First derivative test:
You have to take the derivative of the function and set it equal to zero. Then you solve for the x values (critical points) and set them up into intervals between negative infinity and infinity. Then you plug in values between those intervals into the first derivative to find max or mins or if the graph is increasing or decreasing.
Second derivative test:
You take the derivative of the function twice and set it equal to zero. You solve for the x values and set them up into intervals between negative infinity and infinity. You plug in numbers between those intervals into the second derivative to see where the graph is concave up, concave down, or where there is a point of inflection.
limit rules:
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.
EXAMPLES:
1. If f(x)=ln((x^2)-1), then f^1(x) *(x^2-1)is really the absolute value of x^2-1
*take derivative
f^1(x)=ln(x^2-1)
(1/x^2-1)(2x)
2x/(x^2)-1
2. d/dx cos^2(x^3)
chain rule!
-2cos(x^3)sin(x^3)(3x^2)
-6x^2cos(x^3)sin(x^3)
3. Let f be a differentiable function such that f(3)=2 and f^1(3)=5. If the tangent line to the graph of f at x=3 is used to find an approximation to a zero of f, that approximation is?
f(3)=2 (3=x, 2=y)
f^1(3)=5 (<--5 slope)
y-2=5(x-3)
y-2=5(3.1-3)
y-2=5(.1)
y=2.5
*approximation=pick the answer choice closest which is 2.6.
I know this is going to sound dumb, buttt i don't remember how to integrate a fraction, i know there is no quotient rule, but i don't know what to do.
Saturday, April 17, 2010
Post #35
Tangent Lines
Tangent lines can be used for many purposes. Finding a tangent line requires only a little knowledge of calculus but a substantial amount of algebra.
Example:
Find the line tangent to the graph y=2x2+4x+6 at x=1.
1. Identify the equation and point of tangency. If not given a y value, plug the x value into the original equation.
y=2x2+4x+6 y=2(1)2
2. Differentiate you equation.
dy/dx=4x+4
3. Plug in x value then solve for dy/dx.
dy/dx=4(1)+4=8
Your dy/dx value is your slope from here on you can create your equation of the tangent line. There are three forms that your equation can be presented in: point-slope, slope-intercept, or standard form. For these purposes, I am using point-slope.
(y-12)=8(x-1) is your final answer.
Another Example:
Find the line tangent to the graph y=x^3 that is parallel to the perpendicular graph of y=-(1/3)x.
1. Find the slope of the perpendicular line of y=-(1/3)x
-1/3 =m Perpendicular m=3
2. Set first derivative =3.
3x^2=3
3. Solve for x.
3x^2=3 x^2=1 x=1 x=-1
4. Plug x back into the original equation to find y values.
X^3 (1)^3 (-1)^3 y=1 y=-1
Tangent lines can be used for many purposes. Finding a tangent line requires only a little knowledge of calculus but a substantial amount of algebra.
Example:
Find the line tangent to the graph y=2x2+4x+6 at x=1.
1. Identify the equation and point of tangency. If not given a y value, plug the x value into the original equation.
y=2x2+4x+6 y=2(1)2
2. Differentiate you equation.
dy/dx=4x+4
3. Plug in x value then solve for dy/dx.
dy/dx=4(1)+4=8
Your dy/dx value is your slope from here on you can create your equation of the tangent line. There are three forms that your equation can be presented in: point-slope, slope-intercept, or standard form. For these purposes, I am using point-slope.
(y-12)=8(x-1) is your final answer.
Another Example:
Find the line tangent to the graph y=x^3 that is parallel to the perpendicular graph of y=-(1/3)x.
1. Find the slope of the perpendicular line of y=-(1/3)x
-1/3 =m Perpendicular m=3
2. Set first derivative =3.
3x^2=3
3. Solve for x.
3x^2=3 x^2=1 x=1 x=-1
4. Plug x back into the original equation to find y values.
X^3 (1)^3 (-1)^3 y=1 y=-1
Post #35
Alright...15 DAYS LEFT!!! + Exams =(
This week is the first week that we actually took an entire AP Exam and then we found out what we would have if that was the real test. So I'll go over things that need to be remembered as well as the things that I needed to look up and know because I FORGOT THEM during the test.
Things to Remember:
If the top exponent is greater than the bottom it’s the limit as it approaches infinity, and if the top exponent is less than the bottom it’s the limit as it approaches zero!
DON’T GET THEM CONFUSED LIKE I DID!
Here’s a trick:
Remember if you’re divided a number by another number and you have a bigger number on top it’s usually not zero. Also, if you’re dividing a number by another number and you have a smaller number on top it’s usually not a big number. Therefore, when the bigger number is on top since it’s a bigger number than zero it’s INFINITY and when the bigger number is on the bottom that means that it’s being lessened so it’s ZERO!
JUST A REMINDER if it's for the x-axis then you solve for y and if it's on the y-axis then you solve for x..and if you plug the y-axis one's in your calculator to graph then you'll have to turn your calculator sideways with the screen on the right to see how it would really look.
AREA:
it's worked the same just they'll give you two equations and if you don't have numbers for the inbetween then you'll set them equal to get them, but if you do then don't worry and keep going. for area there is no pi and no squaring. so you put the integral of the first equation minus the second equation and simplify and solve. then take the integral of it and plug in the numbers inbetween!
