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.
Sunday, January 3, 2010
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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