This week we went over everything we have learned in Calculus so far. One of my major problems is not knowing when and where to plug numbers in sometimes. The one part of calculus i have a good handle on are the first and second derivative tests.
First derivative test:
-take the derivative of the original function
-solve for x (the values will be your critical values)
-set those values up into intervals between negative infinity and infinity
- plug in numbers between the intervals into the function
-this will show you when the function is increasing, decreasing, and you will find max's and mins.
Second derivative test:
-take the derivative of the original function twice
-solve for x values(critical values)
-set up into intervals between infinity and negative infinity
-plug in values between the intervals into the function
-this will show you where the graph is concave up and down, and where there is a point of inflection.
Another concept i have a decent handle on is related rates:
-Write down the given
-figure out which formula they want you to use
-plug in the given
-take the derivative
-solve for the unknown
One of my problems is that i do not always know what the formulas are that they want me to use. I'm going to try to study them over the weekend to see if that will help. Another concept i don't know really well is optimization and linearization. I know for linearization the key word is approximate but i'm not sure what to do to work the problem. I am actually getting better at looking at a graph and finding max and mins and point of inflections, but i still have to work on that too. I am still having trouble doing trig functions by hand even though i remeber my trig chart so that is another thing i'm going to have to work on. If anyone could help me with any of these that would be awesome. Thanks :)
Saturday, January 9, 2010
Post #21
So this week we reviewed and took AP Practice tests, wow, I didn't realize I needed THAT much work! Therefore, I'm kinda at a lost on what to do my blog on because just about everything thing we reviewed I didn't know. SCARYY!!!
What I don't understand is how to know what the problem is asking me. I know no one can teach me that, but I'm going to start looking over all my notes that we've been taking so see if it will start clicking, because going from calculus to MAO I notices that the problems repeat and its just a little different or it just has different numbers, so hopefully that helps!
Let's just go over what we should know for the NON-CALCULATOR PORTION in order to do these problems:
~know the rules that go with the limit going to infinity
• If the exponents are the same then take the coefficients of them!
• If the exponents are not the same then look at which one’s bigger:
O If the top is bigger then it’s infinity
o If the bottom is bigger then it’s zero
~know the rules for the definition of a derivative
~know what a tangent line is and how to find one
~know how to read charts and figure them out
~know what the graphs of equations look like
~know how to take the derivative
• Using the product rule
• Using the quotient rule
• Using all formulas given on the first day
~know how to take the integral
~know that when you’re finding the area you’ll be integrating
~know how to take the average value---1/b-a(integral from a to b or f(x))
~know how to find velocity
~know how to do LRAM, RRAM, and MRAM
I really need to remember the H’(x)= 1/ f ’[H(x)]…
I don’t know how to do anything with optimization or linearization =(
I really feel lost right now…I can’t explain none of these problems except for the limit ones so I really need help on anything you feel that you KNOW…thanks!
~ElliE~
What I don't understand is how to know what the problem is asking me. I know no one can teach me that, but I'm going to start looking over all my notes that we've been taking so see if it will start clicking, because going from calculus to MAO I notices that the problems repeat and its just a little different or it just has different numbers, so hopefully that helps!
Let's just go over what we should know for the NON-CALCULATOR PORTION in order to do these problems:
~know the rules that go with the limit going to infinity
• If the exponents are the same then take the coefficients of them!
• If the exponents are not the same then look at which one’s bigger:
O If the top is bigger then it’s infinity
o If the bottom is bigger then it’s zero
~know the rules for the definition of a derivative
~know what a tangent line is and how to find one
~know how to read charts and figure them out
~know what the graphs of equations look like
~know how to take the derivative
• Using the product rule
• Using the quotient rule
• Using all formulas given on the first day
~know how to take the integral
~know that when you’re finding the area you’ll be integrating
~know how to take the average value---1/b-a(integral from a to b or f(x))
~know how to find velocity
~know how to do LRAM, RRAM, and MRAM
I really need to remember the H’(x)= 1/ f ’[H(x)]…
I don’t know how to do anything with optimization or linearization =(
I really feel lost right now…I can’t explain none of these problems except for the limit ones so I really need help on anything you feel that you KNOW…thanks!
~ElliE~
Friday, January 8, 2010
post 21
REVIEW!!
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
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
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.
i still dont understand related rates so if someone could help me that would be good
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
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
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.
i still dont understand related rates so if someone could help me that would be good
Number 40 on the AP from Friday
40. So we set up the equation in class as
45=50e^(9k)
45/50=e^(9k)
.9=e^(9k)
ln.9 = ln (e^9k)
ln.9 = 9k
k=ln.9/9
Then you set up the second equation as
20 = 50e^(ln.9/9)t
20/50 = e^(ln.9/9)t
.4 = e^(ln.9/9)t
ln.4 = ln e^(ln.9/9)t
ln.4 = (ln.9/9)t
t = 78
Monday, January 4, 2010
Post #20
So the blog wouldn't let me sign in yesterday, but I finally got it working today.
Well, I think I have talked about the first derivative test way too much. So, I guess I will focus on the second derivative test this time.
PROBLEM: Finding the points of inflection for x^3-6x^2+12x.
1. take the derivative:
3x^2-12x+12
2. take the second derivative:
6x-12
3. set = to 0:
6x-12=0 x=2 --->gives possible pts. of inflection
4. set up intervals:
(-infinity,2) u (2,infinity)
5. plug in to SECOND DERIVATIVE:
6(0)-12= -ve --->concave down
6(3)-12= +ve --->concave up
So, x=2 is the point of inflection.
Also, I thought I might review implicit derivatives.
