Before i start talking about Calculus i hope everyone had a great week off! To start this blog off, i'll go over the things i definitely know:
First derivative test:
In the problem, they will give you a function, and they will ask if it is increasing, decreasing, a max or a min. To do this you have to use the first derivative test. you take the derivative of the original function and solve for your x values (critical points). Then you set those points up into intervals and then plug in numbers between those intervals into the derivative. if the answer is positive it is increasing, if it is negative it is decreasing.
Second derivative test:
it will be set up like the first derivative test except they will ask if the function is concave up or concave down. They may also ask where there is a point of inflection. To do this you must use the second derivative test. You take the derivative of the function twice and solve for your critical values and set them up into intervals. You then plug numbers between those intervals into your second derivative to see where there is a change in concavity. If the answer is positive it is concave up; if the answer is negative it is concave down; where there is a change in concavity there is a point of inflection.
Limit Rules:
1. If the degree on top is smaller than the degree on the botton, then the limit is zero.
2. If the degree on top is bigger than the degree on the bottom, then the limit is infinity.
3. If the degree on top is the same as the degree on the bottom, you divide the coefficients to get the limit.
If they give you a limit and ask what is the limit as a h goes to a number, you must take the derivative of the number behind what is in the parenthesis on top, then substitute the number in for any x value.
Some things i still do not understand:
the particle problems
Angle of elevation
related rates
See everyone on Monday :)
Saturday, February 20, 2010
Friday, February 19, 2010
Post #27
I'm going to continue with more frequently missed questions.
Product Rule
Product rule is used when anything two things are being multiplied together.
Product rule states copy the first times the derivative of the second + copy the second times the derivative of the first or uv' + vu'
Example: Find the derivative of t sin (t)
t (cos t) + sin (t) (1) = t cos t + sin t
Example 2: cos x sin x
cos x (cos x) + (sin x) ( - sin x)
cos ^2 x - sin^2 x
Chain rule
Use chain rule when you are finding the derivative with exponents.
To do chain rule, work from the outside in.
Example: cos^2 (x^2)
It helps me to rewrite it as cos (x^2) ^2
2 (cos (x^2)) (-sin (x^2) (2x)
-4x cos (x^2) sin (x^2)
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
Example 2: Integrate cos x/ sin x
The derivative of sin x is cos x so
ln |sin x | + c
Equation of a normal line
To find the equation of a normal line, find the perpendicular slope and plug into the point slope formula.
Example: An equation of the line normal to the graph of y= (3x^2 + 2x) ^1/2 at (2,4) is
First step is to take the derivative and plug in x to find the slope.
1/2 (3x^2 + 2x ) ^-1/2 (6x + 2)
6(2) + 2 / 2 (3 (2)^2 + 2(2) )^1/2
m = 14/ 8 or 7/4
Now take the negative reciprocal which is -4/7
Since you are given a point and now have your slope, just plug in.
y-4 = -4/7 (x -2 )
You can simply that until you get an answer choice.
I still need help on numbers like 30 on the calculator portion of the last test we took. It asks how fast is the distance between the boats increasing and also number 44 that says find the distance traveled.
Product Rule
Product rule is used when anything two things are being multiplied together.
Product rule states copy the first times the derivative of the second + copy the second times the derivative of the first or uv' + vu'
Example: Find the derivative of t sin (t)
t (cos t) + sin (t) (1) = t cos t + sin t
Example 2: cos x sin x
cos x (cos x) + (sin x) ( - sin x)
cos ^2 x - sin^2 x
Chain rule
Use chain rule when you are finding the derivative with exponents.
To do chain rule, work from the outside in.
Example: cos^2 (x^2)
It helps me to rewrite it as cos (x^2) ^2
2 (cos (x^2)) (-sin (x^2) (2x)
-4x cos (x^2) sin (x^2)
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
Example 2: Integrate cos x/ sin x
The derivative of sin x is cos x so
ln |sin x | + c
Equation of a normal line
To find the equation of a normal line, find the perpendicular slope and plug into the point slope formula.
Example: An equation of the line normal to the graph of y= (3x^2 + 2x) ^1/2 at (2,4) is
First step is to take the derivative and plug in x to find the slope.
1/2 (3x^2 + 2x ) ^-1/2 (6x + 2)
6(2) + 2 / 2 (3 (2)^2 + 2(2) )^1/2
m = 14/ 8 or 7/4
Now take the negative reciprocal which is -4/7
Since you are given a point and now have your slope, just plug in.
y-4 = -4/7 (x -2 )
You can simply that until you get an answer choice.
I still need help on numbers like 30 on the calculator portion of the last test we took. It asks how fast is the distance between the boats increasing and also number 44 that says find the distance traveled.
Thursday, February 18, 2010
post 26
here we go
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve. For 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!!!)
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 number 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)
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve. For 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!!!)
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 number 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)
RELATED RATES:
1. identify all variables and equations
2. identify what you are looking for
3. sketch and label
4. write an equations 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 in derivative and solve
Tangent line- they give you a function and an x value. If no y value is given, plug the x value into the original function. To find the slope, you take the derivative of the original function and plug the x value in. Then you set it up into point slope form. y-y1= slope(x-x1). For a normal line, you would take the negative inverse of the slope and solve.
im need help w/ e-integration, tangent lines i can never remember, and thats bout it.Wednesday, February 17, 2010
Post Number Twenty Six
Well, last week at school before Mrs. Robinson’s last day we reviewed some stuff. Needless to say everything actually clicked. I felt like that’s the first time I ever got taught it for some reason, apparently I wasn’t paying attention before. That helped a lot just saying.
