Well, this past week in calculus was .. just two days. Monday we reviewed some more things that we did not understand throughout the year. I can honestly say that I am BEGINNING to understand the stuff that Mrs. Robinson taught us about the picture of a graph, so I think that’s what I’m going to explain.
• If you have f(x):
o Use a solid line to divide where it’s increasing or decreasing: if it’s increasing then the first derivative is positive, and if it’s decreasing then the first derivative is negative.
o Use a dotted line to separate concave up and concave down: if it’s concave up then the second derivative is positive, and if it’s concave down then the second derivative is negative.
o Put a large dot on MAX and MIN: MAXs and MINs mean that it’s a horizontal tangent and the first derivative is zero.
o Put an open circle on points of inflection: points of inflection is where the second derivative is zero.
• If you have f’(x):
o Use a solid line where it’s increasing or decreasing: if it’s increasing then the second derivative is positive, and if it’s decreasing then the second derivative is negative.
o Use a dotted line where it’s above and below the axis: when it’s above then the original is increasing and when it’s below then the original is decreasing.
o Use solid dots on MAX’s and MIN’s then the second derivative is equal to zero and it’s a point of inflection on the derivative.
o Then put open circles on the x intercepts: when the MAX is above the axis then it’s below for the second derivative, and when the MIN is below the axis then it’s above for the second derivative.
So I know that I understand the concept perfectly fine and it’s really simple, it’s just that looking at a graph confuses me so I would need more practice to look at the graph and figure out what is what. I didn’t do so well on my exam so it looks like I’ll be reviewing a lot over the break so I can catch up. THANK GOD IT’S THE HOLIDAYS.
~ElliE~
Friday, December 18, 2009
Tuesday, December 15, 2009
post seventeen
Ok so this week we were preparing for log one and for our exam.
Some things that i understand in calculus include:
Tangent Line:
1. If you are only given a x value, plug into your equation to find the y value.
2. take the derivative of the equation
3. Plug in your x value into the derivative to find the slope
4. Put in point slope
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)
STEPS:
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
These are fairly simple processes and equations to work with. For what i dont know is linierization. I dont even have notes on this
Some things that i understand in calculus include:
Tangent Line:
1. If you are only given a x value, plug into your equation to find the y value.
2. take the derivative of the equation
3. Plug in your x value into the derivative to find the slope
4. Put in point slope
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)
STEPS:
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
These are fairly simple processes and equations to work with. For what i dont know is linierization. I dont even have notes on this
Sunday, December 13, 2009
Post #17
Last week in calculus was review week. We didn't learn anything new so I'm going to review some of the old stuff because it will be on our exam.
Tangent Line:
1. If you are only given a x value, plug into your equation to find the y value.
2. take the derivative of the equation
3. Plug in your x value into the derivative to find the slope
4. Put in point slope
EXAMPLE: y=3x^3 + 4x^2 +5 at x=1
1. 3(1)^3+4(1)^2+5 = 12
2. 9x^2 + 8x
3. 9(1)^2 + 8(1) = 17
4. y-12 = 17 (x-1)
Normal line is the same except you take the negative reciprocal of the slope to plug in.
y-12 = -1/17 (x-1)
Rolle's Theorem:
In order to use Rolle's theorem, the fuction must be continuous and differentiable on the interval given and f(a) = f(b)
Example:
-x^2+14x on [0,14]
The function is continuous and differentiable
f(0) = -(0)^2 + 14(0) = 0
f(14) = -(14)^2 + 14(14) = 0
Therefore Rolle's applies
Now take the derivative: -2x+14
and set equal to zero: -2x+14 = 0
solve for x: x=7 or c=7
Mean value theorem:
The function again has to be continuous and differentiable at the interval given and you plug into f(b) - f(a) / b-a
Example: x^2 on [1,7]
The function is continuous and differentiable
First take the derivative: 2x
Then plug into the formula: f(7) - f(1) / 7-1
14-2/ 7-1 = 12/6 which equals 2.
I am still having troubles with graphs and going from one graph to another. I don't understand how to find increasing/decreasing, concave up/down, maxs, mins, and points of inflection. Help please!
