Well this week was a very fun and exciting week with Thanksgiving and my birthday being yesterday :), i have been working on the packet throughout the week and there are parts on there i am very confused about. For example, the ones that say to solve for dy/dx subject to initial condition, how do you even start those, i do not understand where the initial position comes in at. Another example of some things i am confused about is when they ask to approximate the area under a certain place using different Riemann Sums, is that a form of linearization or is it just plain integration. One other part i am confused about is when they ask to find the area of a region enclosed by two functions and two numbers. For example:
y=x-4 and x=-4y^2+21y+148 between y=-1 and y=5. Do i have to solve the x function for y? or do i have to do something else.
Some of the things i do understand is the Left and Right Riemann Sums: For Left Riemann sums, you find your delta x which can be found by b-a/n and then you times the delta x by [f(a)+ f(a+delta x)+.. f(b-delta x)]
For right Riemann sums, you find the delta x just like left riemann but you use a different formula. For right, you times delta x by [f(a+delta x)+...+f(b)]
Well as far as the packet goes, besides just regular indefinite and definite integrals that basically the whole packet: but here are some things we still have to remember throughout calculus.
Related Rates: this is the one thing i have learned in calculus that i have success with lol. Just remember when dealing with related rates, always write the given. Then you figure out what formula they want you to use. After that, take the derivative of the function and plug in what you are given and solve for the unknown.
Well that's all for this past week folks, if anyone can help me with the problems i have listed above, it will be very helpful. I hope you all have a good weekend :)
Saturday, November 28, 2009
Posts #14 and 15
So, i think i got the wrong idea on this blog thing..so i'm going to do both in one.
My bad..i guess I should ask questions if i didn't understand, or check edline..but anyways, here goes nothing.
For POST #14
Today i'll explain absolute max and mins.
First, you must do the first derivative test which includes taking the derivative and setting it equal to zero, then solving for x.
Second, you must plug in those values, also known as critical values, into original function to get y-values.
Third, plug in endpoints to original function to get more y-values.
Lastly, your highest y-value is an absolute max and lowest y-value is an absolute min.
THINGS YOU SHOULD REMEMBER:
**Absolute max and mins are written as a POINT or simply as a y-value
**They always and only include the y-value
**NEVER involve the x!!! It ONLY involves the y-values!
So, example problem:
Find the absolute max or min of f(x) = 3x^4 - 4x^3 on [-1,2]
12x^3-12x62 = 0
12X^2(x-1)
x = 0,1
(-1, 0) u (0,1) u (1,2)
f`(-.5) = -ve
f`(.5) = -ve
f`(1.5) = +ve
Then plug in endpoints and you recieve the points:
(1, -1)
(-1, 7)
(2, 16)
The absolute min is (1, -1) or y = -1
The absolute max is (2, 16) or y = 16
The thing I do not understand is average speed, velocity, and definition of a derivative. Could anyone help me because it is something that i can never quite grasp..
POST #15
One thing i am comfortable with is Rolle's theorem. In order to classify to be functionally correct with rolles, it must be continuous and differentiable and the y-values MUST match.
Rolles theorem gives conditions that guarantee the existance of an extrema in the interior of a closed interval.
It states, let f be continuous on a closed interval [a,b] and differentiable on the open interval (a,b). If f(a) = f(b) then there is AT LEAST one number, or c, in the interval (a,b) such that f`(c) = 0.
****Y-VALUES MUST MATCH!****
For the function f(x) = x^2 - 3x + 2, show that f`(x) = 0 on some interval.
x^2 - 3x + 2 = 0
(x-2)(x-1)
x-intercepts = [1,2]
f(1) = f(2) = 0
Continuous: check mate
Differentiable: check mate
f`(x) = 2x - 3
2x = 3
x = 3/2
c = 3/2
Using Rolle's Theorem because the function is continuous and differentiable and the y-values match; after the derivative was taken and set equal to zero, one found the c = 3/2.
The newest thing I don't understand and get lost with in integration is how you can not use product and quotient rule. I do not understand how to break it up..so can anyone explain how to break these things up? Also, substitution confuses me because i don't understand when you use substitution and when you break it up..