Now on to calculus…
e integration:
whatever is raised to the e power will be your u and du will be the derivative of u.
I don’t think it was this easy the whole time. I can’t believe I’ve been skipping these problems!
Max’s and Min’s
Just do the first derivative test!
I finally got average rate of change and average value straight.
Average rate of change:
F(b) – f(a)/b-a
Average Value:
1/b-a S f(x) dx
Derivatives of integrals
To find the derivative of an integral, just plug in b the multiply that by the derivative of b.
position, veloctiy, accelaration.
Just right it like that up and down.
If you move down take derivative, if you move up integrate.
RELATED RATES:
1. identify all variables and equations
2. identify what you are looking for
3. sketch and label
4. write an equations 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 in derivative and solve
Sooooo hopefully B-robs pregnancy goes alright.
Good luck and happy mardi gras!
Kbye.
Post 26
So today is Wednesday, not Sunday, but this is my post 26. Better late than never.
Anyway, last week before B-rob left, we went over a few things that were bothering us about the AP tests. We did two days of hard core review, and it was really helpful. Some of the things that were bothering me included limits, and derivatives of integrals when the bounds have variables.
First of all, I was a little confused about how limits worked when the bottom turned out to be zero. One thing you can do for this is to factor and cancel. Also, if when the number is plugged in and you get zero over zero, you can use L’Hopital’s Rule. For this rule, you can take the derivative of the top and the bottom separately and try to plug in until you get a number instead of zero over zero.
For derivatives of integrals I knew the derivative of the integral canceled each other out, but I didn’t know what to do when the bounds had variables in them. I learned that you plug in the b variable then solve. For example if a problem asks to take the derivative of S from 0-x^2 of sin(t), this is how you do it
First, the derivative and the integral cancel each other out, so you just plug in only your b bounds
sin(x^2) then you have to remember to multiply by the derivative of the insde X 2x
= 2xsin(x^2)
Knowing this will help a lot on the AP now.
Some things I still don’t understand include particle problems. We go over these time and time again in class, and I seem to understand them when we’re going over them, but on the test, I don’t know, it seems like they change or I forget. Help?
Anyway, last week before B-rob left, we went over a few things that were bothering us about the AP tests. We did two days of hard core review, and it was really helpful. Some of the things that were bothering me included limits, and derivatives of integrals when the bounds have variables.
First of all, I was a little confused about how limits worked when the bottom turned out to be zero. One thing you can do for this is to factor and cancel. Also, if when the number is plugged in and you get zero over zero, you can use L’Hopital’s Rule. For this rule, you can take the derivative of the top and the bottom separately and try to plug in until you get a number instead of zero over zero.
For derivatives of integrals I knew the derivative of the integral canceled each other out, but I didn’t know what to do when the bounds had variables in them. I learned that you plug in the b variable then solve. For example if a problem asks to take the derivative of S from 0-x^2 of sin(t), this is how you do it
First, the derivative and the integral cancel each other out, so you just plug in only your b bounds
sin(x^2) then you have to remember to multiply by the derivative of the insde X 2x
= 2xsin(x^2)
Knowing this will help a lot on the AP now.
Some things I still don’t understand include particle problems. We go over these time and time again in class, and I seem to understand them when we’re going over them, but on the test, I don’t know, it seems like they change or I forget. Help?
post 26
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.
Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
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)
The terms for the First Derivative Test:
1. Increasing
2. Decreasing
3. Horizontal Tangent
4. Min/Max
Steps:
1. Take the derivative
2. Set it equal to zero
3. Solve for x to get the possible critical points [[can someone clarify this part???? My notebook says it equals critical points AND mins/maxs/horizontal tangents]]
4. Set up intervals with your x value(s)
5. Plug into your first derivative
6. To find an absolute extrema, plug in the values from step 5 into your original function
linearization.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
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 dont understand e integration still...
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.
Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
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)
The terms for the First Derivative Test:
1. Increasing
2. Decreasing
3. Horizontal Tangent
4. Min/Max
Steps:
1. Take the derivative
2. Set it equal to zero
3. Solve for x to get the possible critical points [[can someone clarify this part???? My notebook says it equals critical points AND mins/maxs/horizontal tangents]]
4. Set up intervals with your x value(s)
5. Plug into your first derivative
6. To find an absolute extrema, plug in the values from step 5 into your original function
linearization.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
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 dont understand e integration still...
Monday, February 15, 2010
post 26
For this post i am going to go into detail on the second derivative test to find all possible points of inflection and intervals of concavity. remember, points of inflection only happen where there is a change of concavity.
Example: f'(x)= 6/(x^(2)+3)
First, you have to take the derivative of that, and you have to use the quotient rule, so the beginning of the problem will look like, [(x^(2)+3)(0)-6(2x)]/(x^(2)+3)^(2), which simplifies to -12x/(x^(2)+3)^2 remember, that was just the first derivative.
Second step is to take the derivative of the first derivative, that would make this step called taking the second derivative.
