Well, I feel as though I have reviewed just about everything we’ve learned several times over…except word problems. The free response questions on the practice AP tests that we’ve been doing are essentially just word problems. I’ll go over how to work through a simple related rate problem.
There are steps that you can take to decipher related rate word problems and make them easier to understand. First, you need to identify all of the information the word problem gives, such as variables and what you’re actually trying to find. Once you do that, then you need to find a formula that connects all of your variables and your shape together (such as the area of a triangle or the volume of a cube). After this, solve for your desired rate with respect to t (such as dA/dt for the rate of area changing or dV/dt for the rate of volume changing). And lastly, plug in all of your variables to get your answer.
*Don’t forget to include your units (such as cm/min or ft^3/sec).
Some key words you need to know are:
“rate of change” = derivative
“speed” = derivative
“with respect to” = what’s the bottom variable of the derivative (ex. rate of changing area A with respect to time t : dA/dt)
“area” or “volume” = formula
“cube” or “sphere” or “square” or “triangle” or “circle” = shape
“radius” or “diameter” or “height” or “depth” = possibly what you need to take the derivate of or important components of the formula
“increasing” or “decreasing” = whether the derivative is positive or negative
and any form of units (especially squared or cubic units) = squared means area; cubic means volume
Ex. The radius r of a circle is increasing at a rate of 4 centimeters per minute. Find the rate of change of the area when r = 8 cm.
First of all, we know the following info: r = 8 & dr/dt = 4 cm/min
We also know we are looking for: dA/dt = ?
Second, because we are looking for the rate of change of the AREA, we know where using the area of a circle formula: A = π(r^2)
Now, solve for dA/dt by taking the derivative of the area formula: dA/dt = 2π r dr/dt
**Remember to put dr/dt after you take the derivative of r because it is an implicit derivative.
Finally, plug in your variables: dA/dt = 2π (8)(4) = 64π cm^2/min
***cm is squared because area is always squared, just like volume is always cubed.
Sunday, April 3, 2011
Alaina's blog, 3 April 2011
Something some of us may have forgotten how to do: LRAM Steps: -first you must create a number line with each of the bounds as the end points then you should put markers on the time line for each significant number between the bounds -the next step is to draw a specific number of rectangles as specified by the given n= across the numberline resembeling a bar graph you must -then plug in each significant number into the given equation plug those results into the rectangles to serve as the height of the rectangle -then you must decipher what the width would be considering the number of rectangles needed in relation to the bounds. -then you will multiply each of the widths by the heights to get multiple solutions which you will add together to get the area of the bounded region -keep in mind that for lram you will multiply all widths by heights except for that of the last rectangle. RRAM Steps: -first you must create a number line with each of the bounds as the end points -then you should put markers on the time line for each significant number between the bounds -the next step is to draw a specific number of rectangles as specified by the given n= across the numberline resembeling a bar graph -you must then plug in each significant number into the given equation plug those results into the rectangles to serve as the height of the rectangle -then you must decipher what the width would be considering the number of rectangles needed in relation to the bounds. -then you will multiply each of the widths by the heights to get multiple solutions which you will add together to get the area of the bounded region -keep in mind that for Rram you will multiply all widths by heights except for that of the first rectangle. TRAM Steps: -first you must create a number line with each of the bounds as the end points -then you should put markers on the time line for each significant number between the bounds -the next step is to draw a specific number of trapezoids as specified by the given n= across the numberline resembeling a bar graph -you must then plug in each significant number into the given equation plug those results into the trapezoids to serve as the width of the trapezoid -then you must decipher what the height would be considering the number of rectangles needed in relation to the bounds. -then you will plug in each of the numbers from the widths and heights into 1/2(b1+b2) to then get multiple solutions and then add the solutions all together to get the area when using the trapezodial rule.
4/3/11
Here are some of the rules we learned a reallllly long time ago. I guess a lot of them can be applied to the AP test so it does come in handy :)
THE CONSTANT RULE:
The constant rule: the derivative of a constant is always 0
Ex. 1) d/dx 7 = 0
THE POWER RULE:
The power rule: whenever you take the derivative of something with an exponent, you lose a power. The formula for this shortcut is d/dx (x^n) = nx^(n-1)
Ex. 2) d/dx 3x^2
First, bring the exponent to the front (and if there is a constant in front of the x, multiply the two). Then, subtract one from the exponent.
= 2(3)x^(2-1) = 6x
Ex. 3) d/dx (x^3 + 9x)
When there are multiple terms in an equation, take the derivative of each term individually.
