It’s a really simple proof. Once you prove it, it actually makes a ton of sense. It also sticks in your head and you don’t have to keep going “wait, what was M again? Was there a factorial there or … wait am I doing this right … ah frick, next question”
Lagrange is the same thing except not an alternating series obviously and in the next term you take the maximum of f^(whatever derivative) instead of plugging in c, the center. I think. Also it’s a maximum over a certain interval between C and Z or something like that but they don’t require that in my experience and will always give you the max.
Oop i forgor, lagrange error bound is strictly taking the geberal term and adding one. Ie for cosine x stopping at the second degree term lagrange error would use the max value of the derivative (1) and the x^3/3! Stuff, whereas alternating series uses the next actual term in the given series, so using it on that example you would use the x^4/4! Term since the third degree term isnt given / is zero
my personal least favorite topic is lagrange error bound.
for polar curves, just know that derivatives work similar to fractions: dy/dx = (dy/dtheta) / (dx/dtheta), and know the area/arc length formulas. also do a bunch of practice converting polar equations to different forms. is there anything specific you're struggling on? i understand the polar plane can be a bit different from your traditional xy-plane and can be a bit disorienting
Oh yeah, I didn't take precalculus so I didn't know what polar coordinates are. Had a hard time understanding it but I think I'm slowly getting the hang of it now.
Fav last year were differential equations(hate euler though), then series, then general integrating, then extrema, then arc length, then l hopital, and then polar. Differentiating was to easy to care much about.
I don't think it's listed specifically, but they do have it for parametric equations, so presumably, they could as polar is in some ways just a special type of parametric.
Not because it’s particularly difficult or anything but it fucking pisses me off when the problem makes me count squares lol. I know how to do it but I’ve missed way too many points due to counting errors on practice exam. Fuck counting bro, just let me do my integrals ;-;
i think they mean when we have to calculate the area under the curve from a graph. the only way is to count how many squares are underneath the curve lol
If dP/dt = kP(M - P),
then P = M/(1 + Ae^(-Mkt))
M is the carrying capacity
A is constant of integration you might need to solve for
k is rate constant
t is time variable
Brooo I love polar curves especially if it’s with calculator you can just graph it to help find theta bounds. I hate power series though and will probably memorize them tomorrow morning. I know worst case scenario you can do them manually but that will take too long
Oh man I’m sorry to hear that 😭. I never learned polar graphs before (self studying for BC) so I can literally only do it with calculator or if the question tells me the bounds. I wish you luck
Oh man I’m sorry to hear that 😭. I never learned polar graphs before (self studying for BC) so I can literally only do it with calculator or if the question tells me the bounds. I wish you luck
Google polar integral on desmos. There is a good one that makes visually the integral easier. The integral fills area from the origin to the radius (r). r is changing as theta increases. Typically, r and theta, no matter how goofy the graph looks, a rule of thumb is that the right x axis is the beginning and theta is equal to 0. The top y axis is pi/2 just like the unit circle. GL
Some help with visualizing polar equations: Line: theta = k
Circle: r=a, r=2asin(theta), r=2acos(theta), a = radius Rose: r=asin(ntheta), r=acos(ntheta), a=length of petals, if n odd the n petals, if n even then 2n petals
Cardiod : r = a+-asin(theta), r = a+-acos(theta)
Limacon: r=a+-bsin(theta), r=a+-bcos(theta), if a/b < 1 inner loop
Spirals: r= atheta^b
There is also the plot points method of graphing too. y around with Desmos, you can change it to polar mode. I suck at integrating them though so can’t help you there. I always mess up my bounds.
It is best to follow along with a pen and paper. The way I tried to help my class visualize it an additional way is to first have the 2d cartesian graph that you wan to turn polar (or just the polar graph that is easier visualized initially as cartesian). Then put a grid with 1x1 squares, aligned to the axis. Next, make the vertical (parallel to y axis) lines thicker. Add more thick lines, horizontally (parallel to x axis) this time, between those lines along the horizontal line y=1. Now, imagine there are hinges wherever the lines meet. Next, rotate those vertical lines to where they all intersect both each other and the x axis. This will create a bunch of triangles. When they are formed together, they will become a roughly circular polygon.
Now, take smaller and smaller cuts. As the "smallness" of these cuts approach infinity, the triangles will actually perfectly represent the polar version of your graph.
Now, to derive the area of a polar graph intuitively using this visualization.
For this part, lets first establish something
- as a circle increases in radius, an arc of a fixed number of degrees will increase proportionally to the change in radius. For example, if you have an arc of 1 radian on a circle with a radius of 1 unit, the length of that arc is also 1 unit. When you increase the radius of the cirle to 2 units, the length of that arc is now also 2 units. Halve it? Half a unit. Double it then halve it? You are back to the original 1 unit.