PROBLEM: y^3+y^2-5y-x^2=-4
1. take derivative of both sides: (implicit derivatives have an = sign)
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0
*remember every time you take the derivative of y, you have to note it by dy/dx or y^1
3. solve for dy/dx: (you are going to have to take out a dy/dx when solving)
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
*Also, if you want the slope you must plug in a x and y-value.
So, I would just like to point out that I understand most (definitely not all) things we have learned so far in Calculus. I just seem to forget a step, mess up a derivative, or the problem is just simply tough haha. I think I need to start reviewing everything again because now I'm getting worried about the ap test!
Well, I think I have talked about the first derivative test way too much. So, I guess I will focus on the second derivative test this time.
PROBLEM: Finding the points of inflection for x^3-6x^2+12x.
1. take the derivative:
3x^2-12x+12
2. take the second derivative:
6x-12
3. set = to 0:
6x-12=0 x=2 --->gives possible pts. of inflection
4. set up intervals:
(-infinity,2) u (2,infinity)
5. plug in to SECOND DERIVATIVE:
6(0)-12= -ve --->concave down
6(3)-12= +ve --->concave up
So, x=2 is the point of inflection.
Also, I thought I might review implicit derivatives.
PROBLEM: y^3+y^2-5y-x^2=-4
1. take derivative of both sides: (implicit derivatives have an = sign)
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0
*remember every time you take the derivative of y, you have to note it by dy/dx or y^1
3. solve for dy/dx: (you are going to have to take out a dy/dx when solving)
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
*Also, if you want the slope you must plug in a x and y-value.
So, I would just like to point out that I understand most (definitely not all) things we have learned so far in Calculus. I just seem to forget a step, mess up a derivative, or the problem is just simply tough haha. I think I need to start reviewing everything again because now I'm getting worried about the ap test!
Sunday, January 3, 2010
Post 20
Ok, so we go back to school soon, which is depressing. Now instead of just doing blogs, we actually have to learn stuff and have homework again :/
Today I'll talk about a definition of a derivative, product rule, quotient rule, and how to find tangent lines.
Instantaneous speed, or a definition of a derivative is described with this formula:
limn-0 f(x+h)-f(x)/h
As I said before, you can use this to find instantaneous speed. For example:
If you are asked to find instantaneous speed at x=2 and y=2x^2, you just plug it in to the formula of the definition of a derivative.
limn-0 16(2+h)^2 - 16(2)^2/h
= 16(4 + 4h + h^2) - 64
= 64 + 64h + 16h^2 - 64
= 64 u/sec
Product rule is just another way to find a derivative when something is multiplied together. The formula for this is uv' + vu'. This means to coppy the first term, times the derivative of the second term, plus coppy the second term, times the derivative of the first term. Other than taking a regular derivative with nothing multiplied or devided by it, product rule is pretty easy if the problem isn't huge.
For example:
(x^2)(3x)
= x^2(3) + 3x(2x)
= 3x^2 + 6x^2
= 9x^2
Quotient rule is a way of taking a derivative when it is in a fraction and it cannot be broken up or simplified. The formula for this is uv' - vu'/v^2. This means to coppy the first term down, timesed by the derivative of the second term, minus coppy the second term down, timesed by the derivative of the first term, all over the second term squared. The quotient rule is also pretty easy if the problem isn't huge. It can get confusing with many functions in it. You should not use quotient rule if other things can be done to the derivative to get around it. You should not use a quotient rule if you have a problem such as 5/x^2
For tangent lines, you plug information given by the problem into the point slope formula. You plug the exact numbers in for x and y. To find m, which is your slope, you will need to take the derivative of the two (a derivative is a slope). I used to have problems with tangent line problems. I don't remember why, because they are sooo easy.
Today I'll talk about a definition of a derivative, product rule, quotient rule, and how to find tangent lines.
Instantaneous speed, or a definition of a derivative is described with this formula:
limn-0 f(x+h)-f(x)/h
As I said before, you can use this to find instantaneous speed. For example:
If you are asked to find instantaneous speed at x=2 and y=2x^2, you just plug it in to the formula of the definition of a derivative.
limn-0 16(2+h)^2 - 16(2)^2/h
= 16(4 + 4h + h^2) - 64
= 64 + 64h + 16h^2 - 64
= 64 u/sec
Product rule is just another way to find a derivative when something is multiplied together. The formula for this is uv' + vu'. This means to coppy the first term, times the derivative of the second term, plus coppy the second term, times the derivative of the first term. Other than taking a regular derivative with nothing multiplied or devided by it, product rule is pretty easy if the problem isn't huge.
For example:
(x^2)(3x)
= x^2(3) + 3x(2x)
= 3x^2 + 6x^2
= 9x^2
Quotient rule is a way of taking a derivative when it is in a fraction and it cannot be broken up or simplified. The formula for this is uv' - vu'/v^2. This means to coppy the first term down, timesed by the derivative of the second term, minus coppy the second term down, timesed by the derivative of the first term, all over the second term squared. The quotient rule is also pretty easy if the problem isn't huge. It can get confusing with many functions in it. You should not use quotient rule if other things can be done to the derivative to get around it. You should not use a quotient rule if you have a problem such as 5/x^2
For tangent lines, you plug information given by the problem into the point slope formula. You plug the exact numbers in for x and y. To find m, which is your slope, you will need to take the derivative of the two (a derivative is a slope). I used to have problems with tangent line problems. I don't remember why, because they are sooo easy.
2nd and 3rd holiday posttt!
okay so these two i'm putting together because i went on vacation and was not able to do my second one when it was dueee.
let's go over the lram and that stuffff.