AREA:
y=-4x^2+41x+94 AND y=x-2 inbetween 7 and 1
so graph and then put the top equation over the bottom equation
the integral of (-4x^2+41x+94)-(x-2)dx
simplify: the integral of (-4x^2+40x+96) dx
((-4/3)(x)^3+20(x)^2+96(x)) of 1 and 7
f(7)=(3584/3) f(1)=(344/3)
ANSWER: 1080
VOLUME:
y=the sqr. root of (-2(x)^2-10(x)+48) inbetween 1 and -2
so graph and then put take the equation they give you into the integral and square it
PI times the integral of (-2(x)^2-10(x)+48) dx
PI [(-2/3)(x)^3-5(x)2+48(x)] of 1 and -2
f(1)=(127/3) f(-2)=(-332/3)
ANSWER: (153 PI)
VOLUME BY DISKS
so you know that the formula is pi times the integral of the [function given] squared times dx. well then you gotta know what to do right 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...oh, REMEMBER TO GRAPH..
just look at what they give you...they're should be numbers that they want them inbetween
IF NOT..set the two equations equal to find the bounds
VOLUME BY WASHERS
so you know that 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 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. REMEMBER TO GRAPH!
hope this helps...BECAUSE IT SURELY HELPED ME!!!
~ElliE~
This week is the first week that we actually took an entire AP Exam and then we found out what we would have if that was the real test. So I'll go over things that need to be remembered as well as the things that I needed to look up and know because I FORGOT THEM during the test.
Things to Remember:
If the top exponent is greater than the bottom it’s the limit as it approaches infinity, and if the top exponent is less than the bottom it’s the limit as it approaches zero!
DON’T GET THEM CONFUSED LIKE I DID!
Here’s a trick:
Remember if you’re divided a number by another number and you have a bigger number on top it’s usually not zero. Also, if you’re dividing a number by another number and you have a smaller number on top it’s usually not a big number. Therefore, when the bigger number is on top since it’s a bigger number than zero it’s INFINITY and when the bigger number is on the bottom that means that it’s being lessened so it’s ZERO!
JUST A REMINDER if it's for the x-axis then you solve for y and if it's on the y-axis then you solve for x..and if you plug the y-axis one's in your calculator to graph then you'll have to turn your calculator sideways with the screen on the right to see how it would really look.
AREA:
it's worked the same just they'll give you two equations and if you don't have numbers for the inbetween then you'll set them equal to get them, but if you do then don't worry and keep going. for area there is no pi and no squaring. so you put the integral of the first equation minus the second equation and simplify and solve. then take the integral of it and plug in the numbers inbetween!
AREA:
y=-4x^2+41x+94 AND y=x-2 inbetween 7 and 1
so graph and then put the top equation over the bottom equation
the integral of (-4x^2+41x+94)-(x-2)dx
simplify: the integral of (-4x^2+40x+96) dx
((-4/3)(x)^3+20(x)^2+96(x)) of 1 and 7
f(7)=(3584/3) f(1)=(344/3)
ANSWER: 1080
VOLUME:
y=the sqr. root of (-2(x)^2-10(x)+48) inbetween 1 and -2
so graph and then put take the equation they give you into the integral and square it
PI times the integral of (-2(x)^2-10(x)+48) dx
PI [(-2/3)(x)^3-5(x)2+48(x)] of 1 and -2
f(1)=(127/3) f(-2)=(-332/3)
ANSWER: (153 PI)
VOLUME BY DISKS
so you know that the formula is pi times the integral of the [function given] squared times dx. well then you gotta know what to do right 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...oh, REMEMBER TO GRAPH..
just look at what they give you...they're should be numbers that they want them inbetween
IF NOT..set the two equations equal to find the bounds
VOLUME BY WASHERS
so you know that 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 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. REMEMBER TO GRAPH!
hope this helps...BECAUSE IT SURELY HELPED ME!!!
~ElliE~
Monday, April 12, 2010
Post #34
alright, spring break is over, and it's back to school!...realizing that we only have a few days left i'm beginning to panic over all of the projects that we have due!!!
OH EM GEE!!
so my spring break was good..i can't complain..and i'm missin it already, but here's what we should know for Calculus:
Monday thru Thursday we will be taking AP Tests and we are not allowed to do Corrections for a perfect score, because corrections will be a grade in themselves!..GOOD LUCK TO EVERYONE!
so lets go over a few things:
LINERAZATION
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
LIMIT RULES ARE A MUST !!!! WE SEE THEM ON EVERYTHING!!!
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.
RELATED RATES SHOW UP A LOT TOO..AND THEY TIE INTO A LOT OF THINGS AS WELL:
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
not to mention,.....ANGLE OF ELEVATION:
I'm not very good at those so can anyone explain that
also...OPTIMIZATION:
not so great at those either...HELP please!!
~ElliE~
OH EM GEE!!
so my spring break was good..i can't complain..and i'm missin it already, but here's what we should know for Calculus:
Monday thru Thursday we will be taking AP Tests and we are not allowed to do Corrections for a perfect score, because corrections will be a grade in themselves!..GOOD LUCK TO EVERYONE!
so lets go over a few things:
LINERAZATION
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
LIMIT RULES ARE A MUST !!!! WE SEE THEM ON EVERYTHING!!!
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.
RELATED RATES SHOW UP A LOT TOO..AND THEY TIE INTO A LOT OF THINGS AS WELL:
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
not to mention,.....ANGLE OF ELEVATION:
I'm not very good at those so can anyone explain that
also...OPTIMIZATION:
not so great at those either...HELP please!!
~ElliE~
Sunday, April 11, 2010
Post 34
Tangent lines:
The problem will give you a function and an x value. Sometimes they may give you a y value; if not then you plug the x value into the original function and solve for y to get the y value. Next, you take the derivative of the function and plug in x to get the slope. After that, you plug everything into point-slope form.
First derivative test:
For the first derivative test, you are solving for max and mins and may be trying to see where the graph is increasing and decreasing. You take the derivative of the function and and set it equal to zero and solve for the x values (critical points). Then you set those points up into intervals between negative infinity and infinity. Then, you plug in numbers between those intervals to see if it is positive or negative.