HELP PLEASEEE!
My bad..i guess I should ask questions if i didn't understand, or check edline..but anyways, here goes nothing.
For POST #14
Today i'll explain absolute max and mins.
First, you must do the first derivative test which includes taking the derivative and setting it equal to zero, then solving for x.
Second, you must plug in those values, also known as critical values, into original function to get y-values.
Third, plug in endpoints to original function to get more y-values.
Lastly, your highest y-value is an absolute max and lowest y-value is an absolute min.
THINGS YOU SHOULD REMEMBER:
**Absolute max and mins are written as a POINT or simply as a y-value
**They always and only include the y-value
**NEVER involve the x!!! It ONLY involves the y-values!
So, example problem:
Find the absolute max or min of f(x) = 3x^4 - 4x^3 on [-1,2]
12x^3-12x62 = 0
12X^2(x-1)
x = 0,1
(-1, 0) u (0,1) u (1,2)
f`(-.5) = -ve
f`(.5) = -ve
f`(1.5) = +ve
Then plug in endpoints and you recieve the points:
(1, -1)
(-1, 7)
(2, 16)
The absolute min is (1, -1) or y = -1
The absolute max is (2, 16) or y = 16
The thing I do not understand is average speed, velocity, and definition of a derivative. Could anyone help me because it is something that i can never quite grasp..
POST #15
One thing i am comfortable with is Rolle's theorem. In order to classify to be functionally correct with rolles, it must be continuous and differentiable and the y-values MUST match.
Rolles theorem gives conditions that guarantee the existance of an extrema in the interior of a closed interval.
It states, let f be continuous on a closed interval [a,b] and differentiable on the open interval (a,b). If f(a) = f(b) then there is AT LEAST one number, or c, in the interval (a,b) such that f`(c) = 0.
****Y-VALUES MUST MATCH!****
For the function f(x) = x^2 - 3x + 2, show that f`(x) = 0 on some interval.
x^2 - 3x + 2 = 0
(x-2)(x-1)
x-intercepts = [1,2]
f(1) = f(2) = 0
Continuous: check mate
Differentiable: check mate
f`(x) = 2x - 3
2x = 3
x = 3/2
c = 3/2
Using Rolle's Theorem because the function is continuous and differentiable and the y-values match; after the derivative was taken and set equal to zero, one found the c = 3/2.
The newest thing I don't understand and get lost with in integration is how you can not use product and quotient rule. I do not understand how to break it up..so can anyone explain how to break these things up? Also, substitution confuses me because i don't understand when you use substitution and when you break it up..
HELP PLEASEEE!
Friday, November 27, 2009
Post #15
okay...HAPPY THANKSGIVING!
so on the week off i worked on my packet as much as possible and i'm still having trouble with alot of them for some reason.
anyways...
friday i went to the sjh football game [they lost]
saturday i began working on all kinda stuff for the Georgia trip i was leaving for on monday i went to church and watched the lsu football game which was bad at the end
sunday i continued working on my packet and working of stuff for the Georgia trip...cleaned house and then watched some of the saints game and later went to my mawmaw's house.
monday..woke up early to go to school in order to get the lunches back in time and leave for georgia...and had a lot of fun on the bus to georgia until 1:30am
tuesday..we were in Georgia and went to Stone Mountain Park [all 30 of us]...i hiked down the mountain and stood at the tippy top of it...it was beautiful!
wednesday we began heading home...had fun on the bus and we stopped in new orleans for cafe du monde [idk how to spell that] and went walk by the river...it was good to be almost home...got home, unpacked and had a bad rest of the day..
thursday...HAPPY THANKSGIVING...didn't feel too hungry but the food was great...i stayed with both sides of my family all day
friday..THATS TODAY..i'm continuing to work on my homework [packet, blogs]..then of course my sister has projects to do too so i'm gonna help her with that...then i have to help out somewheres at one PM .. after that..i have no clue, might wanta go to the EDW game and if they win then a party after...saturday and sunday...wow that's too long from now to think of..haha
i hope everone had a good week off...and i'll see everyone monday =( ALREADYYY
~Ellie~
so on the week off i worked on my packet as much as possible and i'm still having trouble with alot of them for some reason.