LRAM is left hand approximation
FORMULA: delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
EXAMPLE: calculate the left Riemann Sum for -4x -5 on the interval [-3, -1] divided into 2 subintervals.
delta x equals: -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
FORMULA: 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
FORMULA: delta x [ f(mid) + f(mid) + .... ]
To find midpoints, you would add the two numbers together then divide by two
----
DIFFERENCE BETWEEN INDEFINITE AND DEFINITE INTERGRALS
Indefinate - only an equation with the intergral symbol; you simply take the derivative backwards, you divide the exponent from the coefficient and add one to the exponent. Also, you MUST place a + c at the end of the equation. Why, you may ask, because you don't know if the beginning equation had a constant at the end of it, so you MUST mark it, or it will be wrong, for these you get an equation as an answerr.
Definate Integrals - almost the same thing except there will be a number at the top and bottom of the integral symbol; the end of your equation will be marked with a dx. You treat it the same as an indefinate integral except you plug in the top number, or the b, into the derivative and then you have to SUBTRACT the plugged in bottom number, a, from the derivative plugged in. You get a numerical answerrr!
EXAMPLE:
3
S x^2 dx
0
= x^3/3 ..... (3)^3/3 - (0)^3/3 ...... = 9-0
= 9
linearization:
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
VOLUME BY WASHERS
FORMULA: pie times the integral of the [top function] squared minus the [bottom function] squared times dx.
If you don't have the inbetween 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.
QUESTIONSS! i have no idea what e integration is, i mean when i see it i might, but just hearing people say it i don't have a clue! anyone up for explaining?
let's go over the lram and that stuffff.
LRAM is left hand approximation
FORMULA: delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
EXAMPLE: calculate the left Riemann Sum for -4x -5 on the interval [-3, -1] divided into 2 subintervals.
delta x equals: -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
FORMULA: 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
FORMULA: delta x [ f(mid) + f(mid) + .... ]
To find midpoints, you would add the two numbers together then divide by two
----
DIFFERENCE BETWEEN INDEFINITE AND DEFINITE INTERGRALS
Indefinate - only an equation with the intergral symbol; you simply take the derivative backwards, you divide the exponent from the coefficient and add one to the exponent. Also, you MUST place a + c at the end of the equation. Why, you may ask, because you don't know if the beginning equation had a constant at the end of it, so you MUST mark it, or it will be wrong, for these you get an equation as an answerr.
Definate Integrals - almost the same thing except there will be a number at the top and bottom of the integral symbol; the end of your equation will be marked with a dx. You treat it the same as an indefinate integral except you plug in the top number, or the b, into the derivative and then you have to SUBTRACT the plugged in bottom number, a, from the derivative plugged in. You get a numerical answerrr!
EXAMPLE:
3
S x^2 dx
0
= x^3/3 ..... (3)^3/3 - (0)^3/3 ...... = 9-0
= 9
linearization:
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
VOLUME BY WASHERS
FORMULA: pie times the integral of the [top function] squared minus the [bottom function] squared times dx.
If you don't have the inbetween 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.
QUESTIONSS! i have no idea what e integration is, i mean when i see it i might, but just hearing people say it i don't have a clue! anyone up for explaining?
Ash's 20th Post
So, school tomorrow! Yay? No.
I hate mornings. -.-
These holidays were relatively productive ^^ I lie. I got nothing accomplished except eating, showering, procrastinating, and sleeping. Oh, and finding my aquarium shrimp I have been looking for, but that's not mathy.
But anyway, on to math.
This isn't relating to something that I learned this year in Calc, yet, but I want to talk about it anyway. (since it mayyy ;] come in handy for an upcoming project)
The Spiral of Archimedes! I learned about this in Math Circle during the summer (totally fun, must re-apply this year). This Spiral, I don't remember what's so amazing about it, but it occurs naturally!! look at sunflower seeds while they are still in the plant, the pattern they make is the Spiral of Archimedes! A Nautilus Shell also is in the shape of the Spiral of Archimedes! Today, this spiral is used in building roller coasters! This could be useful for our future projects...
Anyway, that was my random bit of math for the day!
Now, I want to review something that we actually learned in Calc. Hmm....what can I do that no one hasn't exhausted already? Nothing =/
But I'll do.....IMPLICIT DERIVATIVES!
So, you know what a derivative is, right? *random mumbling is heard* THAT'S RIGHT!
So, implicit derivatives are just that but solving for dy/y'!
Example!
3x^2 + 2y = 0
1. 6x dx/dx + 2dy/dx = 0
2. 2dy/dx = -6x
3. dy/dx = -3x!
(I think...my mind's trying to fall into denial about school tomorrow)
That IS right, isn't it?
As for what I STILL don't get:
Optimization! I'm failing..miserably
I half-way understand washers/disks..I'm not sure what I'm confused about though...
I wish everyone luck in school tomorrow!!
I hate mornings. -.-
These holidays were relatively productive ^^ I lie. I got nothing accomplished except eating, showering, procrastinating, and sleeping. Oh, and finding my aquarium shrimp I have been looking for, but that's not mathy.
But anyway, on to math.
This isn't relating to something that I learned this year in Calc, yet, but I want to talk about it anyway. (since it mayyy ;] come in handy for an upcoming project)
The Spiral of Archimedes! I learned about this in Math Circle during the summer (totally fun, must re-apply this year). This Spiral, I don't remember what's so amazing about it, but it occurs naturally!! look at sunflower seeds while they are still in the plant, the pattern they make is the Spiral of Archimedes! A Nautilus Shell also is in the shape of the Spiral of Archimedes! Today, this spiral is used in building roller coasters! This could be useful for our future projects...
Anyway, that was my random bit of math for the day!
Now, I want to review something that we actually learned in Calc. Hmm....what can I do that no one hasn't exhausted already? Nothing =/
But I'll do.....IMPLICIT DERIVATIVES!
So, you know what a derivative is, right? *random mumbling is heard* THAT'S RIGHT!
So, implicit derivatives are just that but solving for dy/y'!
Example!
