This week in calculus we took our two APs on tuesday and wednesday because the juniors were not at school on Monday. We went over corrections on thursday and friday we gave our sub a going away party. So lets go over some material:
First derivative test:
You are given a function and are either asked to find increasing or decreasing or max and mins. You take the derivative of the function and set equal to zero. You then solve for the x values also known as critical points. You then set those points up into intervals between negative infinity and infinity.
Second derivative test:
You are given a function and are asked to find whether the graph is concave up or down or where there is a point of inflection. You take the derivative of the function twice and solve for the critical points once more. You set the x values up into intervals between negative infinity and infinity.
Tangent lines:
You are given a function and an x value. Sometimes they may give you a y value but if not you can find it by plugging the x value into the original function and solving for y. Then you take the derivative of the function and plug in the x value to get the slope. Then you set everything up into point slope form y-y1=slope(x-x1).
Limits:
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 the top is the same as the degree on the bottom, you divide the coefficients to get the limit.
If they give you a limit problem where there is any letter going to 0 and they have a huge problem with parenthesis in it, you take the derivative of what is behind the parenthesis and plug in for x if needed.
Things i need help with:
integrals with trig functions in them
three question problems
normal lines
integrating problems with e and ln in the same problem
Have a great weekend :)
Saturday, March 20, 2010
Friday, March 19, 2010
Post #31
so this week, monday the juniors weren't there so I worked on corrections.
by the way, when am i going to get those back???
then tuesday we took the AP Non-Calculator Portion...it was almost like last week
&& wednesday we took the AP Calculator Portion...it was almost like last weeks one too!
thursday we got them back and decided to work on corrections, but since [as mentioned above] the test were almost the same, we didn't have very many questions!
today, FRIDAY, we had a going away party for our sub...THE FOOD WAS GREAT!..HA
so, let's go over some easier stuff...
Tangent lines:
The problem will give you a function and an x value. Sometimes they 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.
QUESTIONS???
*i think i need to go over all the trig formulas so i can know them...
*i need to know the formulas for area and volume..etc
*big integrals
*the problems where you have to solve for c
SEE YALL MONDAY
HAVE A GOOD WEEKEND!
~ElliE~
by the way, when am i going to get those back???
then tuesday we took the AP Non-Calculator Portion...it was almost like last week
&& wednesday we took the AP Calculator Portion...it was almost like last weeks one too!
thursday we got them back and decided to work on corrections, but since [as mentioned above] the test were almost the same, we didn't have very many questions!
today, FRIDAY, we had a going away party for our sub...THE FOOD WAS GREAT!..HA
so, let's go over some easier stuff...
Tangent lines:
The problem will give you a function and an x value. Sometimes they 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.
QUESTIONS???
*i think i need to go over all the trig formulas so i can know them...
*i need to know the formulas for area and volume..etc
*big integrals
*the problems where you have to solve for c
SEE YALL MONDAY
HAVE A GOOD WEEKEND!
~ElliE~
Wednesday, March 17, 2010
Post 30
This week we took a few more aps. I must say I'm getting better at them. I do well on the non-calculator portion just because every week we only have time to really thoroughly go through the non-calculator portion. I may start switching it up.
Anyway, last week on the aps we had problems like taking th ederivative of a natural log, chain rule derivatives, mean value theorem, and more integration I didn't know could be done using substitution.
Ok, for taking the derivative of a natural log, it's actually really simple. At the beginning of the year, B-rob gave us a sheet that showed the derivatives of everything that could be derived and natural logs were of course on there. Troughout the time from then to now, I guess I forgot how to do it and though tit was really hard, but it actually isn't. All it is is 1/x
For example:
d/dx ln(ln(2 - cos(x)))
For this one, you are basically doing a chain rule with a natural log. With chain rules, you are to always start with the outside of the problem and work your way in, taking the derivative of each term. So this would start off as
l/ln(2 - cos(x))
Then you'd have to multiply by the derivative of the inside which would be
1/ln(2 - cos(x)) X 1/(2 - cos(x)) X sin (x)
Giving you
sin(x)/(2 - cos(x)) ln(2 - cos(x))
For the mean value theorem, I used to forget to set it equal to the derivative, but the formula is
f(b) - f(a)/b - a = f '(x)
Ok, some things I"m haveing problems with:
Last week on the ap, they asked to give the derivative of f^-1 and saying that the problem is invertible. I really don't know what this means, so I tried to just put 1/the function and take the derivitave in that way, but I never get an answer.
Also, I'm having problems with piecewise functions and rate of change problems
Help please!
Anyway, last week on the aps we had problems like taking th ederivative of a natural log, chain rule derivatives, mean value theorem, and more integration I didn't know could be done using substitution.
Ok, for taking the derivative of a natural log, it's actually really simple. At the beginning of the year, B-rob gave us a sheet that showed the derivatives of everything that could be derived and natural logs were of course on there. Troughout the time from then to now, I guess I forgot how to do it and though tit was really hard, but it actually isn't. All it is is 1/x
For example:
d/dx ln(ln(2 - cos(x)))
For this one, you are basically doing a chain rule with a natural log. With chain rules, you are to always start with the outside of the problem and work your way in, taking the derivative of each term. So this would start off as
l/ln(2 - cos(x))
Then you'd have to multiply by the derivative of the inside which would be
1/ln(2 - cos(x)) X 1/(2 - cos(x)) X sin (x)
Giving you
sin(x)/(2 - cos(x)) ln(2 - cos(x))
For the mean value theorem, I used to forget to set it equal to the derivative, but the formula is
f(b) - f(a)/b - a = f '(x)
Ok, some things I"m haveing problems with:
Last week on the ap, they asked to give the derivative of f^-1 and saying that the problem is invertible. I really don't know what this means, so I tried to just put 1/the function and take the derivitave in that way, but I never get an answer.
Also, I'm having problems with piecewise functions and rate of change problems
Help please!
Tuesday, March 16, 2010
Monday, March 15, 2010
post 30
well, like the other recent weeks, mrs. robinson has been gone and so we have just been doing AP tests during the week and correcting them, then do blogs and wiki's of course... so anyway, here is the week thirty blog, which is about Extreme value theorem, Rolles theorem, and Mean value theorem
EVT: the EVT states that a continuous function on a close interval [a,b], must have both a minimum and a maximum on the interval. However, the max and min can occur at the endpoints.
Rolles theorem gives the conditions that guarantee the existance of an extrema in the interior of a closed inteval.