Second derivative test:
For the second derivative test, you are solving to see whether the graph is concave up, concave down, or where there is a point of inflection in the graph. You take the derivative of the function twice and set it equal to zero and solve for the x values. You set those values up into intervals between negative infinity and infinity. You then plug in numbers between those intervals to see if it is positive or negative. If it is positive, it is concave up. If it is negative it is concave down. Where there is a change in concavity, there is a point of inflection.
What I don't understandd:
The problems where they give you a graph and you have to seperate them into triangles and rectangles to find the area.
Integrals with trig functions in them
Theproblems where they give you two equations and you have two variables (most of the time a and b) and you have to solve for one and solve for the other and then set equal.
The problem will give you a function and an x value. Sometimes they may give you a y value; if not then you plug the x value into the original function and solve for y to get the y value. Next, you take the derivative of the function and plug in x to get the slope. After that, you plug everything into point-slope form.
First derivative test:
For the first derivative test, you are solving for max and mins and may be trying to see where the graph is increasing and decreasing. You take the derivative of the function and and set it equal to zero and solve for the x values (critical points). Then you set those points up into intervals between negative infinity and infinity. Then, you plug in numbers between those intervals to see if it is positive or negative.
Second derivative test:
For the second derivative test, you are solving to see whether the graph is concave up, concave down, or where there is a point of inflection in the graph. You take the derivative of the function twice and set it equal to zero and solve for the x values. You set those values up into intervals between negative infinity and infinity. You then plug in numbers between those intervals to see if it is positive or negative. If it is positive, it is concave up. If it is negative it is concave down. Where there is a change in concavity, there is a point of inflection.
What I don't understandd:
The problems where they give you a graph and you have to seperate them into triangles and rectangles to find the area.
Integrals with trig functions in them
Theproblems where they give you two equations and you have two variables (most of the time a and b) and you have to solve for one and solve for the other and then set equal.
post 34
well dang, i forgot to do my blog last weekend, stupid spring break... well anyways. on to this blog...
substitution:
Substitution takes the place of the derivative rules for problems with a quotient rule and product rule, substitution has a few steps.
1. find u by looking inside the parentheses inside the problem
2. take the derivative of u to find du
3. go into the origional problem and switch out (substitute) the stuff
4. integrate
5. plug in
ok, so ln i kinda understand, ln integration happens when the bottom of a fraction is the u and the top is the derivative of u. so since the derivative of ln(x) is 1/x, the answer is just ln(u)+c like everytime.
e^x integration is pretty decent too, all you have to do for this kind of integration is set your u equal to the exponent of the e. So the derivative of e^x is just e^x times the derivative of x.
substitution:
Substitution takes the place of the derivative rules for problems with a quotient rule and product rule, substitution has a few steps.
1. find u by looking inside the parentheses inside the problem
2. take the derivative of u to find du
3. go into the origional problem and switch out (substitute) the stuff
4. integrate
5. plug in
ok, so ln i kinda understand, ln integration happens when the bottom of a fraction is the u and the top is the derivative of u. so since the derivative of ln(x) is 1/x, the answer is just ln(u)+c like everytime.
e^x integration is pretty decent too, all you have to do for this kind of integration is set your u equal to the exponent of the e. So the derivative of e^x is just e^x times the derivative of x.
34th Post
First off, i hope everyone had a great Easter break. Let's recap on some things before we get back into the swing of school.
First derivative test: You have to take the derivative of the function and set it equal to zero. Then you solve for the x values (critical points) and set them up into intervals between negative infinity and infinity. Then you plug in values between those intervals into the first derivative to find max or mins or if the graph is increasing or decreasing.
Second derivative test: You take the derivative of the function twice and set it equal to zero. You solve for the x values and set them up into intervals between negative infinity and infinity. You plug in numbers between those intervals into the second derivative to see where the graph is concave up, concave down, or where there is a point of inflection.
Tangent lines: You will be given a function and a x value. If no y value is given, plug the x value into the original function to find the y value. Then take the derivative of the function and plug in the x value to solve for the slope. Then you put everything into point-slope form (y-y1=slope(x-x1).
Limit Rules:
If the degree on the top is bigger than the degree on the bottom, the limit is infinity
If the degree on the top is smaller than the degree on the bottom, the limit is zero.
If the degree on the top is the same as the degree on the bottom, you divide the coefficients to get the limit.
Things i still do not understand:
optimization
angles of elevation
related rates
particle problems
inverse of f problems
Have a great week everyone :)
First derivative test: You have to take the derivative of the function and set it equal to zero. Then you solve for the x values (critical points) and set them up into intervals between negative infinity and infinity. Then you plug in values between those intervals into the first derivative to find max or mins or if the graph is increasing or decreasing.
Second derivative test: You take the derivative of the function twice and set it equal to zero. You solve for the x values and set them up into intervals between negative infinity and infinity. You plug in numbers between those intervals into the second derivative to see where the graph is concave up, concave down, or where there is a point of inflection.
Tangent lines: You will be given a function and a x value. If no y value is given, plug the x value into the original function to find the y value. Then take the derivative of the function and plug in the x value to solve for the slope. Then you put everything into point-slope form (y-y1=slope(x-x1).
Limit Rules:
If the degree on the top is bigger than the degree on the bottom, the limit is infinity
If the degree on the top is smaller than the degree on the bottom, the limit is zero.
If the degree on the top is the same as the degree on the bottom, you divide the coefficients to get the limit.
Things i still do not understand:
optimization
angles of elevation
related rates
particle problems
inverse of f problems
Have a great week everyone :)
post 34
spring break was fun. now back to school.
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.
Finding absolute max/min
1. First derivative test
2. Plug critical values into the origonal function to get y-values
3. Plug endpoints into the origional function to get y-values
4. The highest y-value is the absolute maximum
5. The lowest y-value is the absolute minimum
*Remember that absolute maximums or minimums are written as a point, or simply as the y-value
First/second derivative tests.