17th post
alright, we just reviewed this week, so i am going to go over integration. Integration is pretty much the opposite of a derivative.
there are two types of integration, definite integration and indefinite integration.
indefinite integration is easy, you just take the equation, and do this to it, lets say the equation is x^3+2x^2
to take the integral of something, first you have to raise the exponent by one, so lets take the first part of the problem, x^3, and do that to it. it becomes x^4. Now, you just multiple the whole thing by the inverse of the new exponent. so it would become (1/4)x^4.
then you do the same thing to the second part of the problem. 2x^2 becomes 2x^3, then you have to multiply, and it becomes (2/3)x^3
then you put the two parts back together like normal, and it becomes (1/4)x^4+(2/3)x^3.
but we are not done yet!! for indefinite integrals, you always have to add C at the end of the equation, so your final answer becomes: (1/4)x^4+(2/3)x^3+C
and for definite integrals, you just do the same thing, but for definite integrals, they give you a point, and you just plug in the x value from the point, and you do not add the C at the end!!! so lets just use the example problem from indefinite integration, but they'll also give you a point, let's say (2,2) for example. You would integrate it normally, and get (1/4)x^4+(2/3)x^3, then you plug in your x value, which is 2. so then you get (1/4)(2)^4+(2/3)(2)^3, and after that you can just put it into your calculator, which I do not have with me, and you'll just get some number as your answer. So for definite integration, you'll get a a number answer, and for indefinite integration, you'll get an equation.
there are two types of integration, definite integration and indefinite integration.
indefinite integration is easy, you just take the equation, and do this to it, lets say the equation is x^3+2x^2
to take the integral of something, first you have to raise the exponent by one, so lets take the first part of the problem, x^3, and do that to it. it becomes x^4. Now, you just multiple the whole thing by the inverse of the new exponent. so it would become (1/4)x^4.
then you do the same thing to the second part of the problem. 2x^2 becomes 2x^3, then you have to multiply, and it becomes (2/3)x^3
then you put the two parts back together like normal, and it becomes (1/4)x^4+(2/3)x^3.
but we are not done yet!! for indefinite integrals, you always have to add C at the end of the equation, so your final answer becomes: (1/4)x^4+(2/3)x^3+C
and for definite integrals, you just do the same thing, but for definite integrals, they give you a point, and you just plug in the x value from the point, and you do not add the C at the end!!! so lets just use the example problem from indefinite integration, but they'll also give you a point, let's say (2,2) for example. You would integrate it normally, and get (1/4)x^4+(2/3)x^3, then you plug in your x value, which is 2. so then you get (1/4)(2)^4+(2/3)(2)^3, and after that you can just put it into your calculator, which I do not have with me, and you'll just get some number as your answer. So for definite integration, you'll get a a number answer, and for indefinite integration, you'll get an equation.
post #17
Okay so almost everything we did this week was basically in preparation for the exam on tuesdayyy, so i'm just going to go over some things from the past as a review.
Some things that we just learned that are extremely easy form me are finding the volume and area of disks and washers. atleast i know i'll pass some portion of the exams, assuming ofcourse, that they're even on there.
Area between curves:
formula: bSa top equation-bottom equation
example: Find the area of the region enclosed by y=-4x^2+41x+94 and y=x-2 between x=1 and x=7.
1. draw a picture
2. subtract the two equations: (-4x^2+41x+94-(x-2)-don't square, it's not volumee!
(-4x^2+41x+94-(-x+2)
3.combine like terms and integrate: S(-4x^2+41x+94-(-x+2))
(-4/3)x^3+20x^2+96x
4. plug in 1 and 6 and subtract: (3584/3)-(344/3)= 1080
FORMULA FOR DISKS: pi S[r(x)]^2dx, this formula is almost exactly the same as when you are asked to find the area of a disk, however, while finding the area you do not square the radius (the given equation).
FORMULA FOR WASHERS: pi S[(top)^2-(bottom)^2]dx
TANGENT LINES!
Steps:
Take the first der...f ‘(x).
Plug in x to find your slope: m.
Plug x into original functiong..f(x) to find y (if not already given).