Rolles: Let f be continuous on a closed interval [a,b] and differentiable on the open interval (a,b). If f(a)= f(b) then there is at least one number, “c” in (a,b) such the f '(c)=0
MVT: If f is continuous on the closed interval [a,b] and differentiable on the interval (a,b) then there exists a number c in (a,b) such that f '(c)= (f(b)-f(a))/(b-a).
Example: f(x)-x^(2)-3x+2, show that f ' (x)=0 on some interval. (Hint: if they don't give an interval, assume it's x-intercepts)
so first you factor and find that x=1 and x=2
so then, you check and see that f(1)=f(2)=0 so it's continuous and it's differentiable.
So after you've found that, you take the derivative and set it equal to zero.
F '(x)=2x-3
2x-3=0
2x=3
x=3/2
c=3/2. And that's it. You've found c.
----------------
Another example
f(x)=x^(4)-2x^(2) on [-2,2] find all values of c for which f '(c)=0
continuous: yes
differentiable: yes. So let's get started.
First, you have to plug in to see if f(-2) is equal to f(2)
so, f(-2)=(-2)^(4) -2 (-2)^(2)= 8
f(2)= (2)^(4)-2(2)^(2)=8
so that's good. They equal each other. By the way, i'm not sure if i'm typing up all my parentheses correctly, so sorry if I mess that up.
So then you take the derivative and set = to zero
f '(x)=4x^(3)-4x=0
4x(x^(2)-1)=0
x= 0, 1, -1
c= -1, 0, 1. and you're finished again.
okay, so i am good with derivatives and second derivatives and implicits and all of that, but i kinda suck at integrating anything even remotely difficult, so that's the one thing i suppose i could use some help on
EVT: the EVT states that a continuous function on a close interval [a,b], must have both a minimum and a maximum on the interval. However, the max and min can occur at the endpoints.
Rolles theorem gives the conditions that guarantee the existance of an extrema in the interior of a closed inteval.
Rolles: Let f be continuous on a closed interval [a,b] and differentiable on the open interval (a,b). If f(a)= f(b) then there is at least one number, “c” in (a,b) such the f '(c)=0
MVT: If f is continuous on the closed interval [a,b] and differentiable on the interval (a,b) then there exists a number c in (a,b) such that f '(c)= (f(b)-f(a))/(b-a).
Example: f(x)-x^(2)-3x+2, show that f ' (x)=0 on some interval. (Hint: if they don't give an interval, assume it's x-intercepts)
so first you factor and find that x=1 and x=2
so then, you check and see that f(1)=f(2)=0 so it's continuous and it's differentiable.
So after you've found that, you take the derivative and set it equal to zero.
F '(x)=2x-3
2x-3=0
2x=3
x=3/2
c=3/2. And that's it. You've found c.
----------------
Another example
f(x)=x^(4)-2x^(2) on [-2,2] find all values of c for which f '(c)=0
continuous: yes
differentiable: yes. So let's get started.
First, you have to plug in to see if f(-2) is equal to f(2)
so, f(-2)=(-2)^(4) -2 (-2)^(2)= 8
f(2)= (2)^(4)-2(2)^(2)=8
so that's good. They equal each other. By the way, i'm not sure if i'm typing up all my parentheses correctly, so sorry if I mess that up.
So then you take the derivative and set = to zero
f '(x)=4x^(3)-4x=0
4x(x^(2)-1)=0
x= 0, 1, -1
c= -1, 0, 1. and you're finished again.
okay, so i am good with derivatives and second derivatives and implicits and all of that, but i kinda suck at integrating anything even remotely difficult, so that's the one thing i suppose i could use some help on
Sunday, March 14, 2010
Ash's 30th Post
So, after an eventful week which consisted of taking and attempting to correct more AP Tests...I realize: there's only a few weeks left before the AP test. Begin utter freak-mode!
I'm going to ask lots of questions on this one:
Non-Calculator
13. The slope of the tangent line to the graph of 4x^22+cx-2e^y=-2 at x=0 is 4. Give the value of c.
How would I even begin that problem? My only lead: take a derivative. The end? =/
20. Compute lim x->0 (-(4x/sin(2x)) + x/cos2x)
Once again, I have no idea how to go at this.
The answer is zero and the only way I know this is because John tried to push us on the right path...MUCH APPRECIATED! =]
22. Determine 1S2 1/(sqrt(4-t^2))dt
For this...this is a trig identity? The answers all have pi in them, but now do you get pi out of this?
24. A particle's acceration for t>0 is given by a(t)=12t+4 The particle's initial position is 2 and it's velocity at t=1 is 5. What is the position of the particle at t=2?
You take the derivative, right? And then you plug in...what?
People have gone over these time and time again, yet I still do not understand them.
27. Compute the derivative of 0Sx^2 ln(t^2+1)dt
I remember that the derivative basically cancels out the integral...what next? Also, I never really understood how to plug in x^2 for t and all of that, so if anyone could go over that, I'd be very grateful! =]
It's very close to the AP test time and I don't feel like I know this still. I know what to do, mostly, but not how to do it still. When we fill out the keywords sheets (which I need to do for the past...2-3 AP tests...does anyone know where they're at?) I can do that easily! It's just actually computing and calculating the problem...does anyone else still feel like this? I know I asked this a while back on here and a bunch of you guys responded that you felt sort of the same and I was wondering if anyone was STILL kinda stuck...=/
I wish a good luck to everyone on this upcoming week! :D
I'm going to ask lots of questions on this one:
Non-Calculator
13. The slope of the tangent line to the graph of 4x^22+cx-2e^y=-2 at x=0 is 4. Give the value of c.
How would I even begin that problem? My only lead: take a derivative. The end? =/
20. Compute lim x->0 (-(4x/sin(2x)) + x/cos2x)
Once again, I have no idea how to go at this.
The answer is zero and the only way I know this is because John tried to push us on the right path...MUCH APPRECIATED! =]
22. Determine 1S2 1/(sqrt(4-t^2))dt
For this...this is a trig identity? The answers all have pi in them, but now do you get pi out of this?
24. A particle's acceration for t>0 is given by a(t)=12t+4 The particle's initial position is 2 and it's velocity at t=1 is 5. What is the position of the particle at t=2?
You take the derivative, right? And then you plug in...what?
People have gone over these time and time again, yet I still do not understand them.