3x^2 + 2y = 0
1. 6x dx/dx + 2dy/dx = 0
2. 2dy/dx = -6x
3. dy/dx = -3x!
(I think...my mind's trying to fall into denial about school tomorrow)
That IS right, isn't it?
As for what I STILL don't get:
Optimization! I'm failing..miserably
I half-way understand washers/disks..I'm not sure what I'm confused about though...
I wish everyone luck in school tomorrow!!
20th post
hmm, so this is the first post of the new year, but it does not feel any different from all the other posts. oh well... this post is going to be on the second derivative test. The second derivative test is used to find all possible points of inflection and intervals 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
i knew how to do pretty much everything for the exam, so the only thing i would need to know is maybe optimization and just a review of some other stuff
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
i knew how to do pretty much everything for the exam, so the only thing i would need to know is maybe optimization and just a review of some other stuff
Post #20
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.
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))
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.
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))
Post Number Twenty
I’m in complete denial that we have school tomorrow. This blog is the only thing convincing me I do indeed have to go. Anyways happy new year and seniors we’re halfway there : )
I forgot my notebook at school over the holidays of course so I’m just going to spit out stuff basically.
I did most types of integration in my other blogs so now I’ll talk about disks and washers.
Disks are used to find the volume or area of solid objects
Pi bSa [R(x)^2 dx
Washers are used to find the volume or area of objects with holes
Pi bSa (top^2) – (bottom^2) dx
Limits!
The rules:
1. If the degree of the top is larger than the degree of the bottom, the limit approaches infinity.
2. If the degree of the top is smaller than the degree of the bottom, the limit approaches zero.
3. If the degree of the bottom is equal to the degree of the top, the limit is the top coefficient over the bottom coefficient of the largest degree.
Example:
Limx>>infinity (3x^2 –x)/(5x^2 + 4)
The limit is 3/5 because of rule number three.
Related Rates:
1. Pick out all of the variables
2. Pick out all of the equations
3. Pick out what exactly you are looking for
4. Sketch and label a graph
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve
I knew how to do this when we learned but I think I have trouble knowing the formulas I need. It’s all just memory though.
lovely
Linearization:
1. Identify the equation (duh)
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
6. Solve the equation
Tuhhhhh I still suck at optimization, angle of elevation, average speed? And many other things I forgot I ever learned.
So excited to see everyone tomorrow, sikeeeeee
I forgot my notebook at school over the holidays of course so I’m just going to spit out stuff basically.
I did most types of integration in my other blogs so now I’ll talk about disks and washers.
Disks are used to find the volume or area of solid objects
Pi bSa [R(x)^2 dx
Washers are used to find the volume or area of objects with holes
Pi bSa (top^2) – (bottom^2) dx
Limits!
The rules:
1. If the degree of the top is larger than the degree of the bottom, the limit approaches infinity.
2. If the degree of the top is smaller than the degree of the bottom, the limit approaches zero.
3. If the degree of the bottom is equal to the degree of the top, the limit is the top coefficient over the bottom coefficient of the largest degree.
Example:
Limx>>infinity (3x^2 –x)/(5x^2 + 4)
The limit is 3/5 because of rule number three.
Related Rates:
1. Pick out all of the variables
2. Pick out all of the equations
3. Pick out what exactly you are looking for
4. Sketch and label a graph
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve
I knew how to do this when we learned but I think I have trouble knowing the formulas I need. It’s all just memory though.
lovely
Linearization:
1. Identify the equation (duh)
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
6. Solve the equation
Tuhhhhh I still suck at optimization, angle of elevation, average speed? And many other things I forgot I ever learned.
So excited to see everyone tomorrow, sikeeeeee
Post #20
Optimization
Steps:
1. Identify primary and secondary equations
- primary is the one you are maximizing or minimizing
- secondary is the other one
2. Solve secondary equation for one variable and plug into the primary
- If primary equation only has one variable, you can skip this step
3. Take the derivative of primary equation, set equal to zero, then solve for x.
4. Plug into secondary equation to find the other value.
Example:
Find the length and width of a rectangle with a perimeter of 60 meters and a maximum area.
1. Since maximizing the area, the area of a rectangle formula will be the primary. A=lw
The secondary formula will be the perimeter formula because you are given the perimeter.
60 = 2l + 2w
2. Solve secondary equation for one variable:
60 = 2l + 2w
60-2w = 2l
l= 60-2w/2
Then plug back into the primary:
A = 60-2w/2 (w)
3. Multiply the w in to simplify equation:
60w- 2w^2/ 2 = 30w - w^2
Then take the derivative: 30 - 2w
4. Solve for w:
30 = 2w
w = 15
Plug back into secondary to find l:
60= 2l + 2(15)
60 = 2l + 30
30 = 2l
l=15
Implicit Derivatives involve x's and y's
Steps:
1. Take the derivative using same rules of both sides
2. Every time you take the derivative of y, note it with dy/dx or y'.
3. Solve for dy/dx.
Example:
y^3 + y^2 - 5y - x^2 = 4
3y^2 dy/dx + 2y dy/dx - 5 dy/dx -2x = 0
Now solve for dy/dx:
dy/dx = 2x/ 3y^2 + 2y - 5
Example 2:
x^2 + y^2 = 9
2x + 2y dy/dx = 0
dy/dx = -2x/2y
which equals -x/y
I'm still having trouble with using graphs and I forgot how to do linearization so a recap would be splendid.
My 3rd holiday post
My last and thrid blog over the holidays and ugh.. we got school tomorrow im so bummed but were almost finished only like 60 more days or something so thats exiting
Integration
riemann summs...