Using your slope..m and (x,y), plug it into slope-intercept form: (y-y1) = m(x-x1)
okayy something i don't understandd! umm i really don't get how to integrate a fraction, which would be integration by substitution i'm assumine. also some things from the past that would be nice to go over is ofcourse optimization, related rates type stuff. other than that, i think i'm okayy, maybe trapezoidal rule but that's about it! help would be WONDERFULL!
Some things that we just learned that are extremely easy form me are finding the volume and area of disks and washers. atleast i know i'll pass some portion of the exams, assuming ofcourse, that they're even on there.
Area between curves:
formula: bSa top equation-bottom equation
example: Find the area of the region enclosed by y=-4x^2+41x+94 and y=x-2 between x=1 and x=7.
1. draw a picture
2. subtract the two equations: (-4x^2+41x+94-(x-2)-don't square, it's not volumee!
(-4x^2+41x+94-(-x+2)
3.combine like terms and integrate: S(-4x^2+41x+94-(-x+2))
(-4/3)x^3+20x^2+96x
4. plug in 1 and 6 and subtract: (3584/3)-(344/3)= 1080
FORMULA FOR DISKS: pi S[r(x)]^2dx, this formula is almost exactly the same as when you are asked to find the area of a disk, however, while finding the area you do not square the radius (the given equation).
FORMULA FOR WASHERS: pi S[(top)^2-(bottom)^2]dx
TANGENT LINES!
Steps:
Take the first der...f ‘(x).
Plug in x to find your slope: m.
Plug x into original functiong..f(x) to find y (if not already given).
Using your slope..m and (x,y), plug it into slope-intercept form: (y-y1) = m(x-x1)
okayy something i don't understandd! umm i really don't get how to integrate a fraction, which would be integration by substitution i'm assumine. also some things from the past that would be nice to go over is ofcourse optimization, related rates type stuff. other than that, i think i'm okayy, maybe trapezoidal rule but that's about it! help would be WONDERFULL!
Ash's 17th Blog
Wow
Only one more week until we can finally be freed from school for two entire weeks! I'm so excited!!
This week we spent...doesn't stuff? I don't actually remember...so let's skip to Friday
Friday: Beast packet :D
Saturday: ACT + Subway (omg $5 6inch..YUM!) + PJs + Walmart (1.5 hours, with Mal's mom to get EIGHT things...) + New Orleans Hamburger (YUM! SEAFOOD!) + Movies (standing outside trying to decide what movie to watch, New Moon or the Blindside...for...thirty...minutes) + NEW MOON! FINALLY! + Calculus packet again
Sunday: Calculus packet + English packet + Tic Tac Toe + Calculus packet..againn
So..yea...that's all I have to say :)
Hmm...what do I know that I can explain?
Umm...Riemann Sums?
I'm pretty sure that's way over-done, but I have no idea what else to explain! =/
Delta = b-a/n (top number above S minus the bottom number below S divided by the number of intervals)
LRAM: delta(f(a)+f(a+delta)+f(a+2delta)...+f(b-delta))
RRAM: delta(f(a+delta)+f(a+2delta)+...f(b))
MRAM: delta(f(mid)+f(mid)...)
Trapezoidal: delta(x)/2 [ f(a) + 2f(a + delta x) + 2f(a + 2deltax)
+ f(b))
Just don't forget to not go over the amount of intervals you have (n).
Now, I've got major questions
1. Trapezoidal graphs...where can I get the software?
2. What's deltax? is that the same thing as delta..???
3. GRAPHING! AHHH!!
4. Also, for limits, are they asking for the open circle or the closed circle? I always get confused! And each time I change it, I get it wrong..Help!!!
5. Another thing for limits, what do you do if it's over zero? I remember something about it...did she or Alex go over that? Anyway, I can't find those notes and I didn't get it before...=/
Okay, this is my extreme problem:
I've never been able to remember the steps for math
I can do it if I know what I'm doing (makes total sense right?)
But I can't remember the steps
Does ANYONE have a way to do remember?? =/
Only one more week until we can finally be freed from school for two entire weeks! I'm so excited!!
