Well another week down in calculus lol.. let's go over some easy stuff first:
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.
Some things i do not understand:
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
Those problems 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.
Hope you all have a great weekend :)
Saturday, March 6, 2010
Friday, March 5, 2010
Post something
Okay so for my blog this weekend...I don't really know what to post.. My mind is so boggled down atm from doing so many things at once today :o
Anyway, some tips and help on a few things...
If they give you a graph...that CLEARLY consists of triangle and rectangles...and it says find the area under the curve from 0 to some number...All you do is split it up into triangles and rectangle...find the area of each rectangle and triangle (remember, rectangle is length times width and triangle is 1/2 times length times width) and then add all the numbers on the top and subtract any numbers from below the axis from that...and there you go, you just got easy points ;-).
Also, don't forget to be using your trig identities...for example
The problem on the test was some integral of
sin(2x)e^(sin^2(x))
Most of you set your u to sin^2(x) and took the derivative as 2sinxcosx....but you all stopped there! Don't stop, 2sinxcosx is actually in the problem. If you will, jump back into the time space continuum and remember the trig identities test and think...okay...sin...double angle formula...OH, i know. sin(2x)=2sinxcosx. So the whole time it was already there. So you can now proceed on with the problem e^u, integrate it and do whatever definite integral it wants you to do. do not forget to use trig identities.
Also, people people people. It gives you a function...and it asks where is the slope of the tangent line equal to 6...what do you do? Just take the derivative and set equal to 6 and solve...that one was a no-brainer that a LOT of you missed.
Let's see...what else...oh, yeah.
Instantaneous speed: take a derivative, plug in the number. Simple as that. Don't miss this please--this is like zomgfreemonies except its points...and it's on the AP.
Once again...I can not reiterate enough the following:
Original - Position
1st Derivative - Velocity
2nd Derivative - Acceleration
Also, displacement refers to the area under the position curve...so if it gives you a formula for displacement, i am pretty sure you take the derivative of that to get to position...so that will change things.
Original - Displacement
1st Derivative - Position
2nd Derivative - Velocity
3rd Derivative - Acceleration
4th Derivative - Jerk
and so on and so forth....so just remember where you are in the line of things and you will be fine by integrating or deriving... (zomg we integrate and take derivatives? crazy ... i didn't think calculus did that)
Anyway, I'm tired of typing and slightly delirious if you haven't noticed...
ZzZzZZZZzzzzzzzzz............
Anyway, some tips and help on a few things...
If they give you a graph...that CLEARLY consists of triangle and rectangles...and it says find the area under the curve from 0 to some number...All you do is split it up into triangles and rectangle...find the area of each rectangle and triangle (remember, rectangle is length times width and triangle is 1/2 times length times width) and then add all the numbers on the top and subtract any numbers from below the axis from that...and there you go, you just got easy points ;-).
Also, don't forget to be using your trig identities...for example
The problem on the test was some integral of
sin(2x)e^(sin^2(x))
Most of you set your u to sin^2(x) and took the derivative as 2sinxcosx....but you all stopped there! Don't stop, 2sinxcosx is actually in the problem. If you will, jump back into the time space continuum and remember the trig identities test and think...okay...sin...double angle formula...OH, i know. sin(2x)=2sinxcosx. So the whole time it was already there. So you can now proceed on with the problem e^u, integrate it and do whatever definite integral it wants you to do. do not forget to use trig identities.
Also, people people people. It gives you a function...and it asks where is the slope of the tangent line equal to 6...what do you do? Just take the derivative and set equal to 6 and solve...that one was a no-brainer that a LOT of you missed.
Let's see...what else...oh, yeah.
Instantaneous speed: take a derivative, plug in the number. Simple as that. Don't miss this please--this is like zomgfreemonies except its points...and it's on the AP.
Once again...I can not reiterate enough the following:
Original - Position
1st Derivative - Velocity
2nd Derivative - Acceleration
Also, displacement refers to the area under the position curve...so if it gives you a formula for displacement, i am pretty sure you take the derivative of that to get to position...so that will change things.
Original - Displacement
1st Derivative - Position
2nd Derivative - Velocity
3rd Derivative - Acceleration
4th Derivative - Jerk
and so on and so forth....so just remember where you are in the line of things and you will be fine by integrating or deriving... (zomg we integrate and take derivatives? crazy ... i didn't think calculus did that)
Anyway, I'm tired of typing and slightly delirious if you haven't noticed...
ZzZzZZZZzzzzzzzzz............
Post #29
Another week down...
Let’s see what I remember shall we.. if you take the derivative of f^2(x) you treat it as if it was like sine. You do 2f(x) times f’(x) times 1. So it’s almost as if you have (f(x))^2 and you use the chain rule. Bring the 2 to the front, subtract 1, times by the derivative of the inside. And since the derivative of f(x) is f’(x) times the derivative of x, which is 1. These rules seem to help me A LOT!
Lets say you have:
x^2-2 DIVIDED BY 4-x^2 AS THE LIMIT APPROACHES 2
soo..
for this you end up getting two divided by zero. For finite limits though, if you get a number over zero, it’s always infinity. Therefore the answer is INFINITY!
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!
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..WHAT DO YOU DO???
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!
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!
examples:
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)
hope this helps...
~ElliE~
Let’s see what I remember shall we.. if you take the derivative of f^2(x) you treat it as if it was like sine. You do 2f(x) times f’(x) times 1. So it’s almost as if you have (f(x))^2 and you use the chain rule. Bring the 2 to the front, subtract 1, times by the derivative of the inside. And since the derivative of f(x) is f’(x) times the derivative of x, which is 1. These rules seem to help me A LOT!
Lets say you have:
x^2-2 DIVIDED BY 4-x^2 AS THE LIMIT APPROACHES 2
soo..
for this you end up getting two divided by zero. For finite limits though, if you get a number over zero, it’s always infinity. Therefore the answer is INFINITY!
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!
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..WHAT DO YOU DO???
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!
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!
examples:
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)
hope this helps...
~ElliE~
Thursday, March 4, 2010
Tangent Lines
Ok someone asked about this and I couldn't find the blog again... These are what we consider "gimme" questions. They usually don't have complicated derivatives, etc. In an non-implicit problem. You take the derivative of the equation given and then plug in the x-value given to get the slope. Then you plug the slope and x-value into point slope. You will need the y-value to finish the point-slope equation. If it is not given you find it by plugging the x-value into the ORIGINAL equation and solving for y. See several of the problems we have done in earlier APS. We had one in the diagnostic test I believe.
