not know where to go from there. Can someone please help me find some sort of steps to go by for this?

Post #16

Last week in calculus, we learned how to find volume by disk and how to use washers.

Disks is only used with solid objects and the formula is pi interval a, b [R(x)]^2 dx
An example would be: Evaluate the integral that gives the volume of the SOLID formed by revolving the region about the x-axis. y=-x+1
The first step would be to graph the equation which I can't show on here
By graphing it, it will show what interval you are looking at. In this case it is [0,1]
Then plug into your equation: pi integral of (0,1) [-x+1]^2
Next you have to square the equation in order to integrate it: x^2-2x+1
Integrate: 1/3 x^3-2(1/2x^2) +x which simplifies to 1/3x^3 - x^2 +x
Then all you have to do is plug in your intervals: 1/3(1)^3-(1)^2+1-[0] = 1/3 pi.

Washers is used when there is two equations.
The formula is pi integral of (a,b) top^2 - bottom ^2 dx
Example: Find the volumes of the solids generated by revolving the regions bounded by the graphs of the equations about the given lines.
y=6-2x-x^2 , y=x+6 about the x-axis
First step just as in disks is to graph the equations.
Nest you set your equations equal to each other to find the interval you are looking at.
6-2x-x^2=x+6
-3x-x^2 = 0
solve for x: -x(3+x)
x=0 and x=-3 therefore your interval is (-3, 0), 0 on the top because the bigger number is b.
After you find your interval, you have to use your graph to determine which equation is on top. If unsure, you can always plug into your calculator if it is allowed.
In this problem, 6-2x-x^2 is on top so you subtract x+6 from it
(6-2x-x^2)^2 - [x+6] ^2
Then you would have to square each and add like terms
Remember (6-2x-x^2)^2 is solved by multiplying (6-2x-x^2)(6-2x-x^2)
Then you would integrate in and plug your intervals into the integral.
Your answer should be 243/5 pi. Don't forget the pi.

(b) same equations about the line y=3
It's the same steps except you would have to subtract 3 from each equation
6-2x-x^2-3 - [x+6-3]
simplifies to: (3-2x-x^2)^2 - [x+3]^2
The interval would remain the same so on (-3, 0)
After following the same steps as problem a which were square everything, simplify, integrate, then plug in, your answer should be 108/5 pi.

I am having trouble with trig inverse integration, which I think we also learned last week. I have no idea where to start or the what to do at all so if anyone can explain it to me from start to finish I would greatly appreciate it.

Post #16

Hello week 16. Friday during class, we were working on a packet. I was getting the answers right! I tried finishing it today, but I started to have problems with the back page. So, I'm going to talk about what I am getting and what I'm not comfortable doing.

Area between curves:
formula: bSa top equation-bottom equation

example: Find the area of the region enclosed by y=-4x^2+41x+94 and y=x-2 between x=1 and x=7.

1. draw a picture: This would be a parabola facing down with a line going through it.

2. subtract the two equations: (-4x^2+41x+94-(x-2)
(-4x^2+41x+94-(-x+2)

3. put together like terms and integrate: S(-4x^2+41x+94-(-x+2))
(-4/3)x^3+20x^2+96x

4. plug in 1 and 6 and subtract: (3584/3)-(344/3)= 1080


Volume by disks:
formula: pi[R(x)]^2dx

example: Find the volume of the solid obtained by rotating about the x-axis the region under the curve from 0 to 4. square root of(-x^2+6x+27)

1. draw a picture: it would be like a circle once you reflect it

2. plug into formula: square root of(-x^2+6x+27)
pi(0S4)[square root of(-x^2+6x+27)]^2
pi(0S4)(-x^2+6x+27)

3. integrate: pi((-1/3)x^3+3x^2+27x)

4. plug in 0 and 4: pi(404/3)-0= (404/3)pi

Please Help Now :)
Now for Washers..? I am confused a little on this. Is there any problems in the packet that need this or another example anyone can show me? I just need help understanding why and when you use that formula.

WEEK 16

This week we learned how to find volumes of disks and of washers. Let me review how to find area first.

To find area between curves:
The formula you use is b(int)a (top eq.) - (bottom eq.).
If a and b is not already given to you, then you much set the equations equal to each other and solve.
You find which equation is top/bottom by graphing both and simply looking to see which one is on top.
If the area is on the y-axis, then the a and b values need to be set as y-values, and the equations must be solved for x.

Example:
Question: Find the area of the region enclosed by y= -7x^2 + 14x + 152 and y = -4x^2 -x + 2 between x=0 and x=8.