27. Compute the derivative of 0Sx^2 ln(t^2+1)dt
I remember that the derivative basically cancels out the integral...what next? Also, I never really understood how to plug in x^2 for t and all of that, so if anyone could go over that, I'd be very grateful! =]
It's very close to the AP test time and I don't feel like I know this still. I know what to do, mostly, but not how to do it still. When we fill out the keywords sheets (which I need to do for the past...2-3 AP tests...does anyone know where they're at?) I can do that easily! It's just actually computing and calculating the problem...does anyone else still feel like this? I know I asked this a while back on here and a bunch of you guys responded that you felt sort of the same and I was wondering if anyone was STILL kinda stuck...=/
I wish a good luck to everyone on this upcoming week! :D
post 30
bloggin it up yeerd me :)
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 Rates:
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
tangent line
take the derivative of the equation and plug in the x-corrdinates to get your y,
then plug in y to get your slope then plug into the eqn.
MEAN VALUE THEOREM]
If f is continous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c in (a,b) such that F'(c) = f(b) - f(a) / b-a
IMPLICIT DERIVATIVE
while taking a derivative, anytime you take y', put a dy/dx in its place. then solve for dy/dx in terms of x.
i suck at integrals in a fraction. like substitution and all
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 Rates:
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
tangent line
take the derivative of the equation and plug in the x-corrdinates to get your y,
then plug in y to get your slope then plug into the eqn.
MEAN VALUE THEOREM]
If f is continous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c in (a,b) such that F'(c) = f(b) - f(a) / b-a
IMPLICIT DERIVATIVE
while taking a derivative, anytime you take y', put a dy/dx in its place. then solve for dy/dx in terms of x.
i suck at integrals in a fraction. like substitution and all
Post Number Thirty
Third nine weeks is about to end, this is almost it seniors (:
Now for calculus…
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 Rates:
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
Tangent lines?
I always think I know how to do these but never get them right so correct me if I’m wrong.
Take derivative and plug in your x and y to find the slope, then put into point slope.
Normal lines?
Same thing but negative reciprocal
Definition of a derivative?
I have noooooo idea
Integrals of trig functions?
HELPPPPPPPPPPP
Substitution?
Helpppppp more.
I know the steps I just don’t recognize them once I’m working the problem
Mean value theorem?
What is it I can’t find my notes on that…
One more thing, I know how to do average value, but I just never get the correct answer. I don’t know what my problem is, I just suck I guess.
So people should be able to get their comments in this week because I clearly need A LOT of help.
Goodnight world.
Now for calculus…
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 Rates:
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
Tangent lines?
I always think I know how to do these but never get them right so correct me if I’m wrong.
Take derivative and plug in your x and y to find the slope, then put into point slope.
Normal lines?
Same thing but negative reciprocal
Definition of a derivative?
I have noooooo idea
Integrals of trig functions?
HELPPPPPPPPPPP
Substitution?
Helpppppp more.
I know the steps I just don’t recognize them once I’m working the problem
Mean value theorem?
What is it I can’t find my notes on that…
One more thing, I know how to do average value, but I just never get the correct answer. I don’t know what my problem is, I just suck I guess.
So people should be able to get their comments in this week because I clearly need A LOT of help.
Goodnight world.
Post 30
Well its time for the third nine weeks exams. Graduation is coming faster than we think. But it's time for a blog so here we go:
I will start with linearization. The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation
Now I am going to take about the limit rules. The limit rules are:
1. if the highest exponent is the same on the top and bottom then the limit is the top coefficient over the bottom coefficient of the highest exponents.
2. If the highest exponent is on the top then the limit is infinity.
3. But if the highest exponent is on the bottom then the limit is 0.
Next I am going to talk about related rates. The steps for related rates are:
1. Identify all of the variables and equations
2. Identify the things that you are looking for
3. Sketch a graph and then label that graph
4. Create and write an equation using all of the variables
5. Take the derivative of this equation with respect to time
6. Substitute everything back in
7. Solve the equation
Now I am going to talk about angle of elevation. Angle of elevation problems are pretty much worked the same as related rate problems except you are looking for an angle in these instead of a rate.
Can someone give me a brush up on how to work an optimization problem?
I will start with linearization. The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation
Now I am going to take about the limit rules. The limit rules are:
1. if the highest exponent is the same on the top and bottom then the limit is the top coefficient over the bottom coefficient of the highest exponents.
2. If the highest exponent is on the top then the limit is infinity.
3. But if the highest exponent is on the bottom then the limit is 0.
Next I am going to talk about related rates. The steps for related rates are:
1. Identify all of the variables and equations
2. Identify the things that you are looking for
3. Sketch a graph and then label that graph
4. Create and write an equation using all of the variables
5. Take the derivative of this equation with respect to time
6. Substitute everything back in
7. Solve the equation
Now I am going to talk about angle of elevation. Angle of elevation problems are pretty much worked the same as related rate problems except you are looking for an angle in these instead of a rate.
Can someone give me a brush up on how to work an optimization problem?
Post # 30
Well, time has flown by..so fast, i can't believe we're already on week thirty..but anyways, back to actual calculus.
Now, i'm kinda done explaining things I actually understand, so i'm going to halfway explain the things I understand but never finish..so here it goes.
finding a tangent line..
take the derivative of the equation and plug in the x-corrdinates to get your y,
then plug in y to get your slope...
then plug it all into one equation?
definition of a derivative...
take the derivative of the last term at the top..
plug in x for what variable?
or i'm really confused here..
related rates...
find waht the problem is looking for..
list all clues and numbers the problem gives you..
then i forgot how to figure out which equation to use..
These are the major topics that i miss by like one number and always end up getting wrong, i understand integrals now, thanks to alex..so if anyone can help me with these things, i'd appreciate it. I know i didn't say anything i understand, but besides a derivative, I don't have anything..
Now, i'm kinda done explaining things I actually understand, so i'm going to halfway explain the things I understand but never finish..so here it goes.
finding a tangent line..
take the derivative of the equation and plug in the x-corrdinates to get your y,
then plug in y to get your slope...
then plug it all into one equation?
definition of a derivative...
take the derivative of the last term at the top..
plug in x for what variable?
or i'm really confused here..
related rates...
find waht the problem is looking for..
list all clues and numbers the problem gives you..
then i forgot how to figure out which equation to use..
These are the major topics that i miss by like one number and always end up getting wrong, i understand integrals now, thanks to alex..so if anyone can help me with these things, i'd appreciate it. I know i didn't say anything i understand, but besides a derivative, I don't have anything..
30
so this was another week in calculus, and we had more ap test to take. For some stuff that i understand a little bit:
PROBLEM: Finding the points of inflection for x^3-6x^2+12x.