Once again u need to use the quotient rule, so f''(x)={(x2+3)^2-(12)-[(-12x)2(x^(2)+3)2x} all that over (x^(2)+3)^4 then you get a bunch of stuff, then you simplify, then you cancel, so I am just going to type the end answer of the second derivative. Which is, (3)(6)(x+1)(x-1) all over (x^2+3)^3
The possible points of inflection are found in the numerator of the finished second derivative, in this case, if you look, it would be x=1, and x=-1
so then you set up your points, (-infinity, -1) u (-1, 1) u (1, infinity)
then you plug in. f''(-2)= positive value f''(0)=negative value f''(2)=positive value
then you know that your intervals concave up at (-infinity, -1) u (1,infinity) or x<1,>1
and it is concave down at (-1,1) or -1
and you're points of inflection are x=-1, and x=1
that's it for the second derivative test, the only thing i have any problems with is substition... so yeah..
Example: f'(x)= 6/(x^(2)+3)
First, you have to take the derivative of that, and you have to use the quotient rule, so the beginning of the problem will look like, [(x^(2)+3)(0)-6(2x)]/(x^(2)+3)^(2), which simplifies to -12x/(x^(2)+3)^2 remember, that was just the first derivative.
Second step is to take the derivative of the first derivative, that would make this step called taking the second derivative.
Once again u need to use the quotient rule, so f''(x)={(x2+3)^2-(12)-[(-12x)2(x^(2)+3)2x} all that over (x^(2)+3)^4 then you get a bunch of stuff, then you simplify, then you cancel, so I am just going to type the end answer of the second derivative. Which is, (3)(6)(x+1)(x-1) all over (x^2+3)^3
The possible points of inflection are found in the numerator of the finished second derivative, in this case, if you look, it would be x=1, and x=-1
so then you set up your points, (-infinity, -1) u (-1, 1) u (1, infinity)
then you plug in. f''(-2)= positive value f''(0)=negative value f''(2)=positive value
then you know that your intervals concave up at (-infinity, -1) u (1,infinity) or x<1,>1
and it is concave down at (-1,1) or -1
and you're points of inflection are x=-1, and x=1
that's it for the second derivative test, the only thing i have any problems with is substition... so yeah..
Post #26
Sorry for posting this late but the parade yesterday wiped me out.
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))
Rolle's Theorem
To understand Rolle's Theorem, you must first understand the Extreme Value Theorem. The Extreme Value Theorem states that a continuous function on the closed interval [a,b] must have both a maximum and a minimum. They can be on the endpoint a and b though.
So Rolle's Theorem gives the conditions that guarantee the existence of an extrema in the interior of a closed interval.
Rolle's Theorem-If a function is continuous on the closed interval [a,b] and differentiable on the open interval (a,b) and f(a)=f(b), then there is at least one point, designated "c", where the derivative of f(c)=0.
Example: Find all points "c" where the derivative of f(c)=0 f(x)=x^4-2x^2 [-2,2]
1. I first found that f(a) equals f(b).
(-2)^4-(-2)^2=8
(2)^4-2(2)^2=8
2. Once finding that f(a)=f(b), I set the derivative equal to 0 and solved for x.
4x^3-4x=0
4x(x^2-1)=0
4x=0 x^2-1=0
x=-1,0,1
There are three point of extrema within the [-2,2] of the function x^4-2x^2. One at x=-1 another at x=0 and the last at x=1.
Mean Value Theorem
The Mean Value Theorem states that If the function f(x) is continuous 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).
This is essentially the same thing as average speed, which finds the slope of the equation.
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))
Rolle's Theorem
To understand Rolle's Theorem, you must first understand the Extreme Value Theorem. The Extreme Value Theorem states that a continuous function on the closed interval [a,b] must have both a maximum and a minimum. They can be on the endpoint a and b though.
So Rolle's Theorem gives the conditions that guarantee the existence of an extrema in the interior of a closed interval.
Rolle's Theorem-If a function is continuous on the closed interval [a,b] and differentiable on the open interval (a,b) and f(a)=f(b), then there is at least one point, designated "c", where the derivative of f(c)=0.
Example: Find all points "c" where the derivative of f(c)=0 f(x)=x^4-2x^2 [-2,2]
1. I first found that f(a) equals f(b).
(-2)^4-(-2)^2=8
(2)^4-2(2)^2=8
2. Once finding that f(a)=f(b), I set the derivative equal to 0 and solved for x.
4x^3-4x=0
4x(x^2-1)=0
4x=0 x^2-1=0
x=-1,0,1
There are three point of extrema within the [-2,2] of the function x^4-2x^2. One at x=-1 another at x=0 and the last at x=1.
Mean Value Theorem
The Mean Value Theorem states that If the function f(x) is continuous 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).
This is essentially the same thing as average speed, which finds the slope of the equation.
once again
Okay, Review of old stuff..its what I do!!
Complex Derivatives y=ln(e^x) (Chain Rule)
First off one should should identify the steps of your problem. In this case they would be:
1. Natural Log
2. e^x
you problem should be (1/(e^x)).(e^x)'
then you find the derivative of e^x which is e^x . x' (x'=1)
so your final problem should be (1/(e^x)).(e^x)
After this you have to simplify algebraically, giving you (e^x)/(e^x) ,which equals 1.
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve. For 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!!!)
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 number 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!!!)
UM....okay.. I need help on velocity and position problems. I get the whole concept..and general thing. but some of those problems on those AP's..yeah..I see a keyword and skip it...so if you could help me...much appreciated.
and keeping with chelsea's comment, do we have to still do comments and just send them to brob..or what???? Well, have a good holidays.