Lram-left hand approximationdeltax[f(a)+f(a+deltax)+.....f(b+deltax)]
Rram-right hand approximationdeltax[f(a+delta x)+....f(b)]
Mramdeltax[f(mid)+f(mid)+....]
trapezoiddeltax/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 dxEx
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 dxx=1 so (pie)[(1/2)-(1/5)]-03(pie)/10 is your awnser
and for another big problem i have in calulus is optamization
Integration
riemann summs...
Lram-left hand approximationdeltax[f(a)+f(a+deltax)+.....f(b+deltax)]
Rram-right hand approximationdeltax[f(a+delta x)+....f(b)]
Mramdeltax[f(mid)+f(mid)+....]
trapezoiddeltax/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 dxEx
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 dxx=1 so (pie)[(1/2)-(1/5)]-03(pie)/10 is your awnser
and for another big problem i have in calulus is optamization
Post #1 of the New Year :-)
Okay so first off, Happy New Year... second off, ugh. we have school tomorrow.
Anyway...
What can I explain...I'm so at lost as to what i should explain...I've explained like every concept I remember... twice now.
LIMIT RULES:
1. if the degree of the top is larger than the degree of the bottom, the limit approaches infinity
2. if the degree of the bottom is larger than the degree of the tip, the limit approaches zero
3. if the degree of the bottom is equal to the degree of the top, then you make a fraction out of the coefficients in front of the largest degree
So for implicit derivatives...its pretty easy. You are going to be using this when you have like y^2 + x^2 = 4. It will ask for dy/dx. So what you do is you take the derivative like normal...but whenever you take the derivative of y you write dy/dx. So the above would be 2y (dy/dx) + 2x = 0. Now you solve for dy/dx. To do this, minus over 2x and then divide by 2y. So the answer would be
dy/dx = -x/y.
That's pretty easy...
Um let's see...
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
Okay so first derivative refers to slope of position..which is velocity.
Second derivative is slope of velocity...or acceleration.
Third derivative would be the slope of acceleration which I believe is jerk.
Past that, its just the rate of change of whatever was before...so...yeah.
Anyway, think this is around 285 words so...see you guys tomorrow.
Anyway...
What can I explain...I'm so at lost as to what i should explain...I've explained like every concept I remember... twice now.
LIMIT RULES:
1. if the degree of the top is larger than the degree of the bottom, the limit approaches infinity
2. if the degree of the bottom is larger than the degree of the tip, the limit approaches zero
3. if the degree of the bottom is equal to the degree of the top, then you make a fraction out of the coefficients in front of the largest degree
So for implicit derivatives...its pretty easy. You are going to be using this when you have like y^2 + x^2 = 4. It will ask for dy/dx. So what you do is you take the derivative like normal...but whenever you take the derivative of y you write dy/dx. So the above would be 2y (dy/dx) + 2x = 0. Now you solve for dy/dx. To do this, minus over 2x and then divide by 2y. So the answer would be
dy/dx = -x/y.
That's pretty easy...
Um let's see...
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
Okay so first derivative refers to slope of position..which is velocity.
Second derivative is slope of velocity...or acceleration.
Third derivative would be the slope of acceleration which I believe is jerk.
Past that, its just the rate of change of whatever was before...so...yeah.
Anyway, think this is around 285 words so...see you guys tomorrow.
post of holidays..3
IMPLICIT DERIVATIVES:
1. Take the derivative like normal.
2. For each y-term, you put y' or dy/dx behind it.
3. Solve for dy/dy or y'.
36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x
steps to SECOND IMPLICIT DERIVATIVES
1. Take the derivative of the first derivative
2. Put d^2y/dx^2 for dy/dx
3. Simplify
4. Plug in values
so an example using the same equation is:
we take the first implicit derivative first
36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x
You have to identify what you are looking for and what you are given. Not only does this make it easier on you, it's kind of necessary, especially when you want those points on free response questions (or so I'm told). You also have to realize that when you are doing related rates, you have to put dy/dt or dx/dt or whatever whenever you are taking the derivative of some variable in relation to time (hence the t). So given that:
Given xy = 4
you want to know what dy/dt equals given x = 8 and dx/dt=10.
Take the derivative (product rule):
dx/dt y + dy/dt x = 0
Plug in everything:
dy/dt = -10y/8
= -5y/4
1. Take the derivative like normal.
2. For each y-term, you put y' or dy/dx behind it.
3. Solve for dy/dy or y'.
36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x
steps to SECOND IMPLICIT DERIVATIVES
1. Take the derivative of the first derivative
2. Put d^2y/dx^2 for dy/dx
3. Simplify
4. Plug in values
so an example using the same equation is:
we take the first implicit derivative first
36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x
You have to identify what you are looking for and what you are given. Not only does this make it easier on you, it's kind of necessary, especially when you want those points on free response questions (or so I'm told). You also have to realize that when you are doing related rates, you have to put dy/dt or dx/dt or whatever whenever you are taking the derivative of some variable in relation to time (hence the t). So given that:
Given xy = 4
you want to know what dy/dt equals given x = 8 and dx/dt=10.
Take the derivative (product rule):
dx/dt y + dy/dt x = 0
Plug in everything:
dy/dt = -10y/8
= -5y/4
3rd post of the holidays
ok heres the last one of the holidays.
Related Rates
1. Identify all of the variables and equations.
2. Identify what you want to find.
3. Sketch and label.
4. Write an equations involving your variables.
5. Take the derivative with in terms of time.
6. Substitute the derivative in and solve.
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.
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve.
For maxs and mins, to find out if it is a max or a min, you have to use the derivative test. You have to set up intervals and then test them by plugging in points. If you get a positive number, the function is increasing. If you get a negative number, the function is decreasing. If it goes increasing, point, decreasing then it is a max. If it goes decreasing, point, increasing, then it is a min.