Integral of sin^2(x)
Ok so I am a little disappointed to hear that people are using the chain rule in integration. We have NEVER used the chain rule in integration. Your options are substitution and that is all. You can't substitute in this problem because the derivative of sin(x) is not in the integral. However, it is a trig function so you can use identities to manipulate it. For lack of being able to type mathematically here is a link to it already done.
http://wiki.answers.com/Q/Integral_of_sin_squared_x
Monday, March 1, 2010
post 28
Blogs.
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
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*
SUBSTITUTION:
1. Find u & du
2. set u = whatever isn't the derivative
3. take the derivative of u
4. substitute back in
e integration:
very simple.
whatever e is raised to is your u, the derivative of u is du.
:)
IMPLICIT DERIVATIVE:
take derivative like normal, of both sides of the equation.
any derivative taken for y, mark with dy/dx behind it.
solve for dy/dx
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
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*
SUBSTITUTION:
1. Find u & du
2. set u = whatever isn't the derivative
3. take the derivative of u
4. substitute back in
e integration:
very simple.
whatever e is raised to is your u, the derivative of u is du.
:)
IMPLICIT DERIVATIVE:
take derivative like normal, of both sides of the equation.
any derivative taken for y, mark with dy/dx behind it.
solve for dy/dx
post 28
This past week all we did was another set of AP tests and corrections.
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
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.
One problem am having is integrating something such as sin^2(x).
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
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.
One problem am having is integrating something such as sin^2(x).
post 28
all we did this last week was take an AP test, and then make corrections on it. and this three day weekend thing screwed me up, so i'm a day late on my blog, but anyways, i'm going to do this one on implicit derivatives because that's a super easy topic. :)
Implicit derivatives involve both x's and y's, unlike normal derivatives.
1: So, first you have to take the derivative of whatever they give you as you normally would.
2: Whenever you take the derivative of y, you have to note it with dy/dx.
3: Solve for dy/dx
(if you want to find slope plug in an x and y value)
example: y^3+y^2-5y-x^2=-4
First you just take the derivative, but don't forget to not the derivatives of the y's! So you get: 3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0
Then you have to solve for dy/dx, so you get:
dy/dx(3y^2+2y-5)=2x which then is further solved for to get dy/dx=2x/(3y^2+2y-5)
and that's it for that problem, it's done.
Another example:
Okay, let's say you want to find the slope of 3(x^2+y^2)^2=100xy at the point (3,1)
First you take the derivative, which involves all kinds of product and exponent rule...
6(x^2+y^2)(2x+2y(dy/dx))=100(y+x(dy/dx))
then, you need to foil it n stuff, so you get:
12x^3+12x^2(dy/dx)+12xy^2+12y^2(dy/dx))=100y+100x(dy/dx)
then, as usual, you would have to solve for dy/dx:
dy/dx=(-12^3-12xy^2+100y)/(12x^2+12y-100x)
after you solved for dy/dx, you plug in your x and y value from the point given to get your answer, so I think the final answer would be: 1.84 (if I put it in my calculator right)
that is it for implicit derivatives, and they are really easy to identify, it is the exact same thing as a derivative pretty much, just with x's and y's. Just don't forget to plug in the point that some problems will give you at the end, I have forgotten to do it before.
One thing that i still screw up on is MRAM, i don't know what my problem is, even when i look at the formula i still can't solve the problem, so if someone could review that, it would be good.
Implicit derivatives involve both x's and y's, unlike normal derivatives.
1: So, first you have to take the derivative of whatever they give you as you normally would.
2: Whenever you take the derivative of y, you have to note it with dy/dx.
3: Solve for dy/dx
(if you want to find slope plug in an x and y value)
example: y^3+y^2-5y-x^2=-4
First you just take the derivative, but don't forget to not the derivatives of the y's! So you get: 3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0
Then you have to solve for dy/dx, so you get:
dy/dx(3y^2+2y-5)=2x which then is further solved for to get dy/dx=2x/(3y^2+2y-5)
and that's it for that problem, it's done.
Another example:
Okay, let's say you want to find the slope of 3(x^2+y^2)^2=100xy at the point (3,1)
First you take the derivative, which involves all kinds of product and exponent rule...
6(x^2+y^2)(2x+2y(dy/dx))=100(y+x(dy/dx))
then, you need to foil it n stuff, so you get:
12x^3+12x^2(dy/dx)+12xy^2+12y^2(dy/dx))=100y+100x(dy/dx)
then, as usual, you would have to solve for dy/dx:
dy/dx=(-12^3-12xy^2+100y)/(12x^2+12y-100x)
after you solved for dy/dx, you plug in your x and y value from the point given to get your answer, so I think the final answer would be: 1.84 (if I put it in my calculator right)
that is it for implicit derivatives, and they are really easy to identify, it is the exact same thing as a derivative pretty much, just with x's and y's. Just don't forget to plug in the point that some problems will give you at the end, I have forgotten to do it before.
One thing that i still screw up on is MRAM, i don't know what my problem is, even when i look at the formula i still can't solve the problem, so if someone could review that, it would be good.
Ash's 28th Post
So it just hit me that it's Monday and not Sunday...I hate that.
Anyway, I'm going to explain some AP questions that I got right! Yippee!!
Non-Calculator Portion:
1. If g(x) = 1/32x^4-5x^2, find g'(4).
First step: take the derivative
1/8x^3-10x
Second Step: Plug in 4
1/8(4)^3-10(4)
Third Step: Solve
Answer = -32
5. Evaluate lim h->0 [5(1/2+h)^4-5(1/2)^4]/h
First Step: This is a definition of a derivative, so you take the derivative of 5(1/2)^4
4(5)(1/2)^3
Second Step: Solve
Answer = 5/2
Calculator Portion:
34. The graph of y=x^3-2x^2-5x+2 has a local maximum at:
First Step: Plug into your calculator
Second step: 2nd->Calc
Third Step: go to maximum
Forth Step: enter your bounds and then a "guess"
Answer: (-.786,4.209)
39. Find two non-negative numberfs x and y whose sum is 100 and for which (x^2)(y) is a maximum.
First step: Add up answer choices to make sure all equal 100
Second Step: Plug the x and y values into the given formula and see which is the highest number
Answer: x=66.667 and y=33.333
QUESTIONS:
How do you find the domain of a function without a calculator? I've always had trouble with domain and range even in Advanced Math.
How do you work piecewises? Number 7 is an example on the Non-calculator portion: Find k so that f(x) {(x^2-16)/x-4; x *does not equal* 4 and k; x=4 is continuous for all x. a) all real values, b)0, c)16, d)8, e) no real values
How do you take the derivative of a trig function squared? (sin^2(x) or cos^2(x))
Can someone go over the Mean Value Theorem for Derivatives? Number 12: Find a positive value c, for x, that satisfies the conclusion of the MVTD for f(x)=3x^2-5x+1 on [2,5] a)1, b)13/6, c)11/6, d)23/6, e)7/2
Hope everyone enjoys their day off!