Complex Derivatives y=ln(e^x) (Chain Rule)
First off one should should identify the steps of your problem. In this case they would be:
1. Natural Log
2. e^x
you problem should be (1/(e^x)).(e^x)'
then you find the derivative of e^x which is e^x . x' (x'=1)
so your final problem should be (1/(e^x)).(e^x)
After this you have to simplify algebraically, giving you (e^x)/(e^x) ,which equals 1.
First Derivative Test:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve. For 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!!!)
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 number 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!!!)
UM....okay.. I need help on velocity and position problems. I get the whole concept..and general thing. but some of those problems on those AP's..yeah..I see a keyword and skip it...so if you could help me...much appreciated.
and keeping with chelsea's comment, do we have to still do comments and just send them to brob..or what???? Well, have a good holidays.
Comments?
Do we have to do comments this week?
If so, I am having trouble with the problems with volume, related rates, and the find the distance problems. I forgot to post my questions in my blog.
If so, I am having trouble with the problems with volume, related rates, and the find the distance problems. I forgot to post my questions in my blog.
26
formulas for derivatives
d/dx c=0 (c is a #)
d/dx cu=cu' (c is #)
d/dx cx=c (c is a #)
d/dx u+v=u'+v'
d/dx uv=uv'+vu'
d/dx u/v=(vu'-uv')/v^2
d/dx sinx=cosx(x')
d/dx cosx=-sinx(x')
d/dx tanx=sec^2x(x')
d/dx secx=secxtanx(x')
d/dx cscx=-cscxtanx(x')
d/dx cotx=-csc^2x(x')
d/dx lnu= 1/u(u')
d/dx e^u=e^u(u')
***just remember that integration is opposite for all these derivative formulas. ***
Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
Im having a little trouble with intergration of trig functions. I need some help with that.
d/dx c=0 (c is a #)
d/dx cu=cu' (c is #)
d/dx cx=c (c is a #)
d/dx u+v=u'+v'
d/dx uv=uv'+vu'
d/dx u/v=(vu'-uv')/v^2
d/dx sinx=cosx(x')
d/dx cosx=-sinx(x')
d/dx tanx=sec^2x(x')
d/dx secx=secxtanx(x')
d/dx cscx=-cscxtanx(x')
d/dx cotx=-csc^2x(x')
d/dx lnu= 1/u(u')
d/dx e^u=e^u(u')
***just remember that integration is opposite for all these derivative formulas. ***
Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
Im having a little trouble with intergration of trig functions. I need some help with that.
post 27, part 2
ok so i accidentally posted # 27 before i was finished with it. ahah, weird, i know. anyways, here's the rest of post 27.
chain rule:
so it's easier than it seems, right?
we also went over rules for limits approaching 0.
first you plug in, & see what you get. if it ends up somehow with you dividing by zero, (which is 99.9% of the time), then here are the steps you take:
1st try to factor & cancel. if that doesn't work then,
2nd take derivative of top and plug in zero,
exampe: lim sin(x) + 6
x -> 0 x
take derivative of sin(x) = cos(x)
plug in zero, cos(0) = 1. so your answer is 1.
i need help with tan inverse integration. i know she went over it, but i forgot my binder @ school and i don't remember some of it. can someone go over that with me, please? :)
chain rule:
so it's easier than it seems, right?
we also went over rules for limits approaching 0.
first you plug in, & see what you get. if it ends up somehow with you dividing by zero, (which is 99.9% of the time), then here are the steps you take:
1st try to factor & cancel. if that doesn't work then,
2nd take derivative of top and plug in zero,
exampe: lim sin(x) + 6
x -> 0 x
take derivative of sin(x) = cos(x)
plug in zero, cos(0) = 1. so your answer is 1.
i need help with tan inverse integration. i know she went over it, but i forgot my binder @ school and i don't remember some of it. can someone go over that with me, please? :)
Post #26
This week in calculus, we reviewed topics we should know how to do because they show up on every AP.
Limits:
LIm as x-> c The first step in finding a limit, is to plug in c. If that gives you zero at the bottom, it does not mean the limit doesn't exist. Next thing to try is to factor and cancel the top and bottom then plug in c again. If that still gives you zero, you have to use L'Hopital's rule. L'hopital's rule states if lim as x->c = 0/0, take the derivative of the top and the derivative of the bottom and plug in c. If you still get undefined, take the derivative of each again until you don't.
Example:
lim as x-> 4 3(x-4) / x^2 - 16
First step is plug in 4: 3(0) / 0 = 0/0
Factor and cancel: 3(x-4) / (x+4) (x-4) = 3/ (x+4)
Now plug in: 3/ (4+4) = 3/8, the limit is 3/8
If that would not have worked, you could have used L'hopital's rule
Take derivatve: 3/ 2x and plug in: 3/(4)(2) = 3/8
Example 2:
lim as x-> 0 4 sinx cosx - sinx / x^2
Plugging in gives you 0/0.
Factoring leaves you with 0/0.
So the only thing you can do is L'hopital's rule
Top factored: sinx (-sinx) + cosx (cosx) - cosx
Bottom factored: 2x
Still get 0 at the bottom so take derivative again
(-sin x)^2 - cos x
2 (-sin x) (cos x) + sin x
-2 sin x cos x + sin x / 2
Now plug in 0
-2 (0) (1) + (0) / 2 = 0
The limit as x-> 0 is 0.
Derivatives of integrals
To find the derivative of an integral, just plug in b the multiply that by the derivative of b.