1. take the derivative:
3x^2-12x+12
2. take the second derivative:
6x-12
3. set = to 0:
6x-12=0 x=2 --->gives possible pts. of inflection
4. set up intervals:
(-infinity,2) u (2,infinity)
5. plug in to SECOND DERIVATIVE:
6(0)-12= -ve --->concave down
6(3)-12= +ve --->concave up
So, x=2 is the point of inflection.
Also, I thought I might review implicit derivatives.
PROBLEM: y^3+y^2-5y-x^2=-4
1. take derivative of both sides: (implicit derivatives have an = sign)
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0
*remember every time you take the derivative of y, you have to note it by dy/dx or y^1
3. solve for dy/dx: (you are going to have to take out a dy/dx when solving)
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)=2x
dy/dx(3y^2+2y-5)=2x
dy/dx=2x/3y^2+2y-5
*Also, if you want the slope you must plug in a x and y-value.
I need some serious help with intergration. I suck at it
PROBLEM: Finding the points of inflection for x^3-6x^2+12x.
1. take the derivative:
3x^2-12x+12
2. take the second derivative:
6x-12
3. set = to 0:
6x-12=0 x=2 --->gives possible pts. of inflection
4. set up intervals:
(-infinity,2) u (2,infinity)
5. plug in to SECOND DERIVATIVE:
6(0)-12= -ve --->concave down
6(3)-12= +ve --->concave up
So, x=2 is the point of inflection.
Also, I thought I might review implicit derivatives.
PROBLEM: y^3+y^2-5y-x^2=-4
1. take derivative of both sides: (implicit derivatives have an = sign)
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0
*remember every time you take the derivative of y, you have to note it by dy/dx or y^1
3. solve for dy/dx: (you are going to have to take out a dy/dx when solving)
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)=2x
dy/dx(3y^2+2y-5)=2x
dy/dx=2x/3y^2+2y-5
*Also, if you want the slope you must plug in a x and y-value.
I need some serious help with intergration. I suck at it
week 30
For this post I will explain some things that people seem to have a lot of trouble with.
For integrating, here are the formulas for trig inverses.
d/dx arcsin(u) = 1/sqrt(1-u^2) * u'
d/dx arccos(u) = -1/sqrt(1-u^2) * u'
d/dx arctan(u) = 1/(1+u^2) * u'
d/dx arccot(u) = -1/(1+u^2) * u'
d/dx arcsec(u) = 1 / (abs(u))(sqrt(u^2-1)) *u'
d/dx arccsc(u) = -1 / (abs(u))(sqrt(u^2-1)) *u'
TANGENT LINE
1. find the x of your original function (usually given).
2. find the y of your original function (plug in x)
3. take derivative of original function.
4. plug x into derived function.
5. set up slope-intercept form.
NORMAL LINE
same thing as tangent line except adding a step after step 4.
the new step 5: make the slope perpendicular. Do this by flipping the fraction and making it negative.
CHAIN RULE
y = (tan(x))^3 + tan(x^3)
y' = 3(tanx)^2*(secx)^2 = sec(x^3)*3x^2
MEAN VALUE THEOREM]
If f is continous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c in (a,b) such that F'(c) = f(b) - f(a) / b-a
IMPLICIT DERIVATIVE
while taking a derivative, anytime you take y', put a dy/dx in its place. then solve for dy/dx in terms of x.
For integrating, here are the formulas for trig inverses.
d/dx arcsin(u) = 1/sqrt(1-u^2) * u'
d/dx arccos(u) = -1/sqrt(1-u^2) * u'
d/dx arctan(u) = 1/(1+u^2) * u'
d/dx arccot(u) = -1/(1+u^2) * u'
d/dx arcsec(u) = 1 / (abs(u))(sqrt(u^2-1)) *u'
d/dx arccsc(u) = -1 / (abs(u))(sqrt(u^2-1)) *u'
TANGENT LINE
1. find the x of your original function (usually given).
2. find the y of your original function (plug in x)
3. take derivative of original function.
4. plug x into derived function.
5. set up slope-intercept form.
NORMAL LINE
same thing as tangent line except adding a step after step 4.
the new step 5: make the slope perpendicular. Do this by flipping the fraction and making it negative.
CHAIN RULE
y = (tan(x))^3 + tan(x^3)
y' = 3(tanx)^2*(secx)^2 = sec(x^3)*3x^2
MEAN VALUE THEOREM]
If f is continous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c in (a,b) such that F'(c) = f(b) - f(a) / b-a
IMPLICIT DERIVATIVE
while taking a derivative, anytime you take y', put a dy/dx in its place. then solve for dy/dx in terms of x.
post # 30
alright, so this week we took more ap tests. and went over them not only in class but also with alex in fifth hour. it actually did help a lot
FINDING AREA BETWEEN A CURVE:
The formula you use is b(int)a (top eq.) - (bottom eq.).If a and b is not already given to you, then you much set the equations equal to each other and solve.You find which equation is top/bottom by graphing both and simply looking to see which one is on top.If the area is on the y-axis, then the a and b values need to be set as y-values, and the equations must be solved for x.
KEYWORDS:
Local minimum- first derivative, set equal to zero, solve, set up intervals and check
Maximum speed- absolute value of velocity
Linearization- find the equation of the tangent line (APPROXIMATE)
Lim h> 0 + h – definition of derivative- take derivative of what is behind the parenthesis
Maximum acceleration- find the maximum slope
Starting point- original function; zeros of velocity will give you max and mins
Rate- take derivative
AVG VALUE:
1/b-a (int) f(x)
it usually gives you an interval. it will say on interval from [1,4]
1 goes on bottom of integral & 4 goes on top => a = 1 , b = 4
this week we had instantaneous rate of change on our ap tests, something we haven't seen in a while! i know alex went over it in 5th hour the other day, but i missed it. can someone go over that for me?
i also need help with particle problems.
OH and i recently found out that the integral of sin^2(x) is a trig identity. if someone could go over the trig identities that would be VERY helpful
FINDING AREA BETWEEN A CURVE:
The formula you use is b(int)a (top eq.) - (bottom eq.).If a and b is not already given to you, then you much set the equations equal to each other and solve.You find which equation is top/bottom by graphing both and simply looking to see which one is on top.If the area is on the y-axis, then the a and b values need to be set as y-values, and the equations must be solved for x.
KEYWORDS:
Local minimum- first derivative, set equal to zero, solve, set up intervals and check
Maximum speed- absolute value of velocity
Linearization- find the equation of the tangent line (APPROXIMATE)
Lim h> 0 + h – definition of derivative- take derivative of what is behind the parenthesis
Maximum acceleration- find the maximum slope
Starting point- original function; zeros of velocity will give you max and mins
Rate- take derivative
AVG VALUE:
1/b-a (int) f(x)
it usually gives you an interval. it will say on interval from [1,4]
1 goes on bottom of integral & 4 goes on top => a = 1 , b = 4
this week we had instantaneous rate of change on our ap tests, something we haven't seen in a while! i know alex went over it in 5th hour the other day, but i missed it. can someone go over that for me?
i also need help with particle problems.