This week we spent...doesn't stuff? I don't actually remember...so let's skip to Friday
Friday: Beast packet :D
Saturday: ACT + Subway (omg $5 6inch..YUM!) + PJs + Walmart (1.5 hours, with Mal's mom to get EIGHT things...) + New Orleans Hamburger (YUM! SEAFOOD!) + Movies (standing outside trying to decide what movie to watch, New Moon or the Blindside...for...thirty...minutes) + NEW MOON! FINALLY! + Calculus packet again
Sunday: Calculus packet + English packet + Tic Tac Toe + Calculus packet..againn
So..yea...that's all I have to say :)
Hmm...what do I know that I can explain?
Umm...Riemann Sums?
I'm pretty sure that's way over-done, but I have no idea what else to explain! =/
Delta = b-a/n (top number above S minus the bottom number below S divided by the number of intervals)
LRAM: delta(f(a)+f(a+delta)+f(a+2delta)...+f(b-delta))
RRAM: delta(f(a+delta)+f(a+2delta)+...f(b))
MRAM: delta(f(mid)+f(mid)...)
Trapezoidal: delta(x)/2 [ f(a) + 2f(a + delta x) + 2f(a + 2deltax)
+ f(b))
Just don't forget to not go over the amount of intervals you have (n).
Now, I've got major questions
1. Trapezoidal graphs...where can I get the software?
2. What's deltax? is that the same thing as delta..???
3. GRAPHING! AHHH!!
4. Also, for limits, are they asking for the open circle or the closed circle? I always get confused! And each time I change it, I get it wrong..Help!!!
5. Another thing for limits, what do you do if it's over zero? I remember something about it...did she or Alex go over that? Anyway, I can't find those notes and I didn't get it before...=/
Okay, this is my extreme problem:
I've never been able to remember the steps for math
I can do it if I know what I'm doing (makes total sense right?)
But I can't remember the steps
Does ANYONE have a way to do remember?? =/
Posting...#17
Last week in calculus we did not learn anything new. We took a practic ap monday and wensday then we went over it for tuesday and thursday.
Rolle's Theorem: Let f be continuous on a closed interval [a,b] and differentiable on the open interval (a,b) then there exsists a number c in (a,b) such that
f(b)-f(a)
f ' (c)=----------
b-a
Exampmle: F(x)=x^4 -2x^2 on [-2,2] find c f '(c)=0
Check if its continuous : YES
Check if its differentiable : YES
f(-2)=(-2)^4-2(-2)^2=8
F(2)=(2)^4-2(2)^2=8
f(-2)=f(2)
4x^3-4x=0
4x(x^2-1)=0
X=0,1,-1
Awnser C= -1,0,1
And thats it for roles theroem its pretty easy but for what im having trouble with is even easier so it would be great to get help.
I am confused on Critical Points so can some one explain how to do that for me.
Rolle's Theorem: Let f be continuous on a closed interval [a,b] and differentiable on the open interval (a,b) then there exsists a number c in (a,b) such that
f(b)-f(a)
f ' (c)=----------
b-a
Exampmle: F(x)=x^4 -2x^2 on [-2,2] find c f '(c)=0
Check if its continuous : YES
Check if its differentiable : YES
f(-2)=(-2)^4-2(-2)^2=8
F(2)=(2)^4-2(2)^2=8
f(-2)=f(2)
4x^3-4x=0
4x(x^2-1)=0
X=0,1,-1
Awnser C= -1,0,1
And thats it for roles theroem its pretty easy but for what im having trouble with is even easier so it would be great to get help.
I am confused on Critical Points so can some one explain how to do that for me.
Post Number Seventeen
this week we reviewed for our exam, so i'll just go over stuff that i know and do not know as my blog.
Let's start this off on a good note and explain something i DO understand. Average value :)
The formula for average value is 1/b-a(definite integral)f(a)
For example, Find the average value of f(x) = x^2 on [0,5]
1/5-0 S from 0 to 5 x^2 dx
= 1/5 [1/3x^3]
=1/5[125/3 - 0] = 25/3
Now for something i still don't understand and never did: SUBSTITUTION.
help please?
I also understand rolles and mean value theorems, you just follow them and plug in :)
This may be sad, but i am having trouble finding critical numbers and absolute extrema, those kinds of problems. I need a slight review on this if anyone cares.
I know how to do RRAM and LRAM, but am still having trouble with MRAM and mostly TRAM..