Quotient rule
U/v = (v(u)' - u(v)')/ v^2
and thats my last post. some things like tangent line i cant really remember even though its super easy so if anyone has any little steps or tricks to give me holla.
Related Rates
1. Identify all of the variables and equations.
2. Identify what you want to find.
3. Sketch and label.
4. Write an equations involving your variables.
5. Take the derivative with in terms of time.
6. Substitute the derivative in and solve.
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.
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve.
For maxs and mins, to find out if it is a max or a min, you have to use the derivative test. You have to set up intervals and then test them by plugging in points. If you get a positive number, the function is increasing. If you get a negative number, the function is decreasing. If it goes increasing, point, decreasing then it is a max. If it goes decreasing, point, increasing, then it is a min.
Quotient rule
U/v = (v(u)' - u(v)')/ v^2
and thats my last post. some things like tangent line i cant really remember even though its super easy so if anyone has any little steps or tricks to give me holla.
The 3rd post of the holidays
Now I will do my last post of the holidays. I will start with the limit rules. The limit rules are:
1) if the highest exponent is the same on the top and bottom then the limit is the top coefficient over the bottom coefficient of the highest exponents.
2) If the highest exponent is on the top then the limit is infinity.
3) But if the highest exponent is on the bottom then the limit is 0.
Some examples are:
lim->infinity (5x^2-2x)/(4x^2+4x-1) Then the limit is 5/4 because it follows rule 1.
lim->infinity (x^3+6)/(7x^2-3) Then the limit is infinity because it follows rule 2.
lim->infinity (2x^4-6)/(5x^6+7x+9) Then the limit is 0 because it follows rule 3.
Next I will explain 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
Finally I will talk about the trig inverse intergration formulas. 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 another question I have is instantaneous and average speed. I seem to have lost my notes on this so I cannot review on how to do this.
1) if the highest exponent is the same on the top and bottom then the limit is the top coefficient over the bottom coefficient of the highest exponents.
2) If the highest exponent is on the top then the limit is infinity.
3) But if the highest exponent is on the bottom then the limit is 0.
Some examples are:
lim->infinity (5x^2-2x)/(4x^2+4x-1) Then the limit is 5/4 because it follows rule 1.
lim->infinity (x^3+6)/(7x^2-3) Then the limit is infinity because it follows rule 2.
lim->infinity (2x^4-6)/(5x^6+7x+9) Then the limit is 0 because it follows rule 3.
Next I will explain 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
Finally I will talk about the trig inverse intergration formulas. 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 another question I have is instantaneous and average speed. I seem to have lost my notes on this so I cannot review on how to do this.
2nd post for the holidays
Well I finally have my internet back up and working so I will do the last two blogs I have left to do. I will start with 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
Next I will explain the Rolle's and Mean Value Theorem.
Rolle's
In Rolle's f(a)=f(b) or it can not be done. And there is always at least one answer which is "c". But mainly the way to do Rolle's is to start off making sure the equation is continuous and differentiable. Then make sure f(a)=f(b). Next take the derivative and solve for x. And finally pick the answer that falls between the interval given.
Mean Value
First you find the slope and plug into f(b)-f(a)/b-a. Then take the derivative and set it equal to the slope and solve for x. Finally take the answer between the interval given in the problem.
Now I am going to explain one of the first things we learned which is product and quotient rule.
Product rule is copy the first times the derivative of the last plus copy the first times the derivative of the second.
Quotient rule is copy the bottom times derivative of the top minus copy the top times the derivative of the bottom all divided by the bottom squared.
For my question I still cannot finish an optimization problem. I can start the problems I just cannot finish them.
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
Next I will explain the Rolle's and Mean Value Theorem.
Rolle's
In Rolle's f(a)=f(b) or it can not be done. And there is always at least one answer which is "c". But mainly the way to do Rolle's is to start off making sure the equation is continuous and differentiable. Then make sure f(a)=f(b). Next take the derivative and solve for x. And finally pick the answer that falls between the interval given.
Mean Value
First you find the slope and plug into f(b)-f(a)/b-a. Then take the derivative and set it equal to the slope and solve for x. Finally take the answer between the interval given in the problem.
Now I am going to explain one of the first things we learned which is product and quotient rule.
Product rule is copy the first times the derivative of the last plus copy the first times the derivative of the second.
Quotient rule is copy the bottom times derivative of the top minus copy the top times the derivative of the bottom all divided by the bottom squared.
For my question I still cannot finish an optimization problem. I can start the problems I just cannot finish them.
post 19
i forgot bout the bllogs so ima do one now and one tonight hahaa
first off we have tangent lines
1. take f'(x)
2. plug x in to find your slope/m.
3. plug x into f(x)to get y
4. using m and (x,y) plug it into the equation (y-y1)=m(x-x1).
then we can talk bout implicit derivatives
1. take the derivative of both sides
2. everytime you take the derivative of y, note it with dy/dx or y'
3. solve for dy/dx
FIRST AND SECOND DER TEST :)
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
Second Derivative:
first you find the first derivative and solve it for dy/dx by using the steps for the first derivative steps.
you then take the second derivative of the solved equation. Plugging in d^2y/d^2x everytime you take the derivative of y again. and where you have dy/dx you plug in your solved equation for that.
once you have everything plugged in and ready to go you then solve for d^2y/d^2x
IM NOT GOOD AT ANGLES OF ELEVATIOONNNNNNNNN!!!! OR OPTIMIZATION LOL
first off we have tangent lines
1. take f'(x)
2. plug x in to find your slope/m.