OH and i recently found out that the integral of sin^2(x) is a trig identity. if someone could go over the trig identities that would be VERY helpful
Post #30
Indefinite Integration
Integration, the basis of all Calculus classes, is probably the hardest thing ever to learn. Integration, in lame man's terms, is just the opposite of differentiation. That means that instead of going from x to 1, you from x to (1/2)x^2.
First of all, there is a new sign: its an S-shaped symbol that just stands for opposite of derivative, or integral.
Next, there are two types of integrals: Definite and Indefinite.
Indefinite Integrals are just doing the opposite of a derivative.
int x dx (1/2)x^2 +c
int 5 dx 5x +c
int cosx dx sinx +c
See the pattern there. ALWAYS ADD A +c AT THE END OF ALL INDEFINITE INTEGRALS!!!
Definite integrals just involve one more step after this, plugging in an x value or values
int x dx [2,4] (1/2)x^2 (1/2)(4^2)=8 (1/2)(2^2)=2
int x dx [2,4]=6
You have to subtract f(a) from f(b). This gives you the area under the curve of the function on the interval. Integration problems are very very common on the AP exam and all Calculus students will know everything about basic integration before they can pass the exam or even the class.
Substitution is the only trick that integration needs. Substitution takes the place of the chain rule, multiplication rule, and even the quotient rule. The steps to substitute are as follows:
1. Find derivative inside integral.
2. Substitute u for the non-derivative then differentiate u
3. plug back in with original.
Example:
int (x^2+1)(2x) dx
you may think this is impossible but its not using substitution
1. Find derivative inside integral. u=(x^2+1) du=(2x)
2. Integrate u. (1/2)u^2 +c
3. Plug back in (1/2)(x^2+1)^2 +c
Integration, the basis of all Calculus classes, is probably the hardest thing ever to learn. Integration, in lame man's terms, is just the opposite of differentiation. That means that instead of going from x to 1, you from x to (1/2)x^2.
First of all, there is a new sign: its an S-shaped symbol that just stands for opposite of derivative, or integral.
Next, there are two types of integrals: Definite and Indefinite.
Indefinite Integrals are just doing the opposite of a derivative.
int x dx (1/2)x^2 +c
int 5 dx 5x +c
int cosx dx sinx +c
See the pattern there. ALWAYS ADD A +c AT THE END OF ALL INDEFINITE INTEGRALS!!!
Definite integrals just involve one more step after this, plugging in an x value or values
int x dx [2,4] (1/2)x^2 (1/2)(4^2)=8 (1/2)(2^2)=2
int x dx [2,4]=6
You have to subtract f(a) from f(b). This gives you the area under the curve of the function on the interval. Integration problems are very very common on the AP exam and all Calculus students will know everything about basic integration before they can pass the exam or even the class.
Substitution is the only trick that integration needs. Substitution takes the place of the chain rule, multiplication rule, and even the quotient rule. The steps to substitute are as follows:
1. Find derivative inside integral.
2. Substitute u for the non-derivative then differentiate u
3. plug back in with original.
Example:
int (x^2+1)(2x) dx
you may think this is impossible but its not using substitution
1. Find derivative inside integral. u=(x^2+1) du=(2x)
2. Integrate u. (1/2)u^2 +c
3. Plug back in (1/2)(x^2+1)^2 +c
Post #30
AP questions from last week..
2. Find the slope of the tangent line to the graph of f at x=4, given that f(x) = -x^2 += 4(x)^1/2
A. -8 B. -10 C. -9 D. -5 E. -7
To find the slope of a tangent line, take the derivative of the function given, then plug in the x-value given.
-2x + 2x^-1/2 OR -2x + 2 / x^1/2
-2(4) +2 / (4) ^1/2
-8 + 2/2 = -8 + 1 which equals -7.
The answer is E.
If the problems was asking for the equation of the tangent line, you would have to plug x into the original to get a y. Then you would plug x, y, and the slope you just found into point slope form.
-(4)^2 + 4(4)^1/2
-16+8 = -8
y+8 = -7 (x-4)
9. Give the equation of the normal line to the graph of y= 2x (x^2+8)^1/2 + 2 at the point (0,2).
Normal line is the same steps as tangent line except instead of using the slope, you use the perpendicular slope, which is the negative reciprocal of the slope.
First take the derivative which will be product rule
2x ((1/2 (x^2+8)^-1/2 ) (2x)) + (x^2+8)^1/2 (2)
4x^2 / 2 (x^2 +8) ^1/2 + 2 (x^2+8)^1/2
Instead of trying to simplify this, you can just plug in your x value (0) to find the slope.
0 + 2 times the square root of 8
which simplifies to, 4 times the square root of 2.
That would be the slope if you was finding the tangent line, but since you are finding the normal line, the slope is -1/ 4 times the square root of 2.
Now that we have a point and a slope, plug into point slope form.
y-2 = -1/4 times the square root of 2 (x-0)
Multiple each side by 4 square root of 2 to get rid of the fraction
4 square root of 2 y - 8 square root of 2 = -x
OR x + 4 square root of 2 y = 8 square root of 2
Answer choice B.
11. Compute the integral of 4x^2 (x^3+4) ^1/2 dx
u = x^3 +4 du= 3x^2
You have to get rid of the 3 in du, which you do by dividing something by 3 and gain a 4 which you do by multiplying by 4 so the integral is
4/3 S u^1/2
4/3 (2/3 ) u ^3/2
8/9 ( x^3 +4) ^3/2 + c
The answer is C.
I have a few question on the calculator portion if anyone wants to help.
2. I'm not sure what invertible means.
11.
14.
and 16.
2. Find the slope of the tangent line to the graph of f at x=4, given that f(x) = -x^2 += 4(x)^1/2
A. -8 B. -10 C. -9 D. -5 E. -7
To find the slope of a tangent line, take the derivative of the function given, then plug in the x-value given.
-2x + 2x^-1/2 OR -2x + 2 / x^1/2
-2(4) +2 / (4) ^1/2
-8 + 2/2 = -8 + 1 which equals -7.
The answer is E.
If the problems was asking for the equation of the tangent line, you would have to plug x into the original to get a y. Then you would plug x, y, and the slope you just found into point slope form.