= 3x^2 + 9
THE PRODUCT RULE:
The product rule formula: f(x)g’(x) + g(x)f’(x)
When using the product rule, the problem must state that you take the derivative of two separate equations multiplied by each other.
*note: g’(x) or f’(x) means take the derivative of g(x) or f(x)
Ex. 4) (x+3)(x^2+6x)
(x+3)(2x+6) + (x^2+6x)(1)
2x^2 + 6x + 6x + 18 + x^2 + 6x
3x^2 + 18x + 18
= 3(x^2 + 6x +6)
**note: Make sure the answer is as fully simplified as possible.
THE QUOTIENT RULE:
The quotient rule formula: (g(x)f’(x) – f(x)g’(x)) / (g(x))^2
When using the quotient rule, the problem must state that you take the derivative of two equations divided by each other.
***note: if the denominator is just a single term, it is not considered an equation and you do not use the quotient rule; you simply rewrite the problem and take the derivative normally
Ex. 5) (x^3+4) / (x-2)
((x-2)(3x^2) – (x^3+4)(1)) / ((x-2)^2)
((3x^3-6x^2) – (x^3+4)) / ((x-2)^2)
****note: to keep things in simplest terms, do not factor out the bottom
(3x^3 – 6x^2 – x^3 – 4) / ((x-2)^2)
*****note: be sure to distribute the negative throughout the second equation
(2x^3 – 6x^2 – 4) / ((x – 2)^2)
(2(x^3 – 3x^2 – 2)) / ((x – 2)^2)
THE CHAIN RULE:
You use the chain rule when you have a function inside of a function, f(g(x)).
Ex. 6) (x^2 + 3)^4
First, take the exponent of the entire function, bring it to the front, then subtract one from the exponent, and then just copy the inside.
4(x^2 + 3)^3
Now, multiply that by the derivative of the inside equation.
4(x^2 + 3)^3 (2x)
Put in simplest terms.
= 8x(x^2 + 3)^3
You can also use the chain rule with the product rule or quotient rule. If this is the case, then you first need to determine whether the chain rule is on the outside or the inside; this is to figure out what order you do things in. If the chain rule is on the outside, you do that first. If not, you do the product rule or quotient rule first.
Ex. 7) (x + 3)(x^2 – 1)^2
In this problem, the product rule comes first. When taking the derivative of the second equation, use the chain rule.
(x + 3)(2(x^2 – 1)(2x)) + (x^2 – 1)^2 (1)
= 4x(x + 3)(x^2 – 1) + (x^2 – 1)^2
Ex. 8) ((x + 2)/(x – 1))^2
For this problem, the chain rule is on the outside, therefore it goes first.
2((x + 2)/(x – 1)) (((x – 1)(1) – (x + 2)(1)) / (x – 1)^2))
2((x + 2)/(x – 1)) ((x – 1 – x – 2) / (x – 1)^2)
2((x + 2)/(x – 1)) ((-3)/(x – 1)^2)
= -3(2x + 4) / (x – 1) (x – 1)^2
THE CONSTANT RULE:
The constant rule: the derivative of a constant is always 0
Ex. 1) d/dx 7 = 0
THE POWER RULE:
The power rule: whenever you take the derivative of something with an exponent, you lose a power. The formula for this shortcut is d/dx (x^n) = nx^(n-1)
Ex. 2) d/dx 3x^2
First, bring the exponent to the front (and if there is a constant in front of the x, multiply the two). Then, subtract one from the exponent.
= 2(3)x^(2-1) = 6x
Ex. 3) d/dx (x^3 + 9x)
When there are multiple terms in an equation, take the derivative of each term individually.
= 3x^2 + 9
THE PRODUCT RULE:
The product rule formula: f(x)g’(x) + g(x)f’(x)
When using the product rule, the problem must state that you take the derivative of two separate equations multiplied by each other.
*note: g’(x) or f’(x) means take the derivative of g(x) or f(x)
Ex. 4) (x+3)(x^2+6x)
(x+3)(2x+6) + (x^2+6x)(1)
2x^2 + 6x + 6x + 18 + x^2 + 6x
3x^2 + 18x + 18
= 3(x^2 + 6x +6)
**note: Make sure the answer is as fully simplified as possible.
THE QUOTIENT RULE:
The quotient rule formula: (g(x)f’(x) – f(x)g’(x)) / (g(x))^2
When using the quotient rule, the problem must state that you take the derivative of two equations divided by each other.