1. Take a derivative
2. Set it equal to 0
3. Solve for x to find max, min, horizontal tangents, extrema (critical points)
4. Set up intervals using step 3.
5. Plug in first derivate
6. To find an absolute max/min plug values from #5 into original function
7. Check endpoints
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
tangent lines and me dont mix for some reason
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.
Finding absolute max/min
1. First derivative test
2. Plug critical values into the origonal function to get y-values
3. Plug endpoints into the origional function to get y-values
4. The highest y-value is the absolute maximum
5. The lowest y-value is the absolute minimum
*Remember that absolute maximums or minimums are written as a point, or simply as the y-value
First/second derivative tests.
1. Take a derivative
2. Set it equal to 0
3. Solve for x to find max, min, horizontal tangents, extrema (critical points)
4. Set up intervals using step 3.
5. Plug in first derivate
6. To find an absolute max/min plug values from #5 into original function
7. Check endpoints
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
tangent lines and me dont mix for some reason
posting...#34
Easter holidays are over but we barley have any school left only about 22 days left for me so im so exited.
Implicit derivatives are pretty easy.
1.Take the derivative of both sides like you would normally do
2. Everytime the derivative of y is taken it needs to be notated with either y ' or dy/dx
3. Solve for dy/dx or y ' as if you are solving for x.
Linearization: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
Implicit derivatives are pretty easy.
1.Take the derivative of both sides like you would normally do
2. Everytime the derivative of y is taken it needs to be notated with either y ' or dy/dx
3. Solve for dy/dx or y ' as if you are solving for x.
Linearization: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
Related RAtes1.Identify all variables
2. Identify what you are looking for
3. Sketch & label that graph
4. write an equation using all of the variables
5. Take the derivative of this equation
6. Substitute everything back in
7. Solve
2. Identify what you are looking for
3. Sketch & label that graph
4. write an equation using all of the variables
5. Take the derivative of this equation
6. Substitute everything back in
7. Solve
Even though i know the steps to related rates i do not know how to work them.
Post #34
ptimization
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))
Posts 33 and 34
a few things to rememberrrrrrrr
limit rules:
if the degree on the top is bigger than the degree on the bottom the limit is infinity
if the degree on the top is smaller than the degree on the bottom the limit is zero
if the degree on the top is the same as the degree on the bottom divide the coefficients to find the limit
rules with derivatives:
chain Rule
you start from outside in. let's say you have cos(2x)
first take derivative of cos
then of multiply that by 2x
product Rule
copy the first x derivative of the second + copy the second x derivative of the first
quotient Rule
copy the bottom x derivative of the top - copy the top x derivative of the bottom/bottom ^2
average value:
of f(x)=1/x from x=1 to x=e is
1/e-1[ln (absoulte value of e)- ln (absoulte value of 1)]
1/e-1[1-0]
1/e-1
and implicit derivatives:
1. take the derivative of both sides like you would normally do
2. everytime the derivative of y is taken it needs to be notated with either y^1 or dy/dx
3. solve for dy/dx or y^1 as if you are solving for x
derivatives of integrals:
F(x)= integral from 0 to x^2 sin(t)
sin(x^2)(2x) *the 2x is the derivative of the bounds
and my favorite optimization:
1 Identify primary and secondary equations your primary is the one your or maximizing or minimizing and your secondary is the other equation
2. Solve for your secondary variable and plug into your primary equation if your primary only has on variable this isn’t necessary
3. Plug into secondary equation to find the other value check your end points if necessary
finally fromulas that we all forget with our derivatives:
d/dx c=0 (c is a #)
d/dx cu=cu' (c is #)
d/dx cx=c (c is a #)
d/dx u+v=u'+v'
d/dx uv=uv'+vu'
d/dx u/v=(vu'-uv')/v^2
d/dx sinx=cosx(x')
d/dx cosx=-sinx(x')
d/dx tanx=sec^2x(x')
d/dx secx=secxtanx(x')
d/dx cscx=-cscxtanx(x')
d/dx cotx=-csc^2x(x')
d/dx lnu= 1/u(u')
d/dx e^u=e^u(u')
limit rules:
if the degree on the top is bigger than the degree on the bottom the limit is infinity
if the degree on the top is smaller than the degree on the bottom the limit is zero
if the degree on the top is the same as the degree on the bottom divide the coefficients to find the limit
rules with derivatives:
chain Rule
you start from outside in. let's say you have cos(2x)
first take derivative of cos
then of multiply that by 2x
product Rule
copy the first x derivative of the second + copy the second x derivative of the first
quotient Rule
copy the bottom x derivative of the top - copy the top x derivative of the bottom/bottom ^2
average value:
of f(x)=1/x from x=1 to x=e is
1/e-1[ln (absoulte value of e)- ln (absoulte value of 1)]
1/e-1[1-0]
1/e-1
and implicit derivatives:
1. take the derivative of both sides like you would normally do
2. everytime the derivative of y is taken it needs to be notated with either y^1 or dy/dx
3. solve for dy/dx or y^1 as if you are solving for x
derivatives of integrals:
F(x)= integral from 0 to x^2 sin(t)
sin(x^2)(2x) *the 2x is the derivative of the bounds
and my favorite optimization:
1 Identify primary and secondary equations your primary is the one your or maximizing or minimizing and your secondary is the other equation
2. Solve for your secondary variable and plug into your primary equation if your primary only has on variable this isn’t necessary
3. Plug into secondary equation to find the other value check your end points if necessary
finally fromulas that we all forget with our derivatives:
d/dx c=0 (c is a #)
d/dx cu=cu' (c is #)
d/dx cx=c (c is a #)
d/dx u+v=u'+v'
d/dx uv=uv'+vu'
d/dx u/v=(vu'-uv')/v^2
d/dx sinx=cosx(x')
d/dx cosx=-sinx(x')
d/dx tanx=sec^2x(x')
d/dx secx=secxtanx(x')
d/dx cscx=-cscxtanx(x')
d/dx cotx=-csc^2x(x')
d/dx lnu= 1/u(u')
d/dx e^u=e^u(u')
33 and 34?