One more thing i understand is volume by disks, so i'll explain that quickly.
Volume by disks is used only with solid objects. The formula used is pi definite integral [R(x)]^2 dx
This is pretty easy, everything is basically given to you, the only place i usually mess up is in the algebra itself.
Well i'm stressing about the exam but hey, what can i do?
Goodnight Everyone :)
Let's start this off on a good note and explain something i DO understand. Average value :)
The formula for average value is 1/b-a(definite integral)f(a)
For example, Find the average value of f(x) = x^2 on [0,5]
1/5-0 S from 0 to 5 x^2 dx
= 1/5 [1/3x^3]
=1/5[125/3 - 0] = 25/3
Now for something i still don't understand and never did: SUBSTITUTION.
help please?
I also understand rolles and mean value theorems, you just follow them and plug in :)
This may be sad, but i am having trouble finding critical numbers and absolute extrema, those kinds of problems. I need a slight review on this if anyone cares.
I know how to do RRAM and LRAM, but am still having trouble with MRAM and mostly TRAM..
One more thing i understand is volume by disks, so i'll explain that quickly.
Volume by disks is used only with solid objects. The formula used is pi definite integral [R(x)]^2 dx
This is pretty easy, everything is basically given to you, the only place i usually mess up is in the algebra itself.
Well i'm stressing about the exam but hey, what can i do?
Goodnight Everyone :)
Post #17
Calculus Week #17
Anyway so we are supposed to explain a concept and ask a question or two about another. Well to be quite honest, this week I actually learned something I should have known a long time ago.
Trapezoidal!
The formula is really simple and easy to remember...anyway, it's
Let x = (b-a)/n
x/2 [f(a) + 2f(a+x) + 2f(a+2x) ... f(b)]
This beats trying to find the actual area under the curve and wondering which one of the answer choices might be close. Lol...
Anyway, let's see...what else can I explain.
Okay a crash course to help you all on the take-home test:
If it asks for a tangent line, take the derivative, plug in x, get your slope, use that in point-slope form with the point you had with that slope.
If it asks for critical values...set derivative equal to 0 and solve for x.
Make sure to test end points.
Relative maximums or minimums are when you have multiple maxs and mins. The relative ones are the ones that are not absolute. The ones that are absolute are the ones that have either the highest y value or lowest y value.
Remember, if you are allowed in calculator, you can check your integration.
fnInt((equation),x,a,b) for f(x) = (equation) on the interval [a,b].
Or just plug it into y= on the plot then go to SECOND, CALC and go to the last option. Then type in a press ENTER then press b and press ENTER. Voila, a graphed picture of the area under your curve.
For the area between two curves, you do the top equation minus bottom equation. If you have no a,b to plug in, set the two equations equal and solve. Use that as your lower and upper bounds.
For volume by discs of a solid rotated about an axis, do the definite integral from a to b of the equation squared...times pi. Pretty simple.
For volume by washers, it's the same thing as area between two curves...except this time you square both equations as well, as in the above example.
Average value is 1/(b-a) times the definite integral from a to b of the function.
Average rate of change is a slope which is a derivative.
Derivative of position is velocity, derivative of velocity is acceleration.
So first derivative is velocity, second derivative is acceleration.
Let's see what else I can cover...
If they give you acceleration and you take the integral to find the velocity equation, don't forget you need to solve for +c. Usually they will give you some information such as v(1) = 1 would be used to plug in and solve for c.
Anyway, I think I gave a pretty decent crash course for the take home test. Don't forget to use your calculator...you can check almost all of the problems in your calculator so you might as well do it!
Anyway so we are supposed to explain a concept and ask a question or two about another. Well to be quite honest, this week I actually learned something I should have known a long time ago.
Trapezoidal!
The formula is really simple and easy to remember...anyway, it's
Let x = (b-a)/n
x/2 [f(a) + 2f(a+x) + 2f(a+2x) ... f(b)]
This beats trying to find the actual area under the curve and wondering which one of the answer choices might be close. Lol...
Anyway, let's see...what else can I explain.
Okay a crash course to help you all on the take-home test:
If it asks for a tangent line, take the derivative, plug in x, get your slope, use that in point-slope form with the point you had with that slope.