***note: if the denominator is just a single term, it is not considered an equation and you do not use the quotient rule; you simply rewrite the problem and take the derivative normally
Ex. 5) (x^3+4) / (x-2)
((x-2)(3x^2) – (x^3+4)(1)) / ((x-2)^2)
((3x^3-6x^2) – (x^3+4)) / ((x-2)^2)
****note: to keep things in simplest terms, do not factor out the bottom
(3x^3 – 6x^2 – x^3 – 4) / ((x-2)^2)
*****note: be sure to distribute the negative throughout the second equation
(2x^3 – 6x^2 – 4) / ((x – 2)^2)
(2(x^3 – 3x^2 – 2)) / ((x – 2)^2)
THE CHAIN RULE:
You use the chain rule when you have a function inside of a function, f(g(x)).
Ex. 6) (x^2 + 3)^4
First, take the exponent of the entire function, bring it to the front, then subtract one from the exponent, and then just copy the inside.
4(x^2 + 3)^3
Now, multiply that by the derivative of the inside equation.
4(x^2 + 3)^3 (2x)
Put in simplest terms.
= 8x(x^2 + 3)^3
You can also use the chain rule with the product rule or quotient rule. If this is the case, then you first need to determine whether the chain rule is on the outside or the inside; this is to figure out what order you do things in. If the chain rule is on the outside, you do that first. If not, you do the product rule or quotient rule first.
Ex. 7) (x + 3)(x^2 – 1)^2
In this problem, the product rule comes first. When taking the derivative of the second equation, use the chain rule.
(x + 3)(2(x^2 – 1)(2x)) + (x^2 – 1)^2 (1)
= 4x(x + 3)(x^2 – 1) + (x^2 – 1)^2
Ex. 8) ((x + 2)/(x – 1))^2
For this problem, the chain rule is on the outside, therefore it goes first.
2((x + 2)/(x – 1)) (((x – 1)(1) – (x + 2)(1)) / (x – 1)^2))
2((x + 2)/(x – 1)) ((x – 1 – x – 2) / (x – 1)^2)
2((x + 2)/(x – 1)) ((-3)/(x – 1)^2)
= -3(2x + 4) / (x – 1) (x – 1)^2
Friday, April 1, 2011
4/1/11
This week I'm going to go over PARTICULAR SOLUTIONS. It's a really simple concept, but yet it's something that everyone forgets the steps for and cannot do. So to explain the concept...For the problem given to you, you're typically given a derivative already solved, in the form dy/dx = 'something'..And they'll tell you something along the lines of, "Find the particular solution (or they might just say "solution", but particular solution is implied) with initial condition f(#)=#..(which means they give you an x and y value that they want you to plug in after). Okay so here's a simple example explaining all of the steps thoroughly:
Ex. 1) Find the particular solution for 3y^2 + xy + 2x = 6 given that x=0 when y=1.
*Alright, so first you should notice the obvious--that you're not given what the derivative is. So they expect you to derive that equation before you start the particular solution
*So using implicit differentiation, you should get this as your first step when deriving the equation:
6ydy/dx + (x)(dy/dx) + (y)(1) + 2 = 0
which is
6ydy/dx + xdy/dx + y + 2 = 0
Now to simplify this you're going to have to get all the dy/dx's on one side and everything else on the other side:
6dy/dx + xdy/dx = -y-2
Now factor out a dy/dx
dy/dx(6+x) = -y-2
And divide by 6+x
dy/dx = (-y-2)/(6+x)
**Now we can use that to find our particular solution
*So the first step in finding the particular solution is to cross multiply..as in, multiply dy by 6+x and multiply dx by -y-2, and set them equal like this:
(6+x)dy = (-y-2)dx
*Alright now for the next step you're supposed to get the x's with the dx's and the y's with the dy's (by dividing usually)..butttttt, I don't see how that's possible with this equation..I mean I did make up this problem so that could be why, but this usually wouldn't happen on an AP. Soooo, for me to explain how to work the rest of this problem let's just pretend for our derivative we got this instead:
dy/dx = (6+x)/(-y-2)
*Now cross multiply and you should get this:
(-y-2)dy = (6+x)dx
*Now you're going to integrate both sides of the equation and put the "+C" on the right hand side. So after integrating you should have this:
-1/2y^2 - 2y = 6x + 1/2x^2 + C
*Now you're going to solve for C..And to do this you have to plug in the x and y value they gave you (x=0; y=1)
*So plugging in you should have:
-1/2(1)^2 - 2(1) = 6(0) + 1/2(0)^2 + C
C = 5/2
*Now plug your C into your integrated equation and solve for y to get your final answer
-1/2y^2 - 2y = 6x + 1/2x^2 + 5/2
*To solve this for y, you can first get rid of the fractions by multiplying every term by 2..After doing that you should have:
-y^2 - 4y = 12x + x^2 - 5
Rearranging that you have
-y^2 - 4y = x^2 + 12x - 5
Now you can factor out a -y..then you should get this:
-y(y+4) = x^2+12x-5
And divide by y+4
-y = (x^2+12x-5)/(y+4)
Divide by -1
y = - (x^2+12x-5)/(y+4) ..And that's your particular solution!