Example:
f(x) = the integral of sin (t) on [0,x^2] Find f'(x)
sin (x^2) (2x)
f'(x) = 2x sin (x^2)
Definition of a derivatve
To recognize the definition of a derivative, the problem will look like
lim as a variable -> 0 ( + variable) - something / variable
To solve this, just take the derivative and plug in the number used.
Example: lim g -> 0 sin(pi +g) - sin (pi) / g
Take the derivative of sin (x) so cos (x)
now plug in pi - cos (pi) = -1
Example 2: lim as h -> 0 cos ( pi/4 + h) - cos (pi/4) / h
Derivative of cos (x) = -sin (x)
plug in pi / 4 = - sin (pi/4) = - square root of 2 / 2
Limits:
LIm as x-> c The first step in finding a limit, is to plug in c. If that gives you zero at the bottom, it does not mean the limit doesn't exist. Next thing to try is to factor and cancel the top and bottom then plug in c again. If that still gives you zero, you have to use L'Hopital's rule. L'hopital's rule states if lim as x->c = 0/0, take the derivative of the top and the derivative of the bottom and plug in c. If you still get undefined, take the derivative of each again until you don't.
Example:
lim as x-> 4 3(x-4) / x^2 - 16
First step is plug in 4: 3(0) / 0 = 0/0
Factor and cancel: 3(x-4) / (x+4) (x-4) = 3/ (x+4)
Now plug in: 3/ (4+4) = 3/8, the limit is 3/8
If that would not have worked, you could have used L'hopital's rule
Take derivatve: 3/ 2x and plug in: 3/(4)(2) = 3/8
Example 2:
lim as x-> 0 4 sinx cosx - sinx / x^2
Plugging in gives you 0/0.
Factoring leaves you with 0/0.
So the only thing you can do is L'hopital's rule
Top factored: sinx (-sinx) + cosx (cosx) - cosx
Bottom factored: 2x
Still get 0 at the bottom so take derivative again
(-sin x)^2 - cos x
2 (-sin x) (cos x) + sin x
-2 sin x cos x + sin x / 2
Now plug in 0
-2 (0) (1) + (0) / 2 = 0
The limit as x-> 0 is 0.
Derivatives of integrals
To find the derivative of an integral, just plug in b the multiply that by the derivative of b.
Example:
f(x) = the integral of sin (t) on [0,x^2] Find f'(x)
sin (x^2) (2x)
f'(x) = 2x sin (x^2)
Definition of a derivatve
To recognize the definition of a derivative, the problem will look like
lim as a variable -> 0 ( + variable) - something / variable
To solve this, just take the derivative and plug in the number used.
Example: lim g -> 0 sin(pi +g) - sin (pi) / g
Take the derivative of sin (x) so cos (x)
now plug in pi - cos (pi) = -1
Example 2: lim as h -> 0 cos ( pi/4 + h) - cos (pi/4) / h
Derivative of cos (x) = -sin (x)
plug in pi / 4 = - sin (pi/4) = - square root of 2 / 2
post 27
i'm doing my second post for over the holiday's now because i'm going to be in disney world at national danceteam competition :) everyone wish us luck! but anyways, i'm doing it now because i would probably forget about doing it sunday.
alright, so this is just going to be a continuation of post 26, i'm just gonna go over all the helpful stuff that b rob went over with us on tuesday & wednesday.
chain rule :
this helped me a lot actually. so pretty much the rule is work from the outside in, and whenever there is nothing left, that is when you STOP. don't keep differentiating.
example:
cos^2(3x + 2)
ok so remember, work from the outside to the inside. the biggest thing here, is the EXPONENT. so, cos^2(3x + 2) = 2cos(3x + 2),
then the next biggest thing is the trig function. so 2cos(3x + 2) = 2-sin(3x + 2)
after that, you deal with the last thing.. whatever is in paranthesis. 2-sin(3x + 2) = -2sin(3x + 2) times (3)
so your final answer would be -6sin(3x + 2).
alright, so this is just going to be a continuation of post 26, i'm just gonna go over all the helpful stuff that b rob went over with us on tuesday & wednesday.
chain rule :
this helped me a lot actually. so pretty much the rule is work from the outside in, and whenever there is nothing left, that is when you STOP. don't keep differentiating.
example:
cos^2(3x + 2)
ok so remember, work from the outside to the inside. the biggest thing here, is the EXPONENT. so, cos^2(3x + 2) = 2cos(3x + 2),
then the next biggest thing is the trig function. so 2cos(3x + 2) = 2-sin(3x + 2)
after that, you deal with the last thing.. whatever is in paranthesis. 2-sin(3x + 2) = -2sin(3x + 2) times (3)
so your final answer would be -6sin(3x + 2).
post 26
HAPPY VALENTINES DAY! hope everyone had a good one :)
ok so this week in calculus we took 2 more ap tests & then did worksheets on thursday and friday after brob went on maternity leave. we also did a lot of helpful stuff on monday and tuesday. i'll go over some of that
ok so e integration:
S e^3x
all you do is recopy the problem e^3x, then multiply it by the derivative of the exponent.. (3)
also we learned about position, veloctiy, accelaration.
if you are moving down that list.. for example given velocity, asked for acceleration, then take the derivative. DOWN - DERIVATIVE
if you are moving up that list...ex: given velocity asked for position, then integrate. UP - INTEGRATE
also we learned about ln integration. if the top is the derivative of the bottom, then it's ln integration
example:
S 2x/x^2, then your answer would be ln(x^2) + C
also, if you need to substitute anything in, like a negative or a number, just put it in front of the ln.