Well we have had a week off! But I guess it's back to school, so....
REMEMBER:
position
velocity
acceleration
A particle's position is given by s=t^3 - 6t^2 + 9t. What is its acceleration at time t=4? **When going down (from position to acceleration) you take the derivative. Then simpley plug in 4.
3t^2-12t+9
6t-12
6(4)-12= 12
REMEMBER:
related rates; sometimes we forget the steps
1. Identify all variables
2. Identify what you are looking for
3. Sketch & label that graph
4. write an equation using all of the variables
5. Take the derivative of this equation
6. Substitute everything back in
7. Solve
REMEMBER:
formulas; these problems are usually easy when you remember the formula
delta x=b-a/number of subintervals
LRAM-Left hand approximation= delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation= delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation= delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
volume of disks is S (top)^2-(bottom)^2 dx
area of washers is S (top)-(bottom)
REMEMBER:
implicit derivatives; has an = sign
y^3+y^2-5y-x^2=-4
1. take derivative of both sides:
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.
REMEMBER:
key word stuff; when you see a word you should immeditaly think..
1. linearization-->equation of a tangent line
2. if given a table(the sets of points)-->dealing with slope or rram/lram/ect.
3. how many-->integrate
4.(in calculator part)volume problems-->find intersection
AND REMEMBER:
slope field stuff; we are fixing to start with more free response
I'm just going to review how to draw it.
positive slopes is /
negative slope is \
for a zero slope is a horizontal line
for an undefined slope is a vertical line
*Oh and we take a real ap test tomorrow! I almost forgot..good luck.
REMEMBER:
position
velocity
acceleration
A particle's position is given by s=t^3 - 6t^2 + 9t. What is its acceleration at time t=4? **When going down (from position to acceleration) you take the derivative. Then simpley plug in 4.
3t^2-12t+9
6t-12
6(4)-12= 12
REMEMBER:
related rates; sometimes we forget the steps
1. Identify all variables
2. Identify what you are looking for
3. Sketch & label that graph
4. write an equation using all of the variables
5. Take the derivative of this equation
6. Substitute everything back in
7. Solve
REMEMBER:
formulas; these problems are usually easy when you remember the formula
delta x=b-a/number of subintervals
LRAM-Left hand approximation= delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation= delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation= delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
volume of disks is S (top)^2-(bottom)^2 dx
area of washers is S (top)-(bottom)
REMEMBER:
implicit derivatives; has an = sign
y^3+y^2-5y-x^2=-4
1. take derivative of both sides:
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.
REMEMBER:
key word stuff; when you see a word you should immeditaly think..
1. linearization-->equation of a tangent line
2. if given a table(the sets of points)-->dealing with slope or rram/lram/ect.
3. how many-->integrate
4.(in calculator part)volume problems-->find intersection
AND REMEMBER:
slope field stuff; we are fixing to start with more free response
I'm just going to review how to draw it.
positive slopes is /
negative slope is \
for a zero slope is a horizontal line
for an undefined slope is a vertical line
*Oh and we take a real ap test tomorrow! I almost forgot..good luck.
Post 34
Well I am going to talk about some more old stuff to refresh some memories for the AP test.
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
The Riemann sum approximates the area using the rectangles or trapezoids. The Riemanns Sums are:
LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
delta x=b-a/number of subintervals
Also I am going to talk about taking implicit derivatives. The steps for taking implicit derivatives are:
1. Take the derivative of both sides like you would normally do
2. Everytime the derivative of y is taken it needs to be notated with either y ' or dy/dx
3. Solve for dy/dx or y ' as if you are solving for x.
Can someone refrest me on the mean value theorem?
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
The Riemann sum approximates the area using the rectangles or trapezoids. The Riemanns Sums are:
LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
delta x=b-a/number of subintervals
Also I am going to talk about taking implicit derivatives. The steps for taking implicit derivatives are:
1. Take the derivative of both sides like you would normally do
2. Everytime the derivative of y is taken it needs to be notated with either y ' or dy/dx
3. Solve for dy/dx or y ' as if you are solving for x.
Can someone refrest me on the mean value theorem?
post 33
Well the holidays are over and its time to get ready for the AP test so I will start by going over some old things.
The steps for the first derivative test are:
1. Take the derivative of the original function
2. Solve for x (the values will be your critical values)
3. Set those values up into intervals between negative infinity and infinity
4. Plug in numbers between the intervals into the function
5. This will show you when the function is increasing, decreasing, and you will find max's and mins
The limit rules are:
1. if the degree of the top is bigger than the degree of the bottom, the limit is infinity.
2. if the degree of the top is smaller than the degree of the bottom, the limit is 0.
3. if the degree of the top is equal to the degree of the bottom, the limit is the coefficient of the leading term of the top divided by the coefficient of the leading term of the bottom equation.
The steps for optimization are:
1. Identify all quantities
2. Write an equation
3. Reduce equation
4. Determine domain of equation
5. Determine max/min values
The steps to find maximums and minimums are:
1. First derivative test
2. Plug critical values into the original function to get y-values
3. Plug endpoints into the original function to get y-values
4. The highest y-value is the absolute maximum
5. The lowest y-value is the absolute minimum
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
Ash's 34th Post
So, goodbye spring break, hello school =/
I'm so confused for some reason so I guess I'll go over some pretty broad topics
First Derivative:
Step One: Just take the derivative using your derivative formulas
Step Two: Simplify!