Ex. 1) Find the particular solution for 3y^2 + xy + 2x = 6 given that x=0 when y=1.
*Alright, so first you should notice the obvious--that you're not given what the derivative is. So they expect you to derive that equation before you start the particular solution
*So using implicit differentiation, you should get this as your first step when deriving the equation:
6ydy/dx + (x)(dy/dx) + (y)(1) + 2 = 0
which is
6ydy/dx + xdy/dx + y + 2 = 0
Now to simplify this you're going to have to get all the dy/dx's on one side and everything else on the other side:
6dy/dx + xdy/dx = -y-2
Now factor out a dy/dx
dy/dx(6+x) = -y-2
And divide by 6+x
dy/dx = (-y-2)/(6+x)
**Now we can use that to find our particular solution
*So the first step in finding the particular solution is to cross multiply..as in, multiply dy by 6+x and multiply dx by -y-2, and set them equal like this:
(6+x)dy = (-y-2)dx
*Alright now for the next step you're supposed to get the x's with the dx's and the y's with the dy's (by dividing usually)..butttttt, I don't see how that's possible with this equation..I mean I did make up this problem so that could be why, but this usually wouldn't happen on an AP. Soooo, for me to explain how to work the rest of this problem let's just pretend for our derivative we got this instead:
dy/dx = (6+x)/(-y-2)
*Now cross multiply and you should get this:
(-y-2)dy = (6+x)dx
*Now you're going to integrate both sides of the equation and put the "+C" on the right hand side. So after integrating you should have this:
-1/2y^2 - 2y = 6x + 1/2x^2 + C
*Now you're going to solve for C..And to do this you have to plug in the x and y value they gave you (x=0; y=1)
*So plugging in you should have:
-1/2(1)^2 - 2(1) = 6(0) + 1/2(0)^2 + C
C = 5/2
*Now plug your C into your integrated equation and solve for y to get your final answer
-1/2y^2 - 2y = 6x + 1/2x^2 + 5/2
*To solve this for y, you can first get rid of the fractions by multiplying every term by 2..After doing that you should have:
-y^2 - 4y = 12x + x^2 - 5
Rearranging that you have
-y^2 - 4y = x^2 + 12x - 5
Now you can factor out a -y..then you should get this:
-y(y+4) = x^2+12x-5
And divide by y+4
-y = (x^2+12x-5)/(y+4)
Divide by -1
y = - (x^2+12x-5)/(y+4) ..And that's your particular solution!
Sunday, March 27, 2011
Blog #31
I’m going to review how to integrate using substitution. This is used when you have a product or quotient rule. This can’t always be used, but these are the instances in which it can:
S (u)(v)
…or…
S u/v
where v is the derivative of u
Here’s an example:
Given…S (x²+1)(2x) dx
First, identify u and v. Which is the derivative and which is the original?
u = x²+1
v = 2x
…because 2x is the derivative of x²+1…
Okay, next you remove v (2x dx) from the equation and substitute “u” for x²+1.
S u du
*You insert du because v is the derivative of u, hence du.
Now, integrate u.
½ u² + C
Finally, replace “u” with the original.
½ (x²+1)² + C
I’m also going to review the Second Fundamental Theorem of Calculus:
This theorem state that the derivative of an integral results in the original equation.
The formula: d/dx vSa f(x) dx = (f(v)) (v’) where “a” is a number (usually equaling 0) and v is an equation
An example:
d/dx x²S0 (x+1)² dx
= (((x²) +1)²) (2x)
= 2x(x²+1)²
S (u)(v)
…or…
S u/v
where v is the derivative of u
Here’s an example:
Given…S (x²+1)(2x) dx
First, identify u and v. Which is the derivative and which is the original?
u = x²+1
v = 2x
…because 2x is the derivative of x²+1…
Okay, next you remove v (2x dx) from the equation and substitute “u” for x²+1.
S u du
*You insert du because v is the derivative of u, hence du.
Now, integrate u.
½ u² + C
Finally, replace “u” with the original.