Anyway, I'm going to explain some AP questions that I got right! Yippee!!
Non-Calculator Portion:
1. If g(x) = 1/32x^4-5x^2, find g'(4).
First step: take the derivative
1/8x^3-10x
Second Step: Plug in 4
1/8(4)^3-10(4)
Third Step: Solve
Answer = -32
5. Evaluate lim h->0 [5(1/2+h)^4-5(1/2)^4]/h
First Step: This is a definition of a derivative, so you take the derivative of 5(1/2)^4
4(5)(1/2)^3
Second Step: Solve
Answer = 5/2
Calculator Portion:
34. The graph of y=x^3-2x^2-5x+2 has a local maximum at:
First Step: Plug into your calculator
Second step: 2nd->Calc
Third Step: go to maximum
Forth Step: enter your bounds and then a "guess"
Answer: (-.786,4.209)
39. Find two non-negative numberfs x and y whose sum is 100 and for which (x^2)(y) is a maximum.
First step: Add up answer choices to make sure all equal 100
Second Step: Plug the x and y values into the given formula and see which is the highest number
Answer: x=66.667 and y=33.333
QUESTIONS:
How do you find the domain of a function without a calculator? I've always had trouble with domain and range even in Advanced Math.
How do you work piecewises? Number 7 is an example on the Non-calculator portion: Find k so that f(x) {(x^2-16)/x-4; x *does not equal* 4 and k; x=4 is continuous for all x. a) all real values, b)0, c)16, d)8, e) no real values
How do you take the derivative of a trig function squared? (sin^2(x) or cos^2(x))
Can someone go over the Mean Value Theorem for Derivatives? Number 12: Find a positive value c, for x, that satisfies the conclusion of the MVTD for f(x)=3x^2-5x+1 on [2,5] a)1, b)13/6, c)11/6, d)23/6, e)7/2
Hope everyone enjoys their day off!
post 28
all we did this week was correct our ap tests
1. The graph of y=5x^4-x^5 has an inflection point or points at
y=5x^4-x^5
20x^3-5x^3
20x^2(3-x)
3-x=0
-x=-3
divide by -1
x=3
2. d/dx(integral form 0 to x^2)sin^2t(dt)
sin^2(x^2)
2xsin^2(x^2)
3. The average value of f(x)=1/x from x=1 to x=e is
(1/e-1)[ln e -ln 1] *the e and 1 are the absoule value of
(1/e-1)[1-0]
= 1/e-1
4. Find the area under the curve y=2x-x^2 from x=1 to x=2 with n=4 left-endpoint rectangles.
use LRAM
delta x=2-1/4=1/4
1/4[f(1)+(5/4)+f(6/4)+f(7/4)]
1/4(0+.4375+.75+.9375)
1/4(2.125)= 17/32
i forgot houw to do problems (like #37 on calculator part) when a ladder slides down a wall and what is the speed of the bottom sliding out
1. The graph of y=5x^4-x^5 has an inflection point or points at
y=5x^4-x^5
20x^3-5x^3
20x^2(3-x)
3-x=0
-x=-3
divide by -1
x=3
2. d/dx(integral form 0 to x^2)sin^2t(dt)
sin^2(x^2)
2xsin^2(x^2)
3. The average value of f(x)=1/x from x=1 to x=e is
(1/e-1)[ln e -ln 1] *the e and 1 are the absoule value of
(1/e-1)[1-0]
= 1/e-1
4. Find the area under the curve y=2x-x^2 from x=1 to x=2 with n=4 left-endpoint rectangles.
use LRAM
delta x=2-1/4=1/4
1/4[f(1)+(5/4)+f(6/4)+f(7/4)]
1/4(0+.4375+.75+.9375)
1/4(2.125)= 17/32
i forgot houw to do problems (like #37 on calculator part) when a ladder slides down a wall and what is the speed of the bottom sliding out
Post #28
Another week filled with Ap tests and corrections..
I'm just going to go over some of the problems from those exciting test that I ended up getting wrong.
EXAMPLE 1:
Given f(x)=2x^2-7x-10, find the absolute maximum of f(x) on [-1,3].
*I knew I was suppose to take the derivative and plug in the 7/4. What I didn't know was that you have to plug in your endpoints, too.
f(x)=2x^2-7x-10
4x-7=0
4x=7
x=7/4
2(7/4)^2-7(7/4)-10= -129/8
2(-1)^2-7(-1)-10= -1
2(3)^2-7(3)-10= -13
*You want the absolute maximum so whichever one is the greatest-which is -1.
EXAMPLE 2:
A particle's position is given by s=t^3-6t^2+9t. What is the acceleration at time t=4?
*You are going to use PVA, going from position to acceleration which means taking the derivative twice. Then just simply plut in 4.
t^3-6t^2+9t
3t^2-12t+9
6t-12
6(4)-12= 12
EXAMPLE 3:
The volume generated by revolving about the x-axis the region above the curve y=x^3, below the line y=1, and between x=0 and x=1 is
*This is one on the calculator part, so use your calculator. I got this one wrong because I set up the equations wrong.
(pi) the intergralfrom 0 to 1 (1)^2-(x^3)^2
(enter in calculator!) = 6(pi)/7
What I am getting wrong or don't know?
Can someone explain how to do something like: Find the value(s) of dy/dx of x^2y+y^2=5 at y=1.
I'm just going to go over some of the problems from those exciting test that I ended up getting wrong.
EXAMPLE 1:
Given f(x)=2x^2-7x-10, find the absolute maximum of f(x) on [-1,3].
*I knew I was suppose to take the derivative and plug in the 7/4. What I didn't know was that you have to plug in your endpoints, too.
f(x)=2x^2-7x-10
4x-7=0
4x=7
x=7/4
2(7/4)^2-7(7/4)-10= -129/8
2(-1)^2-7(-1)-10= -1
2(3)^2-7(3)-10= -13
*You want the absolute maximum so whichever one is the greatest-which is -1.
EXAMPLE 2:
A particle's position is given by s=t^3-6t^2+9t. What is the acceleration at time t=4?
*You are going to use PVA, going from position to acceleration which means taking the derivative twice. Then just simply plut in 4.
t^3-6t^2+9t
3t^2-12t+9
6t-12
6(4)-12= 12
EXAMPLE 3:
The volume generated by revolving about the x-axis the region above the curve y=x^3, below the line y=1, and between x=0 and x=1 is
*This is one on the calculator part, so use your calculator. I got this one wrong because I set up the equations wrong.