example:
S -2x/x^2, then your answer would be -ln(x^2) + C
LRAM,
find delta x. then multiply delta x times [f(a) + f(a+1) +... + (fb)]
i FINALLY UNDERSTAND IT! haha, and RRAM is the same, it just starts with f(b) and ends with f(a).
i don't understand integrals. i get the basics, but whenever i get to more complex problems, i don't even know where to start. no matter if they have trig functions or not, i just don't get it. i need help,like the worksheet we did on thursday, i only knew 1 out of 12 problems. can someone just help me in general with that? find the hardest one you can and just give an example.
ok so this week in calculus we took 2 more ap tests & then did worksheets on thursday and friday after brob went on maternity leave. we also did a lot of helpful stuff on monday and tuesday. i'll go over some of that
ok so e integration:
S e^3x
all you do is recopy the problem e^3x, then multiply it by the derivative of the exponent.. (3)
also we learned about position, veloctiy, accelaration.
if you are moving down that list.. for example given velocity, asked for acceleration, then take the derivative. DOWN - DERIVATIVE
if you are moving up that list...ex: given velocity asked for position, then integrate. UP - INTEGRATE
also we learned about ln integration. if the top is the derivative of the bottom, then it's ln integration
example:
S 2x/x^2, then your answer would be ln(x^2) + C
also, if you need to substitute anything in, like a negative or a number, just put it in front of the ln.
example:
S -2x/x^2, then your answer would be -ln(x^2) + C
LRAM,
find delta x. then multiply delta x times [f(a) + f(a+1) +... + (fb)]
i FINALLY UNDERSTAND IT! haha, and RRAM is the same, it just starts with f(b) and ends with f(a).
i don't understand integrals. i get the basics, but whenever i get to more complex problems, i don't even know where to start. no matter if they have trig functions or not, i just don't get it. i need help,like the worksheet we did on thursday, i only knew 1 out of 12 problems. can someone just help me in general with that? find the hardest one you can and just give an example.
Study Guide
Here is an excellent Study Guide in my absence. It includes tips on how to operate your calculator and all important formulas for the exams. http://www.elainetron.com/apcalc/apcalc.pdf
Sunday, February 14, 2010
Post #26
Heyyyyyyyyyyy! Happy Valentines Day. Well I'm hyper, and I just remembered about my blog. So here we goooo...
REMEMBERING:
POSITION
VELOCITY
ACCELERATION
If going down (like position to velocity, or velocity to acceleration) take derivative. If going up (going from acceleration to velocity) take integral. Real simple!
FINALLY I UNDERSTAND THIS SIMPLE THING:
It's called the chain rule. Take derivative of exponent, the derivative of the like sin, cos, tan, ect., then the derivative of inside & remember to simplify.
Example:
sin^2(x^2)
4xsin(x^2)cos(x^2)
RELATED RATES:
1. identify all variables and equations
2. identify what you are looking for
3. sketch and label
4. write an equations 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 in derivative and solve
EXAMPLE:
The variables x and y are differentiable functions of t and are related by the equation y=2x^3-x+4. When x=2, dx/dt=-1. Find dy/dt when x=2.
1. All of your equations are given, go straight to derivative.
2. dy/dt=6x^2dx/dt-dx/dt
3. plug in all your given (in order) to find dy/dt
dy/dt=6(2)^2(-1)-(-1)dy/dt= -23
YOU COULD PROBABLY COMMENT ON:
So I don't have all my ap stuff with me, and I don't remember an exact problem I'm having trouble with. Can someone tell me about mean value theorem again? And for some reason I'm still having trouble with piecewise function things, no matter how many times it is explained. Can someone put it in any simpler terms or something?
REMEMBERING:
POSITION
VELOCITY
ACCELERATION
If going down (like position to velocity, or velocity to acceleration) take derivative. If going up (going from acceleration to velocity) take integral. Real simple!
FINALLY I UNDERSTAND THIS SIMPLE THING:
It's called the chain rule. Take derivative of exponent, the derivative of the like sin, cos, tan, ect., then the derivative of inside & remember to simplify.
Example:
sin^2(x^2)
4xsin(x^2)cos(x^2)
RELATED RATES:
1. identify all variables and equations
2. identify what you are looking for
3. sketch and label
4. write an equations 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 in derivative and solve
EXAMPLE:
The variables x and y are differentiable functions of t and are related by the equation y=2x^3-x+4. When x=2, dx/dt=-1. Find dy/dt when x=2.
1. All of your equations are given, go straight to derivative.
2. dy/dt=6x^2dx/dt-dx/dt
3. plug in all your given (in order) to find dy/dt
dy/dt=6(2)^2(-1)-(-1)dy/dt= -23
YOU COULD PROBABLY COMMENT ON:
So I don't have all my ap stuff with me, and I don't remember an exact problem I'm having trouble with. Can someone tell me about mean value theorem again? And for some reason I'm still having trouble with piecewise function things, no matter how many times it is explained. Can someone put it in any simpler terms or something?
Calculus Post
First off, happy valentine's day to all :o
So this week in Calculus....wait, I wasn't there all week! So unfortunately I came down with a pretty nasty sickness that included things such as coughing blood and not being able to move for hours...but anyway, let's not talk about that.
I did go to school on Friday and worked on an integration worksheet for a few minutes, but we got called out to an assembly.
But anyway, I wanted to point out a few things that I think I remember that a lot of you were having problems with.