-(4)^2 + 4(4)^1/2
-16+8 = -8
y+8 = -7 (x-4)
9. Give the equation of the normal line to the graph of y= 2x (x^2+8)^1/2 + 2 at the point (0,2).
Normal line is the same steps as tangent line except instead of using the slope, you use the perpendicular slope, which is the negative reciprocal of the slope.
First take the derivative which will be product rule
2x ((1/2 (x^2+8)^-1/2 ) (2x)) + (x^2+8)^1/2 (2)
4x^2 / 2 (x^2 +8) ^1/2 + 2 (x^2+8)^1/2
Instead of trying to simplify this, you can just plug in your x value (0) to find the slope.
0 + 2 times the square root of 8
which simplifies to, 4 times the square root of 2.
That would be the slope if you was finding the tangent line, but since you are finding the normal line, the slope is -1/ 4 times the square root of 2.
Now that we have a point and a slope, plug into point slope form.
y-2 = -1/4 times the square root of 2 (x-0)
Multiple each side by 4 square root of 2 to get rid of the fraction
4 square root of 2 y - 8 square root of 2 = -x
OR x + 4 square root of 2 y = 8 square root of 2
Answer choice B.
11. Compute the integral of 4x^2 (x^3+4) ^1/2 dx
u = x^3 +4 du= 3x^2
You have to get rid of the 3 in du, which you do by dividing something by 3 and gain a 4 which you do by multiplying by 4 so the integral is
4/3 S u^1/2
4/3 (2/3 ) u ^3/2
8/9 ( x^3 +4) ^3/2 + c
The answer is C.
I have a few question on the calculator portion if anyone wants to help.
2. I'm not sure what invertible means.
11.
14.
and 16.
post # 30
REVIEW!!
The formula for the volume of disks is S (top)^2 - (bottom)^2 dx
The formula for the area of washers is S (top) - (bottom)
The steps are:
1. Draw the graphs of the equations
2. Subtract top graph's equation by the bottom graph's equation(in disks each equation would be squared)
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. Integrate
volume by disks:
the formula is pi times the integral of the [function given] squared times dx. so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one. then graph
volume by washers:
the formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the in between number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. then graph.
LRAM is left hand approximation and the formula is:
delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
Say you are asked to calculate the left Riemann Sum for -4x -5 on the interval [-3, -1] divided into 2 subintervals.
delta x would equal: -1+3 /2 = 2/2 = 1
1[ f(-3) + f(-3 +1)]
1[ f( -3) + f(-2)]
then plug into your equation
RRAM is right hand approximation and the formula is:
delta x [ f(a + delta x) + .... + f(b)]
so using the same example:
1[ f( -2) + f(-1)] and then plug into your equation
MRAM is to calculate the middle and the formula is:
delta x [ f(mid) + f(mid) + .... ]
To find midpoints, you would add the two numbers together then divide by two
In this problem the numbers would be: -3 , -2, -1
-3 + -2/ 2 = -5/2 and -2 + -1 / 2 = -3/2
so 1[f(-5/2) + f(-3/2)] and the plug in
Trapezoidal is different because instead of multiplying by delta x, you multiply by delta x/2 and you also have on more term then your number of subintervals.
The formula is : delta x/2 [f(a) + 2f(a + delta x) + 2f(a+ 2 delta x) + ....f(b)]
For this problem: 1/2 [ f(-3) + 2 f(-2) + f( -1)] and then plug in.
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
e integration:
whatever is raised to the e power will be your u and du will be the derivative of u. For example:
e^2x-1dx
u=2x-1 du=2
rewrite the function as:
1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
related rates:
The steps for related rates are….
1. Pick out all variables
2. Pick out all equations
3. Pick out what you are looking for
4. Sketch a graph and label
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve
limits:
Rules for Limits:…
1. if the degree of top equals the degree of bottom, the answer is the top coefficient over bottom coefficient
2. if top degree is bigger than bottom degree, the answer is positive or negative infinity
2. if top degree is less than bottom degree, the answer is 0
To find area between curves:
The formula you use is b(int)a (top eq.) - (bottom eq.).
If a and b is not already given to you, then you much set the equations equal to each other and solve.
You find which equation is top/bottom by graphing both and simply looking to see which one is on top.
If the area is on the y-axis, then the a and b values need to be set as y-values, and the equations must be solved for x.
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.
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))
Find the volume of the solid formed by revolving the region bounded by the graphs of y=squareroot of x and y=x^2
after graphing in your graphing calculater you find that you need to use washers
so you get =(pie)S(squareroot of x)^2-(x^2)^2 dxx=1 so (pie)[(1/2)-(1/5)]-03(pie)/10 is your awnser
im having trouble with particle acceleration problems so if anyone wants to help......
The formula for the volume of disks is S (top)^2 - (bottom)^2 dx
The formula for the area of washers is S (top) - (bottom)
The steps are:
1. Draw the graphs of the equations
2. Subtract top graph's equation by the bottom graph's equation(in disks each equation would be squared)
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. Integrate
volume by disks:
the formula is pi times the integral of the [function given] squared times dx. so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one. then graph
volume by washers:
the formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the in between number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. then graph.
LRAM is left hand approximation and the formula is:
delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
Say you are asked to calculate the left Riemann Sum for -4x -5 on the interval [-3, -1] divided into 2 subintervals.
delta x would equal: -1+3 /2 = 2/2 = 1
1[ f(-3) + f(-3 +1)]
1[ f( -3) + f(-2)]
then plug into your equation
RRAM is right hand approximation and the formula is:
delta x [ f(a + delta x) + .... + f(b)]
so using the same example:
1[ f( -2) + f(-1)] and then plug into your equation
MRAM is to calculate the middle and the formula is:
delta x [ f(mid) + f(mid) + .... ]
To find midpoints, you would add the two numbers together then divide by two
In this problem the numbers would be: -3 , -2, -1
-3 + -2/ 2 = -5/2 and -2 + -1 / 2 = -3/2
so 1[f(-5/2) + f(-3/2)] and the plug in
Trapezoidal is different because instead of multiplying by delta x, you multiply by delta x/2 and you also have on more term then your number of subintervals.
The formula is : delta x/2 [f(a) + 2f(a + delta x) + 2f(a+ 2 delta x) + ....f(b)]
For this problem: 1/2 [ f(-3) + 2 f(-2) + f( -1)] and then plug in.
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
e integration:
whatever is raised to the e power will be your u and du will be the derivative of u. For example:
e^2x-1dx
u=2x-1 du=2
rewrite the function as:
1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
related rates:
The steps for related rates are….