3. plug x into f(x)to get y
4. using m and (x,y) plug it into the equation (y-y1)=m(x-x1).
then we can talk bout implicit derivatives
1. take the derivative of both sides
2. everytime you take the derivative of y, note it with dy/dx or y'
3. solve for dy/dx
FIRST AND SECOND DER TEST :)
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
Second Derivative:
first you find the first derivative and solve it for dy/dx by using the steps for the first derivative steps.
you then take the second derivative of the solved equation. Plugging in d^2y/d^2x everytime you take the derivative of y again. and where you have dy/dx you plug in your solved equation for that.
once you have everything plugged in and ready to go you then solve for d^2y/d^2x
IM NOT GOOD AT ANGLES OF ELEVATIOONNNNNNNNN!!!! OR OPTIMIZATION LOL
post 20
happy new year! i can't believe it's already 2010. school year is halfway over. crazy, i know. okay, well back to math:
LIMIT RULES:
1 - When the degree of the bottom is GREATER than the degree of the top, the limit is 0.
2 - When the degree of the bottom is LESS than the degree of the top, the limit is infinity.**
3 - When the degree of the bootom is EQUAL to the degree of the top, divide leading coefficients.
implicit derivatives:
1. take the derivative of both sides
2. everytime you take the derivative of y, note it with dy/dx or y'
3. solve for dy/dx
when taking second derivative for implicit derivatives, it's a little different.
1. take first derivative.
2.take second derivative of your first derivative, noting all derivatives of y with d^2y/d^2x
3. everywhere there is dy/dx, plug in first derivative.
4. solve for d^2y/d^2x
HOW TO FIND THE EQUATION OF A TANGENT LINE:
1. take f'(x)
2. plug x in to find your slope/m.
3. plug x into f(x)to get y
4. using m and (x,y) plug it into the equation (y-y1)=m(x-x1).
well, i'll see you tomorrow. back to school... joy
LIMIT RULES:
1 - When the degree of the bottom is GREATER than the degree of the top, the limit is 0.
2 - When the degree of the bottom is LESS than the degree of the top, the limit is infinity.**
3 - When the degree of the bootom is EQUAL to the degree of the top, divide leading coefficients.
implicit derivatives:
1. take the derivative of both sides
2. everytime you take the derivative of y, note it with dy/dx or y'
3. solve for dy/dx
when taking second derivative for implicit derivatives, it's a little different.
1. take first derivative.
2.take second derivative of your first derivative, noting all derivatives of y with d^2y/d^2x
3. everywhere there is dy/dx, plug in first derivative.
4. solve for d^2y/d^2x
HOW TO FIND THE EQUATION OF A TANGENT LINE:
1. take f'(x)
2. plug x in to find your slope/m.
3. plug x into f(x)to get y
4. using m and (x,y) plug it into the equation (y-y1)=m(x-x1).
well, i'll see you tomorrow. back to school... joy
20th post
Goodbye 2009... hello 2010.. i hope everyone's holidays were good. Let's start off with some of the basics.
To remind anyone who does not know how to take a derivative here are some examples.
Let's say you have the function 4x^2 +5x+6. To take the derivative, you would multiply the 4 and the 2 and then subtract the 2 by 1. For the 5x, you would simply get rid of the x, and the 6 would become zero. So the final answer would be 8x +5.
Some other things to remember is the first and second derivative test. When using the first derivative test, you look for max and mins, and increasing and decreasing. To use the first derivative test, you take the derivative of the original function and solve for x. The x values are called critival values. You then plug these critical values into intervals between negative infinity and infinity. To find out whether your function is increasing or decreasing, you plug in numbers between your intervals into the derivative. If the number is positive it is increasing, if then number is negative then it is decreasing.
Another thing to remember is the second derivative test. It is basically the same thing as the first derivative test; however, you are taking the derivative twice and you are looking for different things such as change in concavity and points of inflection. You take the derivative of the original function twice and solve for x to get the critical values and set them up into intervals. You then plug in numbers between the intervals to see whether the function is concave up or concave down. If the number is positive it is concave up, if the number is negative it is concave down. Where there is a difference in concavity, there is a point of inflection.
Some of the things i am still having trouble with is optimization. I just cannot get a grasp on the concept. See all of you at school tomorrow!
To remind anyone who does not know how to take a derivative here are some examples.
Let's say you have the function 4x^2 +5x+6. To take the derivative, you would multiply the 4 and the 2 and then subtract the 2 by 1. For the 5x, you would simply get rid of the x, and the 6 would become zero. So the final answer would be 8x +5.
Some other things to remember is the first and second derivative test. When using the first derivative test, you look for max and mins, and increasing and decreasing. To use the first derivative test, you take the derivative of the original function and solve for x. The x values are called critival values. You then plug these critical values into intervals between negative infinity and infinity. To find out whether your function is increasing or decreasing, you plug in numbers between your intervals into the derivative. If the number is positive it is increasing, if then number is negative then it is decreasing.
Another thing to remember is the second derivative test. It is basically the same thing as the first derivative test; however, you are taking the derivative twice and you are looking for different things such as change in concavity and points of inflection. You take the derivative of the original function twice and solve for x to get the critical values and set them up into intervals. You then plug in numbers between the intervals to see whether the function is concave up or concave down. If the number is positive it is concave up, if the number is negative it is concave down. Where there is a difference in concavity, there is a point of inflection.
Some of the things i am still having trouble with is optimization. I just cannot get a grasp on the concept. See all of you at school tomorrow!
HOLIDAY POST NUMBER THREE
LAST DAY OF VACATION :( :( :(
Okay, more integration.
So, the sign that looks like a big S called the integral symbol and means you do the opposite of a derivative.
There are two types of integration: indefinite and definite.
Indefinite integration is when the answer is an equation.
In indefinite integration all the same properties of derivatives apply.
Definite integration is when the answer is a number.
Definite integration uses the integral [a,b].