Example: -3x^2-4
-6x
Second Derivative:
Step One: Take the first derivative
Step Two: Take the derivative again
Step Three: Simplify
Example: -3x^2-4
-6x
-6
Integration:
Step One: Take the opposite of a derivative
Step Two: Simplify!
Example: S(-3x^2-4)
-x^2-4x+C
Definition of a Derivative:
Step One: Remove from the parenthesis the equation to be derived
Step Two: Perform the first derivative test
Step Three: Plug in the number given
Step Four: Simplify!
Example: lim x->O [(3x^2+0)+(3x^2)]\h
(3x^2)
6x
6(0)
0
Blahh...confusion is just too great to ask any questions this time...
BTW: Good luck to everyone on the Practice AP this week!
I'm so confused for some reason so I guess I'll go over some pretty broad topics
First Derivative:
Step One: Just take the derivative using your derivative formulas
Step Two: Simplify!
Example: -3x^2-4
-6x
Second Derivative:
Step One: Take the first derivative
Step Two: Take the derivative again
Step Three: Simplify
Example: -3x^2-4
-6x
-6
Integration:
Step One: Take the opposite of a derivative
Step Two: Simplify!
Example: S(-3x^2-4)
-x^2-4x+C
Definition of a Derivative:
Step One: Remove from the parenthesis the equation to be derived
Step Two: Perform the first derivative test
Step Three: Plug in the number given
Step Four: Simplify!
Example: lim x->O [(3x^2+0)+(3x^2)]\h
(3x^2)
6x
6(0)
0
Blahh...confusion is just too great to ask any questions this time...
BTW: Good luck to everyone on the Practice AP this week!
Post #34
Optimization
The steps are:
1. Determine everything given in the problem and what you are looking to maximize or minimize. This is stated in the problem.
2. Write a primary equation, which is an equation for what you are maximizing or minimizing.
3. Write a secondary equation using the other information given.
4. Solve the secondary equation in terms of one variable (if necessary) then plug that into the primary equation for that variable.
5. Take derivative of the plugged in primary equation, set the equation equal to zero, then solve for the remaining variable.
6. Plug back into the secondary equation to find the other variable.
EXAMPLE:
We need to enclose a field with a fence. We have 500 feet of fencing material and a building on one side of the field so won't need any fencing. Determine the dimensions of the field that will enclose the largest area.
1. P=500 ft, one side does not need fencing, maximizing area (enclose the largest area)
2. Since we are maximizing area, the primary equation will be the area equation for a rectangle. A=xy
3. The secondary equation will be perimeter since that is given in the problem. 500= x+2y
Note: the formula for perimeter is usually 2x+2y, but since one side does not need fencing, it is only x+2y.
4. Solve for one variable: 500=x +2y x= 500 -2y
Now plug into primary equation: A= 500 - 2y (y)
Distribute the y in: 500y - 2y^2
5. Take derivative: 500 - 4y
Set equal to zero and solve for y; 500 -4y = 0 y = 125
6. Plug y into secondary equation to find x: 500 = x + 2y 500 = x+2(125) x= 250
The dimensions of the largest area are 250 X 125.
An easy problem on the AP is the derivative of an integral.
For these problems, just plug in B into the equation and multiply that by the derivative of B.
Example: If f(x) = The integral of (t^2-1)^1/3 dt on the interval [0, x+1], then f'(-4) =
Plug in B: ((x+1)^2 -1 )^1/3 (1)
Next, all you have to do is plug in -4.
((-4+1)^2 -1 ) ^1/3
((-3)^2 -1 ) ^1/3
(9-1)^1/3
8^1/3 = 2
Easy points.
A review on related rates and angle of elevation would be greatly appreciated.
The steps are:
1. Determine everything given in the problem and what you are looking to maximize or minimize. This is stated in the problem.
2. Write a primary equation, which is an equation for what you are maximizing or minimizing.
3. Write a secondary equation using the other information given.
4. Solve the secondary equation in terms of one variable (if necessary) then plug that into the primary equation for that variable.
5. Take derivative of the plugged in primary equation, set the equation equal to zero, then solve for the remaining variable.
6. Plug back into the secondary equation to find the other variable.
EXAMPLE:
We need to enclose a field with a fence. We have 500 feet of fencing material and a building on one side of the field so won't need any fencing. Determine the dimensions of the field that will enclose the largest area.
1. P=500 ft, one side does not need fencing, maximizing area (enclose the largest area)
2. Since we are maximizing area, the primary equation will be the area equation for a rectangle. A=xy
3. The secondary equation will be perimeter since that is given in the problem. 500= x+2y
Note: the formula for perimeter is usually 2x+2y, but since one side does not need fencing, it is only x+2y.
4. Solve for one variable: 500=x +2y x= 500 -2y
Now plug into primary equation: A= 500 - 2y (y)
Distribute the y in: 500y - 2y^2
5. Take derivative: 500 - 4y
Set equal to zero and solve for y; 500 -4y = 0 y = 125
6. Plug y into secondary equation to find x: 500 = x + 2y 500 = x+2(125) x= 250
The dimensions of the largest area are 250 X 125.
An easy problem on the AP is the derivative of an integral.
For these problems, just plug in B into the equation and multiply that by the derivative of B.
Example: If f(x) = The integral of (t^2-1)^1/3 dt on the interval [0, x+1], then f'(-4) =
Plug in B: ((x+1)^2 -1 )^1/3 (1)
Next, all you have to do is plug in -4.
((-4+1)^2 -1 ) ^1/3
((-3)^2 -1 ) ^1/3
(9-1)^1/3
8^1/3 = 2
Easy points.
