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.
Sunday, March 14, 2010
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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