(pi) the intergralfrom 0 to 1 (1)^2-(x^3)^2
(enter in calculator!) = 6(pi)/7
What I am getting wrong or don't know?
Can someone explain how to do something like: Find the value(s) of dy/dx of x^2y+y^2=5 at y=1.
Sunday, February 28, 2010
Posting...#28
Linierazation:
1.determine the equation
2. f(x)+f(x)dx
3.Figure out your dx
4.Figure out your x
5. Plug in everything
Trapezoidal: multiply by delta x/2-
The formula: 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.
The formula: 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.
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
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
I have been having trouble witht the non calculator test so if anyone has tips thanks
Post #28
This week in Calculus we took AP tests and then corrected them and did the key words packet. Suprisingly, i did semi-well on the AP tests..so i'm going to do the same thing I did last week in case it's good luck.
I'LL JUST REVIEW :)
So, lets start off with implicits.
They are really simple, but came up alot on the last two APs.
So, you know its implicit if you have x's and y's in the equation.
Everytime you take the derivative of y, you must put dy/dx.
So, problem time.
xy + 2x = 10
x(dy/dx) + y(1) = 0
Now, you simplify for dy/dx.
x(dy/dx) = -y
dy/dx = -y/x
As long as you remember your simple algebra rules and things about derivatives..these are really easy.
Next thing I'll explain is...tangent lines.
well half-way.
I know you take the derivative..then plug something in..
then take another derivative..or maybe plug in another number?
That's where i get confused..but i think it's mostly because i forgot how to do it.
and if they ask for a tangent line, and a line parallel to the tangent line
it should be the same thing, right?
Can anyone help me with that..
and a brush up on related rates and optimization.
I'LL JUST REVIEW :)
So, lets start off with implicits.
They are really simple, but came up alot on the last two APs.
So, you know its implicit if you have x's and y's in the equation.
Everytime you take the derivative of y, you must put dy/dx.
So, problem time.
xy + 2x = 10
x(dy/dx) + y(1) = 0
Now, you simplify for dy/dx.
x(dy/dx) = -y
dy/dx = -y/x
As long as you remember your simple algebra rules and things about derivatives..these are really easy.
Next thing I'll explain is...tangent lines.
well half-way.
I know you take the derivative..then plug something in..
then take another derivative..or maybe plug in another number?
That's where i get confused..but i think it's mostly because i forgot how to do it.
and if they ask for a tangent line, and a line parallel to the tangent line
it should be the same thing, right?
Can anyone help me with that..
and a brush up on related rates and optimization.
post 28
This week in calc we took 2 more ap tests then went over them again. and did corrections with the help of john, thanks!
here's a review:
TANGENT LINES:
1. Take the derivative of the equation
2. Plug in the x value, which gives you your slope
3. Use the slope and the point given and plug into slope intercept form
**if a point is not given and only an x value is given plug the x value into the original which will give you a y value. & THER'S YOUR POINT
SUBSTITUTION:
1. Find u & du
2. set u = whatever isn't the derivative
3. take the derivative of u
4. substitute back in
e integration:
very simple.
whatever e is raised to is your u, the derivative of u is du.
:)
IMPLICIT DERIVATIVE:
take derivative like normal, of both sides of the equation.
any derivative taken for y, mark with dy/dx behind it.
solve for dy/dx
i don't know what to do whenever it says what is tangent to the line parallel to something. i know it's probably simple, but i always leave those questions blank!
here's a review:
TANGENT LINES:
1. Take the derivative of the equation
2. Plug in the x value, which gives you your slope
3. Use the slope and the point given and plug into slope intercept form
**if a point is not given and only an x value is given plug the x value into the original which will give you a y value. & THER'S YOUR POINT
SUBSTITUTION:
1. Find u & du
2. set u = whatever isn't the derivative
3. take the derivative of u
4. substitute back in
e integration:
very simple.
whatever e is raised to is your u, the derivative of u is du.
:)
IMPLICIT DERIVATIVE:
take derivative like normal, of both sides of the equation.
any derivative taken for y, mark with dy/dx behind it.
solve for dy/dx
i don't know what to do whenever it says what is tangent to the line parallel to something. i know it's probably simple, but i always leave those questions blank!
Post Number Twenty Eight
I seem to be doing better but I definitely need to memorize some formulas that I should have known a long time ago and such. So here goes nothing..
Riemann Sums:
LRAM RRAM MRAM and TRAM are used for approximations
Uses the interval [a, b], deltax = (b – a)/n
LRAM:
Estimated from the left side, rectangles are drawn from the x-axis and then up to the graph, they then go over.
Deltax [f(a) + f(a + deltax) + … + f(b – deltax)]
RRAM:
Estimated from the right side, rectangles are drawn from the end of the interval of the graph itself.
Deltax [f(a + deltax) + f(b)]
MRAM:
Midpoints of LRAM and RRAM
deltax [f(midpoint1) + f(midpoint2) + f(midpoint n + 1)]
TRAM:
Most accurate of all Riemann sums
deltax/2 [f(a) + 2f(a + deltax) + 2f(a + 2deltax) + … + f(b)]
Tangent Lines:
First take the derivative of f(x)
Then plug in your x to find the slope
Plug x into the original to get your y
After finding your y and your slope, plug into point slope formula:
y – y1 = m(x – x1)
Simple Integration:
There are two types of integration, definite and indefinite.
In definite integration the answer is always a number.
It uses the integral [a, b].
The formula is bSa f(x) dx = f(b) – f(a) = #
Indefinite integration gives you an equation.
All derivative rules apply S x^n dx = x^(n+1)/(n+1) + C
Some things I don’t get are integration problems with square roots and there is an x in front. I have no idea what to do with those.
Also I need help with normal line and bacteria problems.
And I need BIG help with rate of change problems. Ones like find the volume of the cylinder and what not..
Riemann Sums:
LRAM RRAM MRAM and TRAM are used for approximations
Uses the interval [a, b], deltax = (b – a)/n
LRAM:
Estimated from the left side, rectangles are drawn from the x-axis and then up to the graph, they then go over.
Deltax [f(a) + f(a + deltax) + … + f(b – deltax)]
RRAM:
Estimated from the right side, rectangles are drawn from the end of the interval of the graph itself.