anyways...
friday i went to the sjh football game [they lost]
saturday i began working on all kinda stuff for the Georgia trip i was leaving for on monday i went to church and watched the lsu football game which was bad at the end
sunday i continued working on my packet and working of stuff for the Georgia trip...cleaned house and then watched some of the saints game and later went to my mawmaw's house.
monday..woke up early to go to school in order to get the lunches back in time and leave for georgia...and had a lot of fun on the bus to georgia until 1:30am
tuesday..we were in Georgia and went to Stone Mountain Park [all 30 of us]...i hiked down the mountain and stood at the tippy top of it...it was beautiful!
wednesday we began heading home...had fun on the bus and we stopped in new orleans for cafe du monde [idk how to spell that] and went walk by the river...it was good to be almost home...got home, unpacked and had a bad rest of the day..
thursday...HAPPY THANKSGIVING...didn't feel too hungry but the food was great...i stayed with both sides of my family all day
friday..THATS TODAY..i'm continuing to work on my homework [packet, blogs]..then of course my sister has projects to do too so i'm gonna help her with that...then i have to help out somewheres at one PM .. after that..i have no clue, might wanta go to the EDW game and if they win then a party after...saturday and sunday...wow that's too long from now to think of..haha
i hope everone had a good week off...and i'll see everyone monday =( ALREADYYY
~Ellie~
Post #14
okay so this is totally late..ik ik..i thought i had did it..but it turns out i typed it in word on saturday november 21 and then forgot to put it in the blog because of craziness with my Georgia Vaction that was coming up and no one was ready for...
but..we learned about integration all week..
There were two types of integration. the first type was indefinite and the second was definite. you do the same thing for both except that for definite you see two numbers and at the end you have to plug them in and then subtract what you get by each other in order to get your answer. Remember that if it's indefinite then you have to put the plus 'c' but if it's definite then you just should have a number.
Then we learned about the change of variables which means that there is at least two differnet variables in the equation you are solving.
Then we learned about the area between the curves of graphs and the formula is bSa top equation minus the bottom equaion. in order to find a and b you need to set the equations equal to each other and then solve. what you solve for depends on what you're looking for or have in the equations. for instance if the area is on the y axis, then a and b need to be y values so the equations can be solved for x and opposite if the values need to be x values.
i kinda like the e integration the most. when you have the e integration, whatever it's raised to the e power will be your 'u' and 'du' will be the derivative of 'u'. here's an example:
e^4x-1dx
u=4x-1 du=4
now you rewrite the function as:
1/4{ e^u du, therefore
1/4e^4x-1+C is the final answer
Another type of integration that we learned is ln integration. so for this one just rmember that any time you have like [1/x] it'll be ln lxl plus c... these aren't too bad. An example of ln integration would be:
{1/5x-1dx
u= 5x-1 du=5dx
1/5 { 5/5x-1 dx
1/5 ln |5x-1| dx +C
I am not very good at change in variables substitution and area between curves substitution. I kinda understand e and ln substitution. I just do not know how to break things up in the other types of substitution and i have trouble knowing when to plug numbers in and finding where some of the numbers are coming from.
~Ellie~
but..we learned about integration all week..
There were two types of integration. the first type was indefinite and the second was definite. you do the same thing for both except that for definite you see two numbers and at the end you have to plug them in and then subtract what you get by each other in order to get your answer. Remember that if it's indefinite then you have to put the plus 'c' but if it's definite then you just should have a number.
Then we learned about the change of variables which means that there is at least two differnet variables in the equation you are solving.