So, integrating:
Polynomials
Sx^n dx = (x^(n+1))/(n+1) + C
Example:
Sx^3 dx = x^4/4 = 1/4 x^4 + C
For integrating trig., you basically have to know all your formulas of deriving sin, cos, etc. before you can integrate them.
Example:
S sinx dx = -cosx + C
S sec^2x dx = tanx + C
S 2cscx cotx dx = -2cscx + C
To make the integration problem you face a whole lot easier, you can rewrite the problem (if it works).
You can rewrite exponents at the bottom of a fraction to the negative form at the top.
Example:
1 / x^2 = x^-2
So when you put that into an integration problem it would be:
S 1 / x^2 = S x^-2
So then you simplify that to
x^-1 / -1
Which equals
-1/x + C
Another example:
S sinx / cos^2x dx
= S (1 / cosx)(sinx / cosx) dx
= S secx * tanx dx
= secx + C
Definite ingetration uses a and b to find the definite area under a curve/graph.
Formula for definite integration:
b S a f(x) dx = f(b) - f(a) = number
See you all tomorrow!!!
Okay, more integration.
So, the sign that looks like a big S called the integral symbol and means you do the opposite of a derivative.
There are two types of integration: indefinite and definite.
Indefinite integration is when the answer is an equation.
In indefinite integration all the same properties of derivatives apply.
Definite integration is when the answer is a number.
Definite integration uses the integral [a,b].
So, integrating:
Polynomials
Sx^n dx = (x^(n+1))/(n+1) + C
Example:
Sx^3 dx = x^4/4 = 1/4 x^4 + C
For integrating trig., you basically have to know all your formulas of deriving sin, cos, etc. before you can integrate them.
Example:
S sinx dx = -cosx + C
S sec^2x dx = tanx + C
S 2cscx cotx dx = -2cscx + C
To make the integration problem you face a whole lot easier, you can rewrite the problem (if it works).
You can rewrite exponents at the bottom of a fraction to the negative form at the top.
Example:
1 / x^2 = x^-2
So when you put that into an integration problem it would be:
S 1 / x^2 = S x^-2
So then you simplify that to
x^-1 / -1
Which equals
-1/x + C
Another example:
S sinx / cos^2x dx
= S (1 / cosx)(sinx / cosx) dx
= S secx * tanx dx
= secx + C
Definite ingetration uses a and b to find the definite area under a curve/graph.
Formula for definite integration:
b S a f(x) dx = f(b) - f(a) = number
See you all tomorrow!!!
Post #20
Goooooooooooodmorning! Since I just had to wish my sister to have fun on her vacation that she's leaving on today, there is nothing i would enjoy doing more but my homework being that we go to school tomorrow. YAY! (notice the sarcasm).
Anyways, i have another throwback blog because i didn't bring my binder home so i'm taking these lessons straight from the noggin. So here it goes.
The Second Derivative Test involves points of inflection and convavity, which is concave up or concave down. When taking the second derivative you must be aware that points of inflection only happen if there is a change in concavity.
To do the second derivative, first, you must take the derivative of the equation such as
6/(x^2 +3) = ( (x^2+3)(0) - [(6)(2x)] ) / (x^2 + 3)^2
Which gives you (-12x) / ( (x^2 +3) ^2) in it's simpliest terms.
Next, you must take the derivative of this equation.
Which would give you (36(x^2 -1)) / (x^2 +3)^3
This must then be solved to it's simpliest state set equal to zero which is
(36(x+1)(x-1)) / (x^2+3)^3
Now, you have found the critical values: X = +/- 1.
These are only POTENTIAL points of influction.
Next, you must set up intervals (-infinity, -1) u (-1, 1) u (1, infinity)
Now, you must pick a number in each interval and plug it into the second derivative. This will tell you if the interval concaves up or down.
f''(-2) = positive = concave up
f''(0) = negative = concave down
f''(2) = positvie = concave up
Another way to write this is:
at (-infinity, -1) and (1, infinity) = concave up
at (-1, 1) = concave down
points of inflection = x=-1x=1
These are the points of inflection because this is where the concavity changes. It is very important that you know points of inflection ONLY occur with changes of concavity.
I can't really remember any topics to say what i'm not understanding..so, hopefully everyone is having a good last day off, i might cry.
And umm, Juniors, did we have any homework?
Anyways, i have another throwback blog because i didn't bring my binder home so i'm taking these lessons straight from the noggin. So here it goes.
The Second Derivative Test involves points of inflection and convavity, which is concave up or concave down. When taking the second derivative you must be aware that points of inflection only happen if there is a change in concavity.
To do the second derivative, first, you must take the derivative of the equation such as
6/(x^2 +3) = ( (x^2+3)(0) - [(6)(2x)] ) / (x^2 + 3)^2
Which gives you (-12x) / ( (x^2 +3) ^2) in it's simpliest terms.
Next, you must take the derivative of this equation.
Which would give you (36(x^2 -1)) / (x^2 +3)^3
This must then be solved to it's simpliest state set equal to zero which is
(36(x+1)(x-1)) / (x^2+3)^3
Now, you have found the critical values: X = +/- 1.
These are only POTENTIAL points of influction.
Next, you must set up intervals (-infinity, -1) u (-1, 1) u (1, infinity)
Now, you must pick a number in each interval and plug it into the second derivative. This will tell you if the interval concaves up or down.
f''(-2) = positive = concave up
f''(0) = negative = concave down
f''(2) = positvie = concave up
Another way to write this is:
at (-infinity, -1) and (1, infinity) = concave up
at (-1, 1) = concave down
points of inflection = x=-1x=1
These are the points of inflection because this is where the concavity changes. It is very important that you know points of inflection ONLY occur with changes of concavity.
I can't really remember any topics to say what i'm not understanding..so, hopefully everyone is having a good last day off, i might cry.
And umm, Juniors, did we have any homework?
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