Deltax [f(a + deltax) + f(b)]
MRAM:
Midpoints of LRAM and RRAM
deltax [f(midpoint1) + f(midpoint2) + f(midpoint n + 1)]
TRAM:
Most accurate of all Riemann sums
deltax/2 [f(a) + 2f(a + deltax) + 2f(a + 2deltax) + … + f(b)]
Tangent Lines:
First take the derivative of f(x)
Then plug in your x to find the slope
Plug x into the original to get your y
After finding your y and your slope, plug into point slope formula:
y – y1 = m(x – x1)
Simple Integration:
There are two types of integration, definite and indefinite.
In definite integration the answer is always a number.
It uses the integral [a, b].
The formula is bSa f(x) dx = f(b) – f(a) = #
Indefinite integration gives you an equation.
All derivative rules apply S x^n dx = x^(n+1)/(n+1) + C
Some things I don’t get are integration problems with square roots and there is an x in front. I have no idea what to do with those.
Also I need help with normal line and bacteria problems.
And I need BIG help with rate of change problems. Ones like find the volume of the cylinder and what not..
Post #28
This week in calculus, we took the AP again, so I will review some of the questions I missed.
2. The domain of the function f(x) = the square root of 4-x^2 is
A. x < -2 or x>2
B. x/= 2
C. -2
We learned how to do this in advanced math, but I forgot. To find the domain, set what is inside the square root equal to zero then solve for x. Once you have x, set up intervals and then check the intervals to see if you get a real number.
4-x^2=0
x=2, -2
(-infinity, -2) (-2,2) (2, infinity)
the square root of 4-(-3)^2 = imaginary number
the square root of 4-(0)^2 = + or - 2
the square root of 4-(3)^2 = imaginary
Therefore, your domain is -2
7. Find k so that f(x) x^2 - 16 / x-4 x does not =4
k x=4 is continuous for all x.
A. All real values of k make f(x) continuous for all x.
B. 0
C. 16
D. 8
E. There is no real value of k that makes f(x) continuous for all x.
For this problem, take the limit as x->4.
(x+4) (x-4) / (x-4)
The (x-4)s cancel and you are left with x+4
Plug in 4 and you get 4+4 = 8
The answer is D. 8.
12. Find a positive value c, for x, that satisfies the conclusion of the Mean Value Theorem for Derivatives for f(x)=3x^2-5x+1 on the interval [2,5].
A. 1
B. 13/6
C. 11/6
D. 23/6
E. 7/2
The Mean Value states f(b)-f(a) / b-a, so all you have to do is plug into that formula then set that equal to the derivative.
3(5)^2-5(5)+1 - [3(2)^2-5(2) +1] / 5-2
75-25+1 - [12-10+1] / 3
51-3/3
48/3 = 6x - 5
48= 18x - 15
63 = 18x
x = 7/2 E.
45. If f(x) is continuous and differentiable and f(x) = ax^4+5x ; x bx^2-3x; x>2 then b=
A. 0.5
B. 0
C. 2
D. 6
E. There is no value of b.
The first step is to set the equations equal and plug in 2 for x.
a(2)^4 + 5(2) = x(2)^2-3(2)
16a+10 = 4b -6
Then take the derivative of each, plug in 2, and set them equal.
4ax^3 +5 = 2bx -3
4a(2)^3 + 5 = 2b(2) - 3
32a+5 = 4b-3
Now solve each equation for a.
16a+10=4b-6
a= 4b-16/16
a=b-4/4
32a+5=4b-3
a= 4b-8/32
a= b-2/8
Next, set the two equations solved for a equal and solve for b.
b-4/4 = b-2/8
8b-32= 4b - 8
4b=24
b=6 D.
I'm having the most trouble with substitution, and I can use a review on optimization.
2. The domain of the function f(x) = the square root of 4-x^2 is
A. x < -2 or x>2
B. x/= 2
C. -2
We learned how to do this in advanced math, but I forgot. To find the domain, set what is inside the square root equal to zero then solve for x. Once you have x, set up intervals and then check the intervals to see if you get a real number.
4-x^2=0
x=2, -2
(-infinity, -2) (-2,2) (2, infinity)
the square root of 4-(-3)^2 = imaginary number
the square root of 4-(0)^2 = + or - 2
the square root of 4-(3)^2 = imaginary
Therefore, your domain is -2
7. Find k so that f(x) x^2 - 16 / x-4 x does not =4
k x=4 is continuous for all x.
A. All real values of k make f(x) continuous for all x.
B. 0
C. 16
D. 8
E. There is no real value of k that makes f(x) continuous for all x.
For this problem, take the limit as x->4.
(x+4) (x-4) / (x-4)
The (x-4)s cancel and you are left with x+4
Plug in 4 and you get 4+4 = 8
The answer is D. 8.
12. Find a positive value c, for x, that satisfies the conclusion of the Mean Value Theorem for Derivatives for f(x)=3x^2-5x+1 on the interval [2,5].
A. 1
B. 13/6
C. 11/6
D. 23/6
E. 7/2
The Mean Value states f(b)-f(a) / b-a, so all you have to do is plug into that formula then set that equal to the derivative.
3(5)^2-5(5)+1 - [3(2)^2-5(2) +1] / 5-2
75-25+1 - [12-10+1] / 3
51-3/3
48/3 = 6x - 5
48= 18x - 15
63 = 18x
x = 7/2 E.
45. If f(x) is continuous and differentiable and f(x) = ax^4+5x ; x bx^2-3x; x>2 then b=
A. 0.5
B. 0
C. 2
D. 6
E. There is no value of b.
The first step is to set the equations equal and plug in 2 for x.
a(2)^4 + 5(2) = x(2)^2-3(2)
16a+10 = 4b -6
Then take the derivative of each, plug in 2, and set them equal.
4ax^3 +5 = 2bx -3
4a(2)^3 + 5 = 2b(2) - 3
32a+5 = 4b-3
Now solve each equation for a.
16a+10=4b-6
a= 4b-16/16
a=b-4/4
32a+5=4b-3
a= 4b-8/32
a= b-2/8
Next, set the two equations solved for a equal and solve for b.
b-4/4 = b-2/8
8b-32= 4b - 8
4b=24
b=6 D.
I'm having the most trouble with substitution, and I can use a review on optimization.
Post #28
Average Speed
First of all, remember that a slope is the y value, or dy, of a derivative.
Example:
A ball is flung from a little child. It's path is projected as y=4.9t2m in "t" seconds. What is the average speed of the ball from 0 to 3 seconds?