1. Pick out all variables
2. Pick out all equations
3. Pick out what you are looking for
4. Sketch a graph and label
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve
limits:
Rules for Limits:…
1. if the degree of top equals the degree of bottom, the answer is the top coefficient over bottom coefficient
2. if top degree is bigger than bottom degree, the answer is positive or negative infinity
2. if top degree is less than bottom degree, the answer is 0
To find area between curves:
The formula you use is b(int)a (top eq.) - (bottom eq.).
If a and b is not already given to you, then you much set the equations equal to each other and solve.
You find which equation is top/bottom by graphing both and simply looking to see which one is on top.
If the area is on the y-axis, then the a and b values need to be set as y-values, and the equations must be solved for x.
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.
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))
Find the volume of the solid formed by revolving the region bounded by the graphs of y=squareroot of x and y=x^2
after graphing in your graphing calculater you find that you need to use washers
so you get =(pie)S(squareroot of x)^2-(x^2)^2 dxx=1 so (pie)[(1/2)-(1/5)]-03(pie)/10 is your awnser
im having trouble with particle acceleration problems so if anyone wants to help......
30th
this week of course we took more ap test.....
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.
related rates:
-Write down the given
-figure out which formula they want you to use
-plug in the given
-take the derivative
-solve for the unknown
i need help with particle problems and knowing when to intergrate it or take its deriv to get to something like acceleration or velocity, etc.
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.
related rates:
-Write down the given
-figure out which formula they want you to use
-plug in the given
-take the derivative
-solve for the unknown
i need help with particle problems and knowing when to intergrate it or take its deriv to get to something like acceleration or velocity, etc.
AVERAGE SPEED
something i been having trouble with so ill do a example problem
a bag of flower is dropped off of a roof on to the car, what is the average speed during the first two seconds of falling. Given: y=16t^2 to describe the fall.
x=(0,2)- this represents the first two secons.
to find your y's you would plug the first x and then the second x into the given equation.
f(0)=0 and f(2)=16(4)=64
after completin those steps, you plug in to your slope formula.
(y2-y1)/(x2-x1)=(64-0)/(2-0)= 32
the average speed of this particular problem would be 32 m/s.
x=(0,2)- this represents the first two secons.
to find your y's you would plug the first x and then the second x into the given equation.
f(0)=0 and f(2)=16(4)=64
after completin those steps, you plug in to your slope formula.
(y2-y1)/(x2-x1)=(64-0)/(2-0)= 32
the average speed of this particular problem would be 32 m/s.
FIRST DERIVATIVE TEST STEPS AND EXAMPLE
1. take the deriv.
2. set the deriv. equal to zero.
3. solve for the xmax, mins, horizontal tangents and the critical points, (if you don't know how to do this i'll explain in a second.)
4. set up intervals using the step above.
5. plug in to the first deriv. (which is why it's called the 1st deriv. test)
6. plug in values from above to the original function to find the absolute max and min, but only do this if it asks for it.
EXAMPLE:
x^2-6x+8
2x-6
2x-6=0
x=3 is the critical point.
the intervals are (-infinity, 3)u(3,infinity)
.. after you set up the intervals, plug in numbers within the intervals to see whether or not the intervals are increasing or decreasing and if they are a max or min. so in this case you could plug in 2 and 4.
2. set the deriv. equal to zero.
3. solve for the xmax, mins, horizontal tangents and the critical points, (if you don't know how to do this i'll explain in a second.)
4. set up intervals using the step above.
5. plug in to the first deriv. (which is why it's called the 1st deriv. test)
6. plug in values from above to the original function to find the absolute max and min, but only do this if it asks for it.
EXAMPLE:
x^2-6x+8
2x-6
2x-6=0
x=3 is the critical point.
the intervals are (-infinity, 3)u(3,infinity)
.. after you set up the intervals, plug in numbers within the intervals to see whether or not the intervals are increasing or decreasing and if they are a max or min. so in this case you could plug in 2 and 4.
And i still really need help with integrating
its just getting harder
Post something.
So this week in Calculus we took two new AP tests which came from a Texas website...I think...
Anyway, compared to tests we have taken before, these were a bit harder I would say...
Something they really liked to do was tangent inverse and sin inverse integration. For tangent, usually
dx/(1+x^2) will integrate to tan inverse. However, they were doing things such as dx/(1+4x^2).
The trick to this is to notice that 4x^2 can be changed to (2x)^2. Now, using substitution, u = 2x and du = 2. So, accounting for the du of 2 which should be in the top, this now becomes
(1/2)arctan(2x) + C.
Also, for
dx/(sqrt(a^2 - x^2)
this will become arcsin(x/a) + C
So for instance, the integral of dx/(sqrt(4-x^2)) would be arcsin(x/2) + C
Other than that, this test focused a lot on volume.
Something that some people are still a little confused about is when revolving about the y axis.
Basically, whatever you are rotating about is what you need to have your equation in terms of...so if you are rotating about the y axis, the function would be f(y)=y^2+1, If it was rotated about the x axis, it would be sqrt(x-1) = y, These are important steps.
Also, if you change it from f(x) to f(y), and you don't have bounds, you have to set the two f(y)'s equal to each other and solve for your equations.
From here on you will just use the same steps as normal...which is on top, and if its washers, top squared minus bottom squared..
Anyway, those are still a little bit confusing to some people but we will see how it goes...
Gotta go to a birthday party...
Anyway, compared to tests we have taken before, these were a bit harder I would say...
Something they really liked to do was tangent inverse and sin inverse integration. For tangent, usually
dx/(1+x^2) will integrate to tan inverse. However, they were doing things such as dx/(1+4x^2).
The trick to this is to notice that 4x^2 can be changed to (2x)^2. Now, using substitution, u = 2x and du = 2. So, accounting for the du of 2 which should be in the top, this now becomes
(1/2)arctan(2x) + C.
Also, for
dx/(sqrt(a^2 - x^2)
this will become arcsin(x/a) + C
So for instance, the integral of dx/(sqrt(4-x^2)) would be arcsin(x/2) + C
Other than that, this test focused a lot on volume.
Something that some people are still a little confused about is when revolving about the y axis.
Basically, whatever you are rotating about is what you need to have your equation in terms of...so if you are rotating about the y axis, the function would be f(y)=y^2+1, If it was rotated about the x axis, it would be sqrt(x-1) = y, These are important steps.
Also, if you change it from f(x) to f(y), and you don't have bounds, you have to set the two f(y)'s equal to each other and solve for your equations.