Then we learned about the area between the curves of graphs and the formula is bSa top equation minus the bottom equaion. in order to find a and b you need to set the equations equal to each other and then solve. what you solve for depends on what you're looking for or have in the equations. for instance if the area is on the y axis, then a and b need to be y values so the equations can be solved for x and opposite if the values need to be x values.
i kinda like the e integration the most. when you have the e integration, whatever it's raised to the e power will be your 'u' and 'du' will be the derivative of 'u'. here's an example:
e^4x-1dx
u=4x-1 du=4
now you rewrite the function as:
1/4{ e^u du, therefore
1/4e^4x-1+C is the final answer
Another type of integration that we learned is ln integration. so for this one just rmember that any time you have like [1/x] it'll be ln lxl plus c... these aren't too bad. An example of ln integration would be:
{1/5x-1dx
u= 5x-1 du=5dx
1/5 { 5/5x-1 dx
1/5 ln |5x-1| dx +C
I am not very good at change in variables substitution and area between curves substitution. I kinda understand e and ln substitution. I just do not know how to break things up in the other types of substitution and i have trouble knowing when to plug numbers in and finding where some of the numbers are coming from.
~Ellie~
Thursday, November 26, 2009
AHHH..Someone help please!!
I don't know how to graph!! (I know it's in my Advanced Math notebook, but I only found one of the 3!) I had this problem last year because I simply cannot grasp the concept of it! I can't do Area Between Curves! I'm totally stuck on my packet! I beg someone to help me!!
Also, how do you integrate S(-5/x)dx and S(x^-1)dx??
Also, how do you integrate S(-5/x)dx and S(x^-1)dx??
Monday, November 23, 2009
Ash's 14th Post
Ugh, I forgot to post my blog this weekend -.-
Mom's 40th and major planning.
Anyway, Integration, YAY!
I get the simple integration..Just work backwards! =]
For example, say you have (okay, I dunno how to make the cool symbol, so I'm using S)
S(-4x +2)dx and you want to integrate it using indefinite integration.
1. I usually leave a space and tackle my x's first
Since the derivative of __x^2 is 2(__)x, let's write it like this!
___x^2 +___x + C (DON'T FORGET PLUS C!)
2. Take your coefficient and divide by your exponent to get your original coefficient!
(-4/2)x^2+(2/1)x + C
3. Simplify
-2x^2+2x+C
TA DA!
Okay, next: ln Integration *echoes*
Whenever you have a fraction where the top is the derivative of the bottom, you have to deal with ln Integration.
Just take the bottom, put it inside ln| |+C and see what needs to be added.
Example.
S(2/x)dx
1. You can make 2 the derivative of x by multiplying x by 2. (2x)
2S(1/x)dx
2. Now, put the bottom inside ln| |+C and remove the S
2ln|x|+C
THINGS I'M HAVING TROUBLE WITH!!!
1. Can you ALWAYS manipulate a problem to make it solvable by ln Integrations?
Ex: (x^2+x+1)/(x^2+1)
2. How exactly do you do synthetic for ln Integration? I'm a little confused.
On this beast packet:
1. How do you integrate something like these?
a) S(-5/x)dx
b) S -x^-1
c) S x^-1
2. For numbers 16-25, what does "subject to the initial condition...." mean???
3. Can someone PLEASE explain how to do RRAM, LRAM, MRAM, and TRAM???????
4. How do you go about solving number 46? I have no idea where to even begin.
That is all for the moment, if people comment this, I might actually get somewhere ^^
Mom's 40th and major planning.
Anyway, Integration, YAY!
I get the simple integration..Just work backwards! =]
For example, say you have (okay, I dunno how to make the cool symbol, so I'm using S)
S(-4x +2)dx and you want to integrate it using indefinite integration.
1. I usually leave a space and tackle my x's first
Since the derivative of __x^2 is 2(__)x, let's write it like this!
___x^2 +___x + C (DON'T FORGET PLUS C!)
2. Take your coefficient and divide by your exponent to get your original coefficient!
(-4/2)x^2+(2/1)x + C
3. Simplify
-2x^2+2x+C
TA DA!
Okay, next: ln Integration *echoes*
Whenever you have a fraction where the top is the derivative of the bottom, you have to deal with ln Integration.
Just take the bottom, put it inside ln| |+C and see what needs to be added.
Example.
S(2/x)dx
1. You can make 2 the derivative of x by multiplying x by 2. (2x)
2S(1/x)dx
2. Now, put the bottom inside ln| |+C and remove the S
2ln|x|+C
THINGS I'M HAVING TROUBLE WITH!!!