½ (x²+1)² + C
I’m also going to review the Second Fundamental Theorem of Calculus:
This theorem state that the derivative of an integral results in the original equation.
The formula: d/dx vSa f(x) dx = (f(v)) (v’) where “a” is a number (usually equaling 0) and v is an equation
An example:
d/dx x²S0 (x+1)² dx
= (((x²) +1)²) (2x)
= 2x(x²+1)²
blog blog blog blog blOg :)
For this blog I'm going to review Chapter 5..taking derivatives of logs, natural logs, e, exponents and whatnot. *first let's start with natural logs. The rule for taking the derivative of those is to first write the inside as a fraction over 1 and then multiply it by its derivative. So it would be like this: 1/# x d/dx of # So here are some examples like that: 1.) ln(5x+4) =1/5x+4 x 5 =5/5x+4 2.) ln(5x^2+2x+4) =1/(5x^2+2x+4) x 10x+2 =(10x+2)/(5x^2+2x+4) 3.) ln(x^2-16)^1/2 =1/(x^2-16)^1/2 x 1/2(x^2-16)^-1/2 x 2x =x/(x^2-16) *Now let's go over how to derive exponential functions dealing with e. The rule says that all you do to take the derivative is first recopy the function they give you, then multiply it by the exponent's derivative So here are some examples: 1.) f(x)=(4e)^-3x^2 -First instead of recopying the whole thing, you leave the 4 out in front just like you would do when deriving any other equation. Then you recopy the e^-3x^2 -So you should have 4(e^-3x^2 -Now multiply that by the derivative of e^-3x^2 which is -6x -So now you have 4(e^-3x^2 x -6x) = -24xe^-3x^2 2.) f(x)=x^4e^x -Okay first thing you should realize is that this is a product rule. The same rules apply even though there's an e involved..and to save time, the derivative of e^x is e^x -So using the product rule you get: (x^4)(e^x)+(e^x)(4x^3) =x^4e^x + 4x^3e^x -Then you can simplify that^ further by taking out an x^3 and an e^x -So your final answer is: x^3e^x(x+4) *Last, let's go over how to derive logs. You take the inside of the function, put it as a fraction over 1 times ln(of whatever base) times the derivative of the inside..I'm not sure if I explained that right haha, but anyway here's an example of that: 1.) log (base 3) (6x^2+5x+2) =1/(6x^2=5x+2)ln3 x 12x+5 = (12x+5)/(6x^2+5x+2)ln3
Soooo... differentiability. There are four ways to determine whether a function is differentiable. If it is differentiable it pretty much means it isn't continuous at some point in the graph. *Using these steps to figure out if a function is differentiable or not you can: look at the graph, factor out the problem, or simply just by looking at the problem. So here are the four ways to determine if a function is differentiable:
1) If there is a corner
Corners occur when absolute values and/or piecewises are found in the function that you are trying to take the derivative.
An example of a function with a corner is:
1) |3-x| you set what's inside of the absolute value equal to 0 and solve for x, so |3-x| is not differentiable at x=3
**When graphing this function you will see what a corner is(it will be the point where the graph makes a v).
2) If there is a vertical asymptote
Vertical asymptotes occur when x is raised to the (odd/odd)
x^5/9 would be a vertical asymptote, so x would not be differentiable at x=1
**Vertical asymptotes can also be graphed
3) If it's not continuous
If a function is not continuous, then it's not differentiable. Discontinuities can occur if theres an vertical asymptote, removable, or jump.
4) If there's a cusp
A cusp is when you have x^(even/odd)
x^2/7 would not be differentiable at x=0
**When graphing: if there is a point in there graph where there is a dip then it goes back straight, then it would be a cusp.
1) If there is a corner
Corners occur when absolute values and/or piecewises are found in the function that you are trying to take the derivative.
An example of a function with a corner is:
1) |3-x| you set what's inside of the absolute value equal to 0 and solve for x, so |3-x| is not differentiable at x=3
**When graphing this function you will see what a corner is(it will be the point where the graph makes a v).
2) If there is a vertical asymptote
Vertical asymptotes occur when x is raised to the (odd/odd)
x^5/9 would be a vertical asymptote, so x would not be differentiable at x=1
**Vertical asymptotes can also be graphed
3) If it's not continuous
If a function is not continuous, then it's not differentiable. Discontinuities can occur if theres an vertical asymptote, removable, or jump.
4) If there's a cusp
A cusp is when you have x^(even/odd)
x^2/7 would not be differentiable at x=0
**When graphing: if there is a point in there graph where there is a dip then it goes back straight, then it would be a cusp.
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