Solution: You first graph both functions and see that the first equation given is on top.
So you set up your equation as 8(int)0 (-7x^2 + 14x + 152) - (-4x^2 -x + 2).
You then distribute the negative and combine like terms to get: 8(int)0 (-3x^2 + 15x +150).
You then intergrate and get: -x^3 + (15/2)(x)^2 + 150x 80. You then plug in f(8) - f(0) to get 1168 units squared.

So now on to volume.
First of all: Disks are solid objects with one equation bounded by the equation and usually an axis. Washers are objects bounded by two equations.

Formula for Disk:

(Pi) b(int)a [R(x)]^2.

Formula for Washers:

(Pi) b(int)a (top eq.)^2 - (bottom eq.)^2 dx.


Something I do not get from this week is how to solve equations when you have to shift.

An Example would be number 13 from the 7.2 homework worksheet.

the directions say to find the volumes of the regions bounded by both graphs and the lines given.

y = x^2 , y =4x - x^2

a) the x-axis
b) the y-axis

Overall, I think I am starting to get the process of working these types of problems a little better, but just need a little extra practice and maybe somebody to re-explain the shifting problems.

16th post

This week in calculus i started off being confused by Volume of Disks and Washers, but on Friday, Mrs. Robinson gave us a packet to practice these concepts with and i must say that the packet helped me tremendously.

Volume by Disks:

when dealing with disks, your fourmula will be pie times the function they give you squared. for example, lets say that you are given the formula x^2 +2x and they want the volume by disks, you would square the function which would give you 4x^3 +6x^2. Then you would integrate the function and plug in your bounds... sometimes they may give you the bounds so all you would have to do is plug them in.. otherwise you will either have to set the equations equal to each other or look at the graph if you are given one.

Washers:

volume by washers is basically the same concept except that it is not just one formula squared, it is pie times the top equation squared minus the bottom equation squared. You figure out which formula is the top and bottom by graphing the equation by hand or on your calculator, once you figure that out, you then square the functions and integrate them and solve. Again, the same rules apply to the bounds.

Sometimes, they will not ask you for the volume, instead they will ask you for the area of the disk or washer, if that is the case then you have to use a different function. When asked for the area, there is no pie involved and the formula is the top equation minus the bottom equation. You find your bounds the same way as the volume problems. So you would integrate the functions as usual and plug in the bounds.

Some of the things i am having problems with is the graphs. I can plug them in on my calculator and see what they look like, but i don't understand what to shade. If anyone can help me graph my equations that would be a huge help. Thanks!

post 16

volume by disks:

the formula is pi times the integral of the [function given] squared times dx. so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one. then graph

volume by washers:

the formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the in between number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. then graph.

LRAM is left hand approximation and the formula is:
delta x [f(a) + f( delta x +a) .... + f( delta x - b)]

Say you are asked to calculate the left Riemann Sum for -4x -5 on the interval [-3, -1] divided into 2 subintervals.

delta x would equal: -1+3 /2 = 2/2 = 1
1[ f(-3) + f(-3 +1)]
1[ f( -3) + f(-2)]
then plug into your equation

RRAM is right hand approximation and the formula is:
delta x [ f(a + delta x) + .... + f(b)]
so using the same example:
1[ f( -2) + f(-1)] and then plug into your equation

MRAM is to calculate the middle and the formula is:
delta x [ f(mid) + f(mid) + .... ]
To find midpoints, you would add the two numbers together then divide by two
In this problem the numbers would be: -3 , -2, -1
-3 + -2/ 2 = -5/2 and -2 + -1 / 2 = -3/2
so 1[f(-5/2) + f(-3/2)] and the plug in

Trapezoidal is different because instead of multiplying by delta x, you multiply by delta x/2 and you also have on more term then your number of subintervals.
The formula is : delta x/2 [f(a) + 2f(a + delta x) + 2f(a+ 2 delta x) + ....f(b)]
For this problem: 1/2 [ f(-3) + 2 f(-2) + f( -1)] and then plug in.

Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in

e integration:

whatever is raised to the e power will be your u and du will be the derivative of u. For example:

e^2x-1dx
u=2x-1 du=2
rewrite the function as:
1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.

related rates:

The steps for related rates are….


1. Pick out all variables
2. Pick out all equations
3. Pick out what you are looking for
4. Sketch a graph and label
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve

limits:

Rules for Limits:…
1. if the degree of top equals the degree of bottom, the answer is the top coefficient over bottom coefficient
2. if top degree is bigger than bottom degree, the answer is positive or negative infinity
2. if top degree is less than bottom degree, the answer is 0

this washers and disks stuff is really confusing me I know how to graph but after that im lost.. so help please!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!