1. Can you ALWAYS manipulate a problem to make it solvable by ln Integrations?
Ex: (x^2+x+1)/(x^2+1)
2. How exactly do you do synthetic for ln Integration? I'm a little confused.
On this beast packet:
1. How do you integrate something like these?
a) S(-5/x)dx
b) S -x^-1
c) S x^-1
2. For numbers 16-25, what does "subject to the initial condition...." mean???
3. Can someone PLEASE explain how to do RRAM, LRAM, MRAM, and TRAM???????
4. How do you go about solving number 46? I have no idea where to even begin.
That is all for the moment, if people comment this, I might actually get somewhere ^^
Posting...#14
Well I'm posting this Tuesday because i have been in laffy visiting my family all weekend and when i got home i been going to the hospital to see my aunt thats in ICU :( so i kind of forgot about my blog cause i been having alot of things on my mind.
Well anyways in week 14 we learned how to do some different things involving integrations. And we had two test the first one I think i did real good on but the second one i bombed i don't know why i just didn't know how to do most of it.
So about the second tes i understand definite integrations but i don't understand how to find stuff using deffinite integration such as area etc. Thats where i got confused on the test i don't know how to find that. So if anyone can help me. thanks
I do know how to do indefinite so a few examples are:
Its basically the opposite from a derivative.
You also need to remember to put plus c behind every awnser
1: S(x+7)dx=(x^@/2)+7x +C
2: S(2x-3x^2)dx=x^2-x^3+C
Well i got to get started on my packet soon because that is big and I'm not that good. Also if anyone can tell me what you all did in class at school Friday that would help alot and if its something new can yall post the notes i missed and one example thanks alot.
Goodbye calc
Well anyways in week 14 we learned how to do some different things involving integrations. And we had two test the first one I think i did real good on but the second one i bombed i don't know why i just didn't know how to do most of it.
So about the second tes i understand definite integrations but i don't understand how to find stuff using deffinite integration such as area etc. Thats where i got confused on the test i don't know how to find that. So if anyone can help me. thanks
I do know how to do indefinite so a few examples are:
Its basically the opposite from a derivative.
You also need to remember to put plus c behind every awnser
1: S(x+7)dx=(x^@/2)+7x +C
2: S(2x-3x^2)dx=x^2-x^3+C
Well i got to get started on my packet soon because that is big and I'm not that good. Also if anyone can tell me what you all did in class at school Friday that would help alot and if its something new can yall post the notes i missed and one example thanks alot.
Goodbye calc
14th post
so this is the 14th week of calculus and im sorry if i posted on a Monday but I didn’t have access to a computer this weekend..
substitution:
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.
methods of integration:
The first formula you need to know is x=(b-a)/n [a,b] with n subintervals. 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)]
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 thedegree 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
substitution:
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.
methods of integration:
The first formula you need to know is x=(b-a)/n [a,b] with n subintervals. 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)]
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 thedegree 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
Sunday, November 22, 2009
Post #14
This week in calculus, we learned average value, substitution, e integration, and ln integration. Finding an average value is the same steps as finding a definite interval except you have to multiply the integral by 1/b-a. Example:
Find the average value of f(x)=x on [0,3]
1/ 3-0 = 1/3
1/3[1/2x^2]
then plug in 1/3 [ 1/2 (3)^2 - 1/2 (0) ^2] = 3/2
ln integration is really simple, you just have to recognize it. For ln integration, the top has to be the derivative of the bottom of the fraction.
Examples:
the integral of 1/x dx= ln |x| + c
the integral of 4/x dx = 4 ln |x| +c
the integral of 1/4x-1
u = 4x-1 du= 4 dx
1/4 the integral of 1/x du = 1/4 ln |4x-1| + c
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
Example:
(x^3 +1) (3x^2) dx
u= x^3 +1 du= 3x^2 dx
integral of u du
1/2 u ^2 + c
1/2 (x^3 + 1) ^2 + c
I think I am mostly confused with substitution with variables. I usually get lost after I solve for x. I don't know what to do next or what to plug in to. I am still having trouble with graphs so I am struggling sketching the graph for area between curves, but if the graph is sketch I know how to solve it so if anyone can help me with graphing without using my calculator please do.
Hope you all have a good holiday!
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