A review on related rates and angle of elevation would be greatly appreciated.
Post Number Thirty Three and Post Number Thirty Four
These are my blogs for spring break, I think 33 and 34..
Hope everyone had a great Easter!
I finally remembered Mean Value Theorem:
F(x) must be continuous on a closed interval and differentiable on the open interval.
f '(c) = [f(b) - f(a)] / (b-a)
Optimization:
1. Identify all quantities
2. Write an equation
3. Reduce equation
4. Determine domain of equation
5. Determine max/min values
Finding absolute max/min:
1. First derivative test
2. Plug critical values into the original function to get y-values
3. Plug endpoints into the original function to get y-values
4. The highest y-value is the absolute maximum
5. The lowest y-value is the absolute minimum
Volume by disks:
Pi bSa [R(x)]^2 dx
Volume by washers:
pi bSa (top equation)^2 – (bottom equation)^2 dx
After doing the substitution worksheet the other day, I finally feel more comfortable:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
Limit Rules:
1. if the degree of the top is bigger than the degree of the bottom, the limit is infinity.
2. if the degree of the top is smaller than the degree of the bottom, the limit is 0.
3. if the degree of the top is equal to the degree of the bottom, the limit is the coefficient of the leading term of the top divided by the coefficient of the leading term of the bottom equation.
Although I know the steps for related rates I always get mixed up during the problems:
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
PVA
Remember, it goes down. Moving down the list, take a derivative, such as changing from position to velocity. Moving up the list, integrate, as if changing from acceleration to velocity. If going from position to acceleration, take the derivative twice. If going from acceleration to position, integrate twice.
Easy enough, right?
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.
Although I know what to do a lot of times, I have a lot of trouble putting my knowledge to use. I often get mixed up and lost in the middle of problems and don’t know what to do next. Some things I guess I’ll just never understand.
Oh and I won’t be in class tomorrow, so let me know what goes on!
Hope everyone had a great Easter!
I finally remembered Mean Value Theorem:
F(x) must be continuous on a closed interval and differentiable on the open interval.
f '(c) = [f(b) - f(a)] / (b-a)
Optimization:
1. Identify all quantities
2. Write an equation
3. Reduce equation
4. Determine domain of equation
5. Determine max/min values
Finding absolute max/min:
1. First derivative test
2. Plug critical values into the original function to get y-values
3. Plug endpoints into the original function to get y-values
4. The highest y-value is the absolute maximum
5. The lowest y-value is the absolute minimum
Volume by disks:
Pi bSa [R(x)]^2 dx
Volume by washers:
pi bSa (top equation)^2 – (bottom equation)^2 dx
After doing the substitution worksheet the other day, I finally feel more comfortable:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
Limit Rules:
1. if the degree of the top is bigger than the degree of the bottom, the limit is infinity.
2. if the degree of the top is smaller than the degree of the bottom, the limit is 0.
3. if the degree of the top is equal to the degree of the bottom, the limit is the coefficient of the leading term of the top divided by the coefficient of the leading term of the bottom equation.
Although I know the steps for related rates I always get mixed up during the problems:
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
PVA
Remember, it goes down. Moving down the list, take a derivative, such as changing from position to velocity. Moving up the list, integrate, as if changing from acceleration to velocity. If going from position to acceleration, take the derivative twice. If going from acceleration to position, integrate twice.
Easy enough, right?
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.
Although I know what to do a lot of times, I have a lot of trouble putting my knowledge to use. I often get mixed up and lost in the middle of problems and don’t know what to do next. Some things I guess I’ll just never understand.
Oh and I won’t be in class tomorrow, so let me know what goes on!
posts 33 and 34
happy late easter.
Rolle's Theorem
F(x) must be continuous on a closed intervale. It must be differentiable on the open interval. And f(a) must equal f(b). If all this is true, then there is at least one number 'c' within (a,b) such that f'(c) = 0.
Mean Value Theorem
Same as Rolle's such that: must be continous and differentiable.Different because: f '(c) = [f(b) - f(a)] / (b-a).
EVT
a continuous function on a closed interval [a,b] must have both a mnimmum and a maximum on the interval
Optimization(steps from packet)
1. Identify all quantities
2. Write an equation
3. Reduce equation
4. Determine domain of equation
5. Determine max/min values
Finding absolute max/min
1. First derivative test
2. Plug critical values into the origonal function to get y-values
3. Plug endpoints into the origional function to get y-values
4. The highest y-value is the absolute maximum
5. The lowest y-value is the absolute minimum
*Remember that absolute maximums or minimums are written as a point, or simply as the y-value
First/second derivative tests.
1. Take a derivative
2. Set it equal to 0
3. Solve for x to find max, min, horizontal tangents, extrema (critical points)
4. Set up intervals using step 3.
5. Plug in first derivate
6. To find an absolute max/min plug values from #5 into original function
7. Check endpoints
Volume by disks is used to find the volume of a solid object
The formula for this is: pi S a-b [R(x)]^2 dx
Volume by washers is used to find volume of object with a hole
pi S a-b top^2-bottom^2 dx
*you will always have two equations.
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
ANGLE OF ELEVATION:
HELP.
Rolle's Theorem
F(x) must be continuous on a closed intervale. It must be differentiable on the open interval. And f(a) must equal f(b). If all this is true, then there is at least one number 'c' within (a,b) such that f'(c) = 0.
Mean Value Theorem
Same as Rolle's such that: must be continous and differentiable.Different because: f '(c) = [f(b) - f(a)] / (b-a).