1. Set up equations and intervals: (f(b)-f(a))/(b-a) 4.9t^2 [0,3]
2. Plug in a and b values for t: f(b)=4.9(3)2=44.1 f(a)=4.9(0)2=0
3. Plug into main equation and solve: (44.1-0)/(3-0)=14.7m/s
Average speed is used for many different things, from finding the speed at which a cannonball was launched out of a cannon from how fast a cheetah runs in a straight line trying to catch it's prey. The concept behind average speed is a fairly simple concept that many people understand right away. You're basically finding the slope of the equation using calculus and algebra. If I ask someone what the average speed of a ball from [3,4] if it's path was graphed as y=x.
y=(4) y=(3) (4-3)/(4-3)=1
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))
First of all, remember that a slope is the y value, or dy, of a derivative.
Example:
A ball is flung from a little child. It's path is projected as y=4.9t2m in "t" seconds. What is the average speed of the ball from 0 to 3 seconds?
1. Set up equations and intervals: (f(b)-f(a))/(b-a) 4.9t^2 [0,3]
2. Plug in a and b values for t: f(b)=4.9(3)2=44.1 f(a)=4.9(0)2=0
3. Plug into main equation and solve: (44.1-0)/(3-0)=14.7m/s
Average speed is used for many different things, from finding the speed at which a cannonball was launched out of a cannon from how fast a cheetah runs in a straight line trying to catch it's prey. The concept behind average speed is a fairly simple concept that many people understand right away. You're basically finding the slope of the equation using calculus and algebra. If I ask someone what the average speed of a ball from [3,4] if it's path was graphed as y=x.
y=(4) y=(3) (4-3)/(4-3)=1
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 #28
Okayyyy, so this week was pretty normal even with brandi not there.
we took the non-calc portion monday and the calculator portion tuesday, did our key word sheets and worked on corrections for the remainder of the week.
Ln Integration
This is only used with fractions and when the top is the derivative of the bottom
Example: Integrate 2x / x^2 + 9
The derivative of x^2 + 9 is 2x which is the top of the fraction so the integral is
ln x^2 + 9 + c
Equation of a tangent line:
Take the derivative and plug in the x value.
If you are not given a y value, plug into the original equation to get the y value.
then plug those numbers into point slope form: y − y1 = m(x − x1)
Second Derivative Test:
1. Take the derivative of the first derivative.
2. Set the second 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 in the second derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph)
things i need help with: i don't have my packets infront of me right now but problems like the one that asked for the domain.. i think it was like the first or second problem, and the problem that asked for two non-negative numbers, it was one of the last problems... anyways, i have no idea how to do them. and someone wanna go over MVT?
we took the non-calc portion monday and the calculator portion tuesday, did our key word sheets and worked on corrections for the remainder of the week.
Ln Integration
This is only used with fractions and when the top is the derivative of the bottom
Example: Integrate 2x / x^2 + 9
The derivative of x^2 + 9 is 2x which is the top of the fraction so the integral is
ln x^2 + 9 + c
Equation of a tangent line:
Take the derivative and plug in the x value.
If you are not given a y value, plug into the original equation to get the y value.
then plug those numbers into point slope form: y − y1 = m(x − x1)
Second Derivative Test:
1. Take the derivative of the first derivative.
2. Set the second 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 in the second derivative for x.
6. Solve. For positive numbers, the graph of the derivative is above the x-axis. For negative numbers, the graph of the derivative is below the x-axis. The numbers for x are your points of inflection. (Points of Inflection are only if there is a shift in the graph)
things i need help with: i don't have my packets infront of me right now but problems like the one that asked for the domain.. i think it was like the first or second problem, and the problem that asked for two non-negative numbers, it was one of the last problems... anyways, i have no idea how to do them. and someone wanna go over MVT?
A Post.
In an attempt to further explain Reimann Sums, I shall first give the formula for each Middle RAM, Left RAM, Right RAM, Trapezoidal RAM then explain each in layman’s terms, all while attempting an example problem pertaining to all of the aforementioned sums.
*One must remember that these sums are all ways of estimating the area under a curve (hence the fact that this is under integration). Particular ones work better at estimating certain problems. However, as a general rule, trapezoidal is the most accurate when, and only when, the delta x is not zero.
LRAM-left hand approximation
When doing this, a trick is to start with the furthest left interval (hence the LEFT Reimann sum) and add delta x until right before you reach your b interval term and substitute those numbers into the original function. Then sum all of those values and multiply by delta x. Simply stated:
x [f(x+x) + … + f(b-x)]
RRAM-right hand approximation
This procedure is the same as the left except, you guessed it, you start with the farthest right interval or b!
*note, this formula may not be the same as in your notes, but since addition is commutative, you can start with whatever you like.
x[f(b) + f(b+x)…f(a+x)
MRAM-middle hand approximation
The trick to finding the middle sum is having the capability of finding all of the midpoints in the given interval.
x[f(mid) +…]
You are going to find the numbers in which you find the midpoints of by adding delta x from a to b. (will explain later)
TRAM-trapezoidal approximation
For this particular one, remember that delta x is over two and that you should multiply all of the numbers besides the functions of your intervals by 2.
x[f(a) + 2f(a+x) + 2f(a+2x) +…f(b)]
These are my examples for the above explanation of the four sums.
Given f(x) = x^2 - 3 with 3 subintervals on the interval [1,4], find the LRAM, MRAM, RRAM, and TRAM.
Beginning, one must first find delta x. Delta x is found by subtracting b minus a and dividing by n (subintervals).
So in this case:
x = 4-1/3 = 1
LRAM:
1[f(1) + f(2) + f(3)]
1[-2 + 1 + 6]
5
RRAM:
1[f(2) + f(3) + f(4)]
1[1 + 6 + 13]
20
TRAM:
½ [ f(1) + 2f(2) + 2f(3) + f(4)]
½[ -2 + 2+ 12 + 13]
½ [-1] = -½
MRAM:
1 2 3 4
3/2 5/2 7/2
1[f(3/2 + f(5/2) + f(7/2)]
So after fully breaking down practically step for step the process of finding MRAM, LRAM, RRAM, and TRAM, do you believe you have it?
*One must remember that these sums are all ways of estimating the area under a curve (hence the fact that this is under integration). Particular ones work better at estimating certain problems. However, as a general rule, trapezoidal is the most accurate when, and only when, the delta x is not zero.
LRAM-left hand approximation
When doing this, a trick is to start with the furthest left interval (hence the LEFT Reimann sum) and add delta x until right before you reach your b interval term and substitute those numbers into the original function. Then sum all of those values and multiply by delta x. Simply stated:
x [f(x+x) + … + f(b-x)]
RRAM-right hand approximation
This procedure is the same as the left except, you guessed it, you start with the farthest right interval or b!