If it asks for critical values...set derivative equal to 0 and solve for x.
Make sure to test end points.
Relative maximums or minimums are when you have multiple maxs and mins. The relative ones are the ones that are not absolute. The ones that are absolute are the ones that have either the highest y value or lowest y value.
Remember, if you are allowed in calculator, you can check your integration.
fnInt((equation),x,a,b) for f(x) = (equation) on the interval [a,b].
Or just plug it into y= on the plot then go to SECOND, CALC and go to the last option. Then type in a press ENTER then press b and press ENTER. Voila, a graphed picture of the area under your curve.
For the area between two curves, you do the top equation minus bottom equation. If you have no a,b to plug in, set the two equations equal and solve. Use that as your lower and upper bounds.
For volume by discs of a solid rotated about an axis, do the definite integral from a to b of the equation squared...times pi. Pretty simple.
For volume by washers, it's the same thing as area between two curves...except this time you square both equations as well, as in the above example.
Average value is 1/(b-a) times the definite integral from a to b of the function.
Average rate of change is a slope which is a derivative.
Derivative of position is velocity, derivative of velocity is acceleration.
So first derivative is velocity, second derivative is acceleration.
Let's see what else I can cover...
If they give you acceleration and you take the integral to find the velocity equation, don't forget you need to solve for +c. Usually they will give you some information such as v(1) = 1 would be used to plug in and solve for c.
Anyway, I think I gave a pretty decent crash course for the take home test. Don't forget to use your calculator...you can check almost all of the problems in your calculator so you might as well do it!
Post 17
Ok, so last week we pretty much just reviewed for the upcoming exams. Mrs. Robinson gave us two practice APs, which I failed. They mostly went over things we learned recently like integrals. The first one was much much better than the second one, which is pretty pathetic considering the second one was calculator allowed. We were also given the take home portion of the exam. This portion is only twenty multiple choice questions and is also calculator allowed. The one we will take on exam did will include multiple choice and essay, both of which calculators will not be allowed. Today I was going through the take home portion and I realized I forgot a lot of things we did in the beginning of the year. I thought I knew how to do some problems, but my answers were coming out wrong.
Anyway, two weeks ago we learned how to find volume of a curve. This is done by rotating the curve about an axis or line to make it a solid object. We did this by using both disks and washers. Disks are used when there is only one equation and the object rotated turns out to be a solid. When there are two equations, washers are used since there will be a hole in the figure.
The formula for disks is:
pi S a-b [R(x)]^2 dx
This is basically a definite integral where you plug in the function you are given and square it. Many people forget to square it. This completley changes everything, so don't forget.
The formula for washers is:
pi S a-b top^2-bottom^2 dx
This is also a definite integral where you have to square the functions. I like these because they're very similar to area under a curve, and I also liked those. The only difference is to square both functions given and put a pi on the outside of the integral.
Most of the things we learned recently I'm fine with, but I came to realize I'm confused with a lot of old things. I get average value, average rate of change, and other various things with those words arranged in different ways. I never really caught onto average rate of change, and I still don't know what it is. I also get confused with linerizaton. When I look at the questions, I never know what to do with them. Can someone explain?
Anyway, two weeks ago we learned how to find volume of a curve. This is done by rotating the curve about an axis or line to make it a solid object. We did this by using both disks and washers. Disks are used when there is only one equation and the object rotated turns out to be a solid. When there are two equations, washers are used since there will be a hole in the figure.
The formula for disks is:
pi S a-b [R(x)]^2 dx
This is basically a definite integral where you plug in the function you are given and square it. Many people forget to square it. This completley changes everything, so don't forget.
The formula for washers is:
pi S a-b top^2-bottom^2 dx
This is also a definite integral where you have to square the functions. I like these because they're very similar to area under a curve, and I also liked those. The only difference is to square both functions given and put a pi on the outside of the integral.
Most of the things we learned recently I'm fine with, but I came to realize I'm confused with a lot of old things. I get average value, average rate of change, and other various things with those words arranged in different ways. I never really caught onto average rate of change, and I still don't know what it is. I also get confused with linerizaton. When I look at the questions, I never know what to do with them. Can someone explain?
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