When working an integration problem, a lot of times you need to use substitution. Substitution is useful (at least in my eyes) ESPECIALLY when you see an item in the problem and it's derivative is in the problem somewhere else...let me elaborate a bit more.
In the following problem, you should quickly recognize what your u is and that the derivative of it is in the problem also.
Integral of (1/xln(x)).
Now, when we did this problem in class, a lot of people started to try to go to the log base a of u rule and all of that. First, with these types of problems I always try to think if I can split it up...well in this one you can...so let's do that in our head really quick...well since you can't see it, it is
(1/x)(1/ln(x)). So in this problem you should notice two things. The derivative of ln(x) is (1/x). So really we can rewrite this problem as 1/u where u is lnx. So to integrate 1/u, it's ln|u| + c so the answer to our problem is ln|ln(x)| + c.
If you split up problems visually so that you can better recognize the u and the derivative of u, the problems become way easier.
Anyway, hope that helps someone.
So this week in Calculus....wait, I wasn't there all week! So unfortunately I came down with a pretty nasty sickness that included things such as coughing blood and not being able to move for hours...but anyway, let's not talk about that.
I did go to school on Friday and worked on an integration worksheet for a few minutes, but we got called out to an assembly.
But anyway, I wanted to point out a few things that I think I remember that a lot of you were having problems with.
When working an integration problem, a lot of times you need to use substitution. Substitution is useful (at least in my eyes) ESPECIALLY when you see an item in the problem and it's derivative is in the problem somewhere else...let me elaborate a bit more.
In the following problem, you should quickly recognize what your u is and that the derivative of it is in the problem also.
Integral of (1/xln(x)).
Now, when we did this problem in class, a lot of people started to try to go to the log base a of u rule and all of that. First, with these types of problems I always try to think if I can split it up...well in this one you can...so let's do that in our head really quick...well since you can't see it, it is
(1/x)(1/ln(x)). So in this problem you should notice two things. The derivative of ln(x) is (1/x). So really we can rewrite this problem as 1/u where u is lnx. So to integrate 1/u, it's ln|u| + c so the answer to our problem is ln|ln(x)| + c.
If you split up problems visually so that you can better recognize the u and the derivative of u, the problems become way easier.
Anyway, hope that helps someone.
Posting...#26
Optimization
1 Identify primary and secondary equations your primary is the one your or maximizing or minimizing and your secondary is the other equation
2. Solve for your secondary variable and plug into your primary equation if your primary only has on variable this isn’t necessary
3. Plug into secondary equation to find the other value check your end points if necessary
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-1dxu=2x-1 du=2
rewrite the function as:1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
I need help with angle of elevation
1 Identify primary and secondary equations your primary is the one your or maximizing or minimizing and your secondary is the other equation
2. Solve for your secondary variable and plug into your primary equation if your primary only has on variable this isn’t necessary
3. Plug into secondary equation to find the other value check your end points if necessary
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-1dxu=2x-1 du=2
rewrite the function as:1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
I need help with angle of elevation
Post #26
OFF FOR THE ENTIRE WEEK..THANK GOODNESS!!!
&& Happy Valentines Day!
So this past week in calculus we started out reviewing everything we have problems with! THAT HELPED SOOOOO MUCH..i'm beginning to get some stuff now! =)
the rest of the week we worked on practice worsheets that may have been graded!
can someone explain how to do the integration problems on that worksheet..i knew a few but some where tricky and i didn't know how to start it.
can someone tell me all the RAM fomulas..like LRAM, RRAM, MRAM, & TRAM
what i finally understand is the chain rule:
ex: sin^2(x^2)...the derivative of this is
4xsin(x^2)cos(x^2)
ex: sec^2(2x^3-4x)...the derivative of this is
2(6x^2-4)[sec(2x^3-4x)tan(2x^3-4x)][sec(2x^3-4x)]
CRAZY I KNOW...but it's just taking the derivative of an exponent, then the derivative of the other part, then the derivative of that and simplifing
LN INTEGRATION..
when you look at an integral make sure to look at the bottom first...see if it's derivative is the top and if so then it's ln(bottom) PLUS C
remember:
POSITION--original
VELOCITY--1st derivative
ACCELERATION--2nd derivative
so if you have position and need velocity of acceleration you'll have to take the first or second derivative...HOWEVER..if you have velocity and need position, or if you have acceleration and need either velocity or position you'll have to integrate..
remember to take the derivative or integegrate however many steps down or up it is from what you have already
hope this helps...BE SAFE, and have a GREAT WEEK OFF!!!
~ElliE~
&& Happy Valentines Day!
So this past week in calculus we started out reviewing everything we have problems with! THAT HELPED SOOOOO MUCH..i'm beginning to get some stuff now! =)
the rest of the week we worked on practice worsheets that may have been graded!
can someone explain how to do the integration problems on that worksheet..i knew a few but some where tricky and i didn't know how to start it.
can someone tell me all the RAM fomulas..like LRAM, RRAM, MRAM, & TRAM
what i finally understand is the chain rule:
ex: sin^2(x^2)...the derivative of this is
4xsin(x^2)cos(x^2)
ex: sec^2(2x^3-4x)...the derivative of this is
2(6x^2-4)[sec(2x^3-4x)tan(2x^3-4x)][sec(2x^3-4x)]
CRAZY I KNOW...but it's just taking the derivative of an exponent, then the derivative of the other part, then the derivative of that and simplifing
LN INTEGRATION..
when you look at an integral make sure to look at the bottom first...see if it's derivative is the top and if so then it's ln(bottom) PLUS C
remember:
POSITION--original
VELOCITY--1st derivative
ACCELERATION--2nd derivative
so if you have position and need velocity of acceleration you'll have to take the first or second derivative...HOWEVER..if you have velocity and need position, or if you have acceleration and need either velocity or position you'll have to integrate..
remember to take the derivative or integegrate however many steps down or up it is from what you have already
hope this helps...BE SAFE, and have a GREAT WEEK OFF!!!