*note, this formula may not be the same as in your notes, but since addition is commutative, you can start with whatever you like.
x[f(b) + f(b+x)…f(a+x)
MRAM-middle hand approximation
The trick to finding the middle sum is having the capability of finding all of the midpoints in the given interval.
x[f(mid) +…]
You are going to find the numbers in which you find the midpoints of by adding delta x from a to b. (will explain later)
TRAM-trapezoidal approximation
For this particular one, remember that delta x is over two and that you should multiply all of the numbers besides the functions of your intervals by 2.
x[f(a) + 2f(a+x) + 2f(a+2x) +…f(b)]
These are my examples for the above explanation of the four sums.
Given f(x) = x^2 - 3 with 3 subintervals on the interval [1,4], find the LRAM, MRAM, RRAM, and TRAM.
Beginning, one must first find delta x. Delta x is found by subtracting b minus a and dividing by n (subintervals).
So in this case:
x = 4-1/3 = 1
LRAM:
1[f(1) + f(2) + f(3)]
1[-2 + 1 + 6]
5
RRAM:
1[f(2) + f(3) + f(4)]
1[1 + 6 + 13]
20
TRAM:
½ [ f(1) + 2f(2) + 2f(3) + f(4)]
½[ -2 + 2+ 12 + 13]
½ [-1] = -½
MRAM:
1 2 3 4
3/2 5/2 7/2
1[f(3/2 + f(5/2) + f(7/2)]
So after fully breaking down practically step for step the process of finding MRAM, LRAM, RRAM, and TRAM, do you believe you have it?
28th post
This week in calculus we took two more AP test and surprisingly i did not to bad on these.... anyways lets go over some concepts:
Tangent line:
You take the derivative of the function and plug in the x value to get the slope. If no y value is given, you plug in x into the original function and solve. Then you set the equation into point slope form y-y1=slope(x-x1).
First derivative test:
You use this test to find max and min and to tell if the function is increasing or decreasing. You take the derivative of the function given and solve for the x values (critical points), then you set those points up into intervals between negative infinity and infinity. You then take numbers in between those intervals and plug them into the first derivative.
Second derivative test:
You use this test to find if the function is concave up, concave down, or if there is a point of inflection. You take the derivative of the function twice and solve for the critical points. Then you set those points up into intervals between negative infinity and infinity. You then plug points in between those intervals into the second derivative.
Limit Rules:
If the degree on top is bigger than the degree on the bottom, the limit is infinity
If the degree on top is smaller than the degree on the bottom, the limit is zero.
If the degree on top is the same as the degree on the bottom, you divide the coefficients to get the limit.
h Limits:
If they give you a limit where they have a letter going to a number, you take the derivative of whatever is behind the parenthesis on the top of the fraction. Then you plug in the number where there is an x.
Some things i am having trouble with:
When there is an integral and they have an x in front of a square root.
When you have tan in the integral but for some reason you have to change it to sin.
The questions that have three parts to it, i need helpful hints for those lol.
Hope you all have a great day off... See ya on Tuesday :)
Tangent line:
You take the derivative of the function and plug in the x value to get the slope. If no y value is given, you plug in x into the original function and solve. Then you set the equation into point slope form y-y1=slope(x-x1).
First derivative test:
You use this test to find max and min and to tell if the function is increasing or decreasing. You take the derivative of the function given and solve for the x values (critical points), then you set those points up into intervals between negative infinity and infinity. You then take numbers in between those intervals and plug them into the first derivative.
Second derivative test:
You use this test to find if the function is concave up, concave down, or if there is a point of inflection. You take the derivative of the function twice and solve for the critical points. Then you set those points up into intervals between negative infinity and infinity. You then plug points in between those intervals into the second derivative.
Limit Rules:
If the degree on top is bigger than the degree on the bottom, the limit is infinity
If the degree on top is smaller than the degree on the bottom, the limit is zero.
If the degree on top is the same as the degree on the bottom, you divide the coefficients to get the limit.
h Limits:
If they give you a limit where they have a letter going to a number, you take the derivative of whatever is behind the parenthesis on the top of the fraction. Then you plug in the number where there is an x.
Some things i am having trouble with:
When there is an integral and they have an x in front of a square root.
When you have tan in the integral but for some reason you have to change it to sin.
The questions that have three parts to it, i need helpful hints for those lol.
Hope you all have a great day off... See ya on Tuesday :)
post #28
Tangent lines were on there too and the steps for finding the tangent line are:
1. Take the derivative of the equation like normal
2. Plug in the x value which gives you your slope
3. Use the slope you get and the point given and plug into slope intercept form (y-y1)=slope(x-x1)
*If a point is not given and only an x value is given plug the x value into the original which will give you a y value creating a point.
absolute maxs and mins:
1. Find critical values by using the first derivative test: take the derivative set equal to zero and solve for the variable.
2. Plug both intervals as well as the critical values into your original equation
3. Be sure to keep the value after substitution with each appropriate number.
4. To determine which constitute a max and which constitute a min, you must look and point out those of which have the lowest values-min- and those of the highest value-max.
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
Example:
Integrate (x^2+1) (2x) dx
Since you can't integrate product rule, you know you have to use substitution.
u=x^2 + 1 du= 2x dx
Integrate: u du
1/2 u ^2
Plug u back in
1/2 ( x^2+1)^2 + c
Example 2:
Integrate x (x^2+1)^2 dx
u= x^2 +1 du= 2x
Since you are missing a 2 in the problem, you have to add a 1/2 to get rid of the 2
1/2 S u^2 du
1/2 (1/3) u^3
1/6 (x^2 +1) ^3 + c
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.
implicit derivatives:
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
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation involving your variables. (you can only have one unknown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
i am still having problems with the particle equations on our AP's so if someone could work one of those for me that would be suuppper helpful!!!!!!
1. Take the derivative of the equation like normal
2. Plug in the x value which gives you your slope
3. Use the slope you get and the point given and plug into slope intercept form (y-y1)=slope(x-x1)
*If a point is not given and only an x value is given plug the x value into the original which will give you a y value creating a point.
absolute maxs and mins:
1. Find critical values by using the first derivative test: take the derivative set equal to zero and solve for the variable.
2. Plug both intervals as well as the critical values into your original equation
3. Be sure to keep the value after substitution with each appropriate number.
4. To determine which constitute a max and which constitute a min, you must look and point out those of which have the lowest values-min- and those of the highest value-max.
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
Example:
Integrate (x^2+1) (2x) dx
Since you can't integrate product rule, you know you have to use substitution.
u=x^2 + 1 du= 2x dx
Integrate: u du
1/2 u ^2
Plug u back in
1/2 ( x^2+1)^2 + c
Example 2:
Integrate x (x^2+1)^2 dx
u= x^2 +1 du= 2x
Since you are missing a 2 in the problem, you have to add a 1/2 to get rid of the 2
1/2 S u^2 du
1/2 (1/3) u^3
1/6 (x^2 +1) ^3 + c
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.
implicit derivatives:
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
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation involving your variables. (you can only have one unknown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
i am still having problems with the particle equations on our AP's so if someone could work one of those for me that would be suuppper helpful!!!!!!