From here on you will just use the same steps as normal...which is on top, and if its washers, top squared minus bottom squared..
Anyway, those are still a little bit confusing to some people but we will see how it goes...
Gotta go to a birthday party...
Post #30
So this week we continued taking AP practice TESTS. They were kinda hard, but we’ve learned a lot! After the non-calculator portion Monday, we took the calculator portion on Tuesday. We got them back on Wednesday and worked out the entire two test for 3 days and [for me] it helped to know new tricks and step by step instructions. BIG THANKS goes to JOHN!!!
Okay, well let’s go over some stuff shall we:
related rates:
Steps:
1. Identify all variables and equations
2. Identify what you are looking for
3. Make a sketch and label
4. Write an equation(s) involving your variables (only have 1 unknown)
5. Take the derivative with respect to TIME!
6. Substitute in the Derivative and solve
limits:
Rule #1 - When the degree (exponent) of the bottom is GREATER than the degree of the top, the limit is Zero.
Rule #2 - When the degree (exponent) of the bottom is SMALLER than the degree of the top, the limit is infinity. (positive or negative)
Rule #3 - When the degrees are equal, the limit is the coeffecients.
linierazation:
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
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
Second Derivative:
first you find the first derivative and solve it for dy/dx by using the steps for the first derivative steps.
you then take the second derivative of the solved equation. Plugging in d^2y/d^2x everytime you take the derivative of y again. and where you have dy/dx you plug in your solved equation for that.
once you have everything plugged in and ready to go you then solve for d^2y/d^2x
Intermediate Value Theorem:
1. if f is continuous on [a,b] and k is any number between f(a)and f(b), then there is at least 1 number c when f(c)=k.
* basically you cannot skip any y value
HOW TO FIND THE EQUATION OF A TANGENT LINE: [I always miss this one or do it wrong for some reason…so I went look up the directions of how to solve them.]
1. take f1(x)
2. plug x in to find your slope m
3. plug x into f(x)to get y
4. using m and (x,y) plug it into the equation (y-y1)=m(x-x1).
Can anyone explain implicit derivatives when there’s like, 4x-2xy+3y^2….the 2xy kinda gets me off…
Oh, and the questions that give me a graph then I have to find the areas which means I have to break it all up and then put all the areas together…can we go over that?
~ElliE~
Okay, well let’s go over some stuff shall we:
related rates:
Steps:
1. Identify all variables and equations
2. Identify what you are looking for
3. Make a sketch and label
4. Write an equation(s) involving your variables (only have 1 unknown)
5. Take the derivative with respect to TIME!
6. Substitute in the Derivative and solve
limits:
Rule #1 - When the degree (exponent) of the bottom is GREATER than the degree of the top, the limit is Zero.
Rule #2 - When the degree (exponent) of the bottom is SMALLER than the degree of the top, the limit is infinity. (positive or negative)
Rule #3 - When the degrees are equal, the limit is the coeffecients.
linierazation:
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
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
Second Derivative:
first you find the first derivative and solve it for dy/dx by using the steps for the first derivative steps.
you then take the second derivative of the solved equation. Plugging in d^2y/d^2x everytime you take the derivative of y again. and where you have dy/dx you plug in your solved equation for that.
once you have everything plugged in and ready to go you then solve for d^2y/d^2x
Intermediate Value Theorem:
1. if f is continuous on [a,b] and k is any number between f(a)and f(b), then there is at least 1 number c when f(c)=k.
* basically you cannot skip any y value
HOW TO FIND THE EQUATION OF A TANGENT LINE: [I always miss this one or do it wrong for some reason…so I went look up the directions of how to solve them.]
1. take f1(x)
2. plug x in to find your slope m
3. plug x into f(x)to get y
4. using m and (x,y) plug it into the equation (y-y1)=m(x-x1).
Can anyone explain implicit derivatives when there’s like, 4x-2xy+3y^2….the 2xy kinda gets me off…
Oh, and the questions that give me a graph then I have to find the areas which means I have to break it all up and then put all the areas together…can we go over that?
~ElliE~
POST #30
AREA BETWEEN A CURVE:
FORMULA b(int)a (top eq.) - (bottom eq.): If a and b is not already given to you, then you much set the equations equal to each other and solve.Find which equation is top/bottom by graphing both and simply looking to see which one is on top.If the area is on the y-axis, then the a and b values need to be set as y-values, and the equations must be solved for x.
TANGENT LINE
1. find the x of your original function (usually given).
2. find the y of your original function (plug in x)
3. take derivative of original function.
4. plug x into derived function.
5. set up slope-intercept form.
NORMAL LINE
SAME FIST 4 STETPS..
STEP 5: make the slope perpendicular by flipping the fraction and making it negative.
PROMBLEM FROM AP: Compute the integral of 4x^2 (x^3+4) ^1/2 dx
u = x^3 +4 du= 3x^2
Get rid of the 3 in du, which you do by dividing something by 3 and gain a 4 which you do by multiplying by 4:
4/3 S u^1/2
4/3 (2/3 ) u ^3/2
8/9 ( x^3 +4) ^3/2 + c
ANSWER = C.
INTERMEDIATE VALUE THEOREM:
If f is continuous on [a,b] and k is any number between f(a)and f(b), then there is at least 1 number c when f(c)=k.
i don't understand the area under the curve problems on the AP test, my answer never comes out right.
FORMULA b(int)a (top eq.) - (bottom eq.): If a and b is not already given to you, then you much set the equations equal to each other and solve.Find which equation is top/bottom by graphing both and simply looking to see which one is on top.If the area is on the y-axis, then the a and b values need to be set as y-values, and the equations must be solved for x.
TANGENT LINE
1. find the x of your original function (usually given).
2. find the y of your original function (plug in x)
3. take derivative of original function.
4. plug x into derived function.
5. set up slope-intercept form.
NORMAL LINE
SAME FIST 4 STETPS..
STEP 5: make the slope perpendicular by flipping the fraction and making it negative.
PROMBLEM FROM AP: Compute the integral of 4x^2 (x^3+4) ^1/2 dx
u = x^3 +4 du= 3x^2
Get rid of the 3 in du, which you do by dividing something by 3 and gain a 4 which you do by multiplying by 4:
4/3 S u^1/2
4/3 (2/3 ) u ^3/2
8/9 ( x^3 +4) ^3/2 + c
ANSWER = C.
INTERMEDIATE VALUE THEOREM:
If f is continuous on [a,b] and k is any number between f(a)and f(b), then there is at least 1 number c when f(c)=k.
i don't understand the area under the curve problems on the AP test, my answer never comes out right.
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