~ElliE~
26th post
This week we went over some of the problems everyone was having trouble with.. here are some things i have a good handle on:
1st derivative test:
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. Then you plug numbers in between those intervals into the derivative to see if a function is increasing, decreasing, a max, or a min.
For the second derivative test, you take the derivative of the function two times and solve for the critical points like in the first derivative. You set up intervals the same way and plug numbers into the second derivative to see if it is concave up, concave down, or a point of inflection.
For e integration, the integral is always what e is raised to.
for ln integration, the top has to be the derivative of the bottom.
If there is a +1 at the bottom of a fraction, it is tan inverse.
Anything else if most likely substitution.
Tangent line- they give you a function and an x value. If no y value is given, plug the x value into the original function. To find the slope, you take the derivative of the original function and plug the x value in. Then you set it up into point slope form. y-y1= slope(x-x1). For a normal line, you would take the negative inverse of the slope and solve.
Things i have problems with:
One of the major problems i am having is the problems that give you three statements and ask you which ones are true. I know they are probably the easiest but i am struggling with them. Another thing i am having trouble with is substitution and LRAM, RRAM, MRAM, and TRAM, if anyone has an easy way of remembering the formulas that would be a great help. Hope you all have a safe and happy holiday :)
1st derivative test:
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. Then you plug numbers in between those intervals into the derivative to see if a function is increasing, decreasing, a max, or a min.
For the second derivative test, you take the derivative of the function two times and solve for the critical points like in the first derivative. You set up intervals the same way and plug numbers into the second derivative to see if it is concave up, concave down, or a point of inflection.
For e integration, the integral is always what e is raised to.
for ln integration, the top has to be the derivative of the bottom.
If there is a +1 at the bottom of a fraction, it is tan inverse.
Anything else if most likely substitution.
Tangent line- they give you a function and an x value. If no y value is given, plug the x value into the original function. To find the slope, you take the derivative of the original function and plug the x value in. Then you set it up into point slope form. y-y1= slope(x-x1). For a normal line, you would take the negative inverse of the slope and solve.
Things i have problems with:
One of the major problems i am having is the problems that give you three statements and ask you which ones are true. I know they are probably the easiest but i am struggling with them. Another thing i am having trouble with is substitution and LRAM, RRAM, MRAM, and TRAM, if anyone has an easy way of remembering the formulas that would be a great help. Hope you all have a safe and happy holiday :)
POST 26, HOLIDAYSSS!
i'm just going to go over some easy stuff that everyone should remember, but some tend to forgetttt.
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.
for help, uhhh i really just need help with the whole particle equations and related rates.
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.
for help, uhhh i really just need help with the whole particle equations and related rates.
Ash's 26th Post
First off: Hurray no school!
Second: ...?
I'm finally understanding SOME things!!
Since Definition of Derivatives are on EVERY test, let's go over that, shall we?
This is the way the problem looks
lim (equation and ()) - (equation or ())
()->0 ()
You take the derivative of (equation and ()) then plug in 0! :)
Example!!
lim cos(z)-z
z->0 z
=> cos(z)
derivative = -sin(z)
=> -sin(0)
= 0
I felt like an idiot after B-Rob explained it to me...
Okay, somethings I'm still having trouble with are
Integration Fractions: Yes, they were explained to me, and I do get them! But! That sheet we had to work on the other day was, in my opinion, really hard and I apparently need more practice with them?
Word Problems: Hmm....rate especially gets me!! I know it's a derivative, but I have no idea how to do it. Is that the problem with the formula with the e's, h's, k's, and t's? What IS that formula?
I'm not going to be in town all next week (I'm going to my grandmother's and she lives in the stone age and doesn't have internet last I checked -.-)
I was going to do my comments, but none of the blogs are posted, so I'll do them when I get back?
Have a great holiday everyone!!
Second: ...?
I'm finally understanding SOME things!!
Since Definition of Derivatives are on EVERY test, let's go over that, shall we?
This is the way the problem looks
lim (equation and ()) - (equation or ())
()->0 ()
You take the derivative of (equation and ()) then plug in 0! :)
Example!!
lim cos(z)-z
z->0 z
=> cos(z)
derivative = -sin(z)
=> -sin(0)
= 0
I felt like an idiot after B-Rob explained it to me...
Okay, somethings I'm still having trouble with are
Integration Fractions: Yes, they were explained to me, and I do get them! But! That sheet we had to work on the other day was, in my opinion, really hard and I apparently need more practice with them?
Word Problems: Hmm....rate especially gets me!! I know it's a derivative, but I have no idea how to do it. Is that the problem with the formula with the e's, h's, k's, and t's? What IS that formula?
I'm not going to be in town all next week (I'm going to my grandmother's and she lives in the stone age and doesn't have internet last I checked -.-)
I was going to do my comments, but none of the blogs are posted, so I'll do them when I get back?
Have a great holiday everyone!!
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