EVT
a continuous function on a closed interval [a,b] must have both a mnimmum and a maximum on the interval
Optimization(steps from packet)
1. Identify all quantities
2. Write an equation
3. Reduce equation
4. Determine domain of equation
5. Determine max/min values
Finding absolute max/min
1. First derivative test
2. Plug critical values into the origonal function to get y-values
3. Plug endpoints into the origional function to get y-values
4. The highest y-value is the absolute maximum
5. The lowest y-value is the absolute minimum
*Remember that absolute maximums or minimums are written as a point, or simply as the y-value
First/second derivative tests.
1. Take a derivative
2. Set it equal to 0
3. Solve for x to find max, min, horizontal tangents, extrema (critical points)
4. Set up intervals using step 3.
5. Plug in first derivate
6. To find an absolute max/min plug values from #5 into original function
7. Check endpoints
Volume by disks is used to find the volume of a solid object
The formula for this is: pi S a-b [R(x)]^2 dx
Volume by washers is used to find volume of object with a hole
pi S a-b top^2-bottom^2 dx
*you will always have two equations.
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
ANGLE OF ELEVATION:
HELP.
April 11th blog
So this blog I really don't know what exactly to say...
Something I did think about this week was the thing where you have a cross section...basically it gives you an equation, such as y=x^2+2 and then it says that the cross-section is a square, or some other shape, and it wants you to find the volume. All you do is take the area of the square, or whatever shape...so in this case it is s^2. Then, you plug in your equation for your variable, so it becomes (x^2+2)^2. Then, you integrate this from whatever bounds and you have your volume. The only thing to remember is that if you are doing a square, you will NOT have to times by Pi. Also, they should never give you one with like a rectangle, perhaps, because that has two variables. It will always be in terms of one variable as far as I know...
I really don't know what else to explain anymore...I'm pretty sure I've explained like every concept before...lol.
Limit Rules:
1. if the degree of the top is bigger than the degree of the bottom, the limit is infinity.
2. if the degree of the top is smaller than the degree of the bottom, the limit is 0.
3. if the degree of the top is equal to the degree of the bottom, the limit is the coefficient of the leading term of the top divided by the coefficient of the leading term of the bottom equation.
The above rules are really really easy...they also apply to horizontal asymptotes..
Speaking of, to find vertical asymptotes, you set the bottom equal to 0 and solve for x. (bottom of a fraction).
Removeables are when you factor the top and you factor the bottom and they have a term that is the same and can be canceled out.
Anyway, I don't know what else to write about...
Have a good last day of vacation!
Something I did think about this week was the thing where you have a cross section...basically it gives you an equation, such as y=x^2+2 and then it says that the cross-section is a square, or some other shape, and it wants you to find the volume. All you do is take the area of the square, or whatever shape...so in this case it is s^2. Then, you plug in your equation for your variable, so it becomes (x^2+2)^2. Then, you integrate this from whatever bounds and you have your volume. The only thing to remember is that if you are doing a square, you will NOT have to times by Pi. Also, they should never give you one with like a rectangle, perhaps, because that has two variables. It will always be in terms of one variable as far as I know...
I really don't know what else to explain anymore...I'm pretty sure I've explained like every concept before...lol.
Limit Rules:
1. if the degree of the top is bigger than the degree of the bottom, the limit is infinity.
2. if the degree of the top is smaller than the degree of the bottom, the limit is 0.
3. if the degree of the top is equal to the degree of the bottom, the limit is the coefficient of the leading term of the top divided by the coefficient of the leading term of the bottom equation.
The above rules are really really easy...they also apply to horizontal asymptotes..
Speaking of, to find vertical asymptotes, you set the bottom equal to 0 and solve for x. (bottom of a fraction).
Removeables are when you factor the top and you factor the bottom and they have a term that is the same and can be canceled out.
Anyway, I don't know what else to write about...
Have a good last day of vacation!
A post.
Practice AP questions...
This question asks for you to approximate the value of y at x=3.1, given that the curve x^3+xtan(y)=27 through (3,0).
1. Implicit differentiation.
x^3+xtan(y)=27
3x^2 + xsec^2(y)dy + tan(y) = 0
2. Separate dy from the rest of your function.
3x^2+xsec^2(y)dy+tan(y)=0
xsec^2(y)dy=(-3x^2)-(tan(y))
dy=((-3x^2)-(tan(y)))/(x(sec^2(y)))
3. Plug in your x and y values.
dy=((-3x^2)-(tan(y)))/(x(sec^2(y)))
dy=((-3(3^2))-(tan(0)))/(3(sec^2(0)))
dy=((-3(9))-(0))/(3(1))
dy=(-27)/3
dy=-9
4. Create your equation for the tangent line.
(y-0)=(-9)(x-3)
5. Plug in 3.1
(y-0)=(-9)(x-3)
y=(-9)((3.1)-3)
y=(-9)(.1)
y=(-0.9)
So, no you found your approximated y value.
bye bye.
This question asks for you to approximate the value of y at x=3.1, given that the curve x^3+xtan(y)=27 through (3,0).
1. Implicit differentiation.
x^3+xtan(y)=27
3x^2 + xsec^2(y)dy + tan(y) = 0
2. Separate dy from the rest of your function.
3x^2+xsec^2(y)dy+tan(y)=0
xsec^2(y)dy=(-3x^2)-(tan(y))
dy=((-3x^2)-(tan(y)))/(x(sec^2(y)))
3. Plug in your x and y values.
dy=((-3x^2)-(tan(y)))/(x(sec^2(y)))
dy=((-3(3^2))-(tan(0)))/(3(sec^2(0)))
dy=((-3(9))-(0))/(3(1))
dy=(-27)/3
dy=-9
4. Create your equation for the tangent line.
(y-0)=(-9)(x-3)
5. Plug in 3.1
(y-0)=(-9)(x-3)
y=(-9)((3.1)-3)
y=(-9)(.1)
y=(-0.9)
So, no you found your approximated y value.
bye bye.
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