Calculus Post
Okay before I start my blog I want to address something...
Just because B-Rob is gone, it does not mean that we are done practicing for the AP. Even if the AP is not important to you guys, I know it is important to some of us taking the course...so
Please, make sure you are doing your work and being quiet in class for others so they can get help with questions and finish out their corrections.
Anyway, as to what I will explain...
Some people are still confusing a few things about position, velocity, and acceleration.
The order is, like I said, position, velocity, and acceleration. So, if you are given the velocity function and it says to find the distance traveled from 1 to 5, for example, you would integrate the velocity function from 1 to 5. If you are given velocity and are asked to find the acceleration function, you would take the derivative of velocity. These questions are almost always a give-a-way and are really easy points to get.
When you are doing integration, you really need to have a mindset or an eye that looks for something in the problem that has it's derivative elsewhere. It doesn't have to be the exact derivative but it just needs to be similar.
Also, don't forget that you can manipulate the equation in whatever way possible before taking the integral...
For example, you wouldn't leave e^x(e^(3x) and try to integrate that. You would first combine the two by adding their exponents to get e^(4x) and then integrate that. Manipulating the problem will save you heaps of time.
Also for all of those problems where it gives you like a piecewise-function and it says to make this function continous and differentiable, what value of a or b or whatever variable they give you...
The way you do this is plug in the bounds that it's around..it will usually be by a number...
Then you take the derivative and plug in the bounds you just plugged in....and now you have two equations which is a system of equations. Now all you need to do is solve the system by whatever method you like best. I usually find when you have two variables...just solving each in terms of one variable works best because you can just set them equal and solve for the other one, then just plug that answer back in one of the others to get the other variable.
Anyway, enjoy your day off.
Just because B-Rob is gone, it does not mean that we are done practicing for the AP. Even if the AP is not important to you guys, I know it is important to some of us taking the course...so
Please, make sure you are doing your work and being quiet in class for others so they can get help with questions and finish out their corrections.
Anyway, as to what I will explain...
Some people are still confusing a few things about position, velocity, and acceleration.
The order is, like I said, position, velocity, and acceleration. So, if you are given the velocity function and it says to find the distance traveled from 1 to 5, for example, you would integrate the velocity function from 1 to 5. If you are given velocity and are asked to find the acceleration function, you would take the derivative of velocity. These questions are almost always a give-a-way and are really easy points to get.
When you are doing integration, you really need to have a mindset or an eye that looks for something in the problem that has it's derivative elsewhere. It doesn't have to be the exact derivative but it just needs to be similar.
Also, don't forget that you can manipulate the equation in whatever way possible before taking the integral...
For example, you wouldn't leave e^x(e^(3x) and try to integrate that. You would first combine the two by adding their exponents to get e^(4x) and then integrate that. Manipulating the problem will save you heaps of time.
Also for all of those problems where it gives you like a piecewise-function and it says to make this function continous and differentiable, what value of a or b or whatever variable they give you...
The way you do this is plug in the bounds that it's around..it will usually be by a number...
Then you take the derivative and plug in the bounds you just plugged in....and now you have two equations which is a system of equations. Now all you need to do is solve the system by whatever method you like best. I usually find when you have two variables...just solving each in terms of one variable works best because you can just set them equal and solve for the other one, then just plug that answer back in one of the others to get the other variable.
Anyway, enjoy your day off.
Post #28
yeah, so how about that dance last night huh!
so this week we took AP tests monday and tuesday, that will go in our binders under Princeton, number 2! .. right?
so here's what i know how to do:
number 1 - it's asking for g'(4) so, take the derivative and plug in 4 for x
number 2 - it's asking for the domain for a function so, set the inside equal to 0, solve for x, set up intervals, plug in to see possibilities, and then see what works!
number 12 - it's asking about the mean value theorem, so you use the formula: f(b)-f(a) divided by b-a and set that equal to the derivative, then solve for x!
number 13 - you plug in the point given and what you get when you take the derivative and solve for x is your absolute maximum
EXAMPLE:
*NUMBER 1 - g(x)=1/32x^4-5x^2
g'(x)=1/8x^3-10x
g'(4)=1/8(4)^3-40= -32
*NUMBER 12 - f(b)-f(a) divided by b-a all equal to f'(x)
b=5 a=3 f(b)=54 f(a)=3 f(x)=3x^2-5x+1 f'(x)=6x-5
51-3 divided by 5-3 = 48/3 = 16
set equal to derivative - 6x-5
16=6x-5 - - - solve for x
x=7/2
so here's what i don't quite understand:
number 6 - it's taking the integral, and i have the work for it but i'm not sure if i can do that on my own, can someone explain in detail?
number 11 - how do i get that to become sin?
number 24 - JUST EXPLAIN
GOOD NEWS IS THAT WE'RE OFF TOMORROW!!!
SEE EVERYONE TUESDAY!
~ElliE~
so this week we took AP tests monday and tuesday, that will go in our binders under Princeton, number 2! .. right?
so here's what i know how to do:
number 1 - it's asking for g'(4) so, take the derivative and plug in 4 for x
number 2 - it's asking for the domain for a function so, set the inside equal to 0, solve for x, set up intervals, plug in to see possibilities, and then see what works!
number 12 - it's asking about the mean value theorem, so you use the formula: f(b)-f(a) divided by b-a and set that equal to the derivative, then solve for x!
number 13 - you plug in the point given and what you get when you take the derivative and solve for x is your absolute maximum
EXAMPLE:
*NUMBER 1 - g(x)=1/32x^4-5x^2
g'(x)=1/8x^3-10x
g'(4)=1/8(4)^3-40= -32
*NUMBER 12 - f(b)-f(a) divided by b-a all equal to f'(x)
b=5 a=3 f(b)=54 f(a)=3 f(x)=3x^2-5x+1 f'(x)=6x-5
51-3 divided by 5-3 = 48/3 = 16
set equal to derivative - 6x-5
16=6x-5 - - - solve for x
x=7/2
so here's what i don't quite understand:
number 6 - it's taking the integral, and i have the work for it but i'm not sure if i can do that on my own, can someone explain in detail?
number 11 - how do i get that to become sin?
number 24 - JUST EXPLAIN
GOOD NEWS IS THAT WE'RE OFF TOMORROW!!!
SEE EVERYONE TUESDAY!
~ElliE~
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