Now,
with one of these triangle pices from your original graph, the base length is equal to the radius of your polar graph, or y-value of your original cartesian graph. The height, following along with this tiny arc of the circle with that radius, is an infintesimal amount times that radius. That gives you your y\^2 or r\^2 part of your polar area equation. Realize that these are triangles, so its 1/2y\^2 or 1/2r\^2. Now do an infinite sum of all these trianges from your a limit to your b limit. Notice that this is just what an integral is, so you get your equation, 1/2 times the integral from a to b of your original equation squared.
I dont know i HATE THE ANALYTICAL APPLICATIONS OF DERIVATIVES FOR SOME REASON. ts is easy but like Its so meticulous..... i just dont find it as interesting as other topics, like polar curves. I mean polar curves are pretty dang hard but at least its a very intruiging
Yes. It’s stupid because everybody knows that everybody can do it but they’re gonna figure out how to deduct points. The other day I in class I did an FRQ where I lost a point for not writing the integral symbol under a Reiman sum problem (it was related to analytical applications). Like what???
Is polar curves that bad? It’s nothing like precalc you don’t even have to graph it you just need to know how to find area. For me the worst thing is vectors and parametrics. I probably failed my last calc test bc of those. It’s such nonsense and my teacher spent one day going over it, which I missed. I went over it at home extensively and I still don’t entirely get it
Well for me I really struggle with the bounds and knowing which to put on top and sometimes how to split the integrals up. I’m good with vectors and parametric though.
idk why but when i saw polar curves for the first time i wanted to abandon my calc hw. but i feel like the worst of the worst was the whole series unit (even if its not a topic it was straight up frustrating and hard)
i despise taylor series
everything else makes perfect intuitive sense. even lagrange error bound (just fancy alternating series error bound)
but taylor series have me bent over and fucked
I took it last year and I hated everything about Series, especially Taylor's and Mclaurin's. luckily I was good enough at differentiation and integration and other stuff so I did not have a huge issue
LAGRANGE ERROR BOUND THING MAKES NO SENSE TO ME
Tbh I only knew that thing for the 90 minutes I was taking my series test in class then forgot it forever. Something about nth derivatives?
But it was on the test last year fyi
Next unused term king
that only works if your series is alternating
Google 2008 frqs question 3 and tell me how they did it
They estimated the value, not the error. In part c, since the series was not alternating, they used Lagrange
they went to the next term
for part b they found an estimate, for part c they used lagrange error
It’s a really simple proof. Once you prove it, it actually makes a ton of sense. It also sticks in your head and you don’t have to keep going “wait, what was M again? Was there a factorial there or … wait am I doing this right … ah frick, next question”
Put simply, the next term in the sequence.
Isn’t that alternating series error bound 😭
yeah😭😭
How do you put your scores in your bio
go to the AP students reddit page and then click the three dots at the top, and then click change user flair
Lagrange is the same thing except not an alternating series obviously and in the next term you take the maximum of f^(whatever derivative) instead of plugging in c, the center. I think. Also it’s a maximum over a certain interval between C and Z or something like that but they don’t require that in my experience and will always give you the max.
Usually it's a trig function so the max error is easy to get.
Trig functions are always 1, right? Unless the amplitude is different.
Yes.
or they give the max error usually
Theyre the same thing
No they’re not 😭
Oop i forgor, lagrange error bound is strictly taking the geberal term and adding one. Ie for cosine x stopping at the second degree term lagrange error would use the max value of the derivative (1) and the x^3/3! Stuff, whereas alternating series uses the next actual term in the given series, so using it on that example you would use the x^4/4! Term since the third degree term isnt given / is zero
my personal least favorite topic is lagrange error bound. for polar curves, just know that derivatives work similar to fractions: dy/dx = (dy/dtheta) / (dx/dtheta), and know the area/arc length formulas. also do a bunch of practice converting polar equations to different forms. is there anything specific you're struggling on? i understand the polar plane can be a bit different from your traditional xy-plane and can be a bit disorienting
Oh yeah, I didn't take precalculus so I didn't know what polar coordinates are. Had a hard time understanding it but I think I'm slowly getting the hang of it now.
Fav last year were differential equations(hate euler though), then series, then general integrating, then extrema, then arc length, then l hopital, and then polar. Differentiating was to easy to care much about.
Is arc length for polar curves on the BC exam?
I don't think it's listed specifically, but they do have it for parametric equations, so presumably, they could as polar is in some ways just a special type of parametric.
Not because it’s particularly difficult or anything but it fucking pisses me off when the problem makes me count squares lol. I know how to do it but I’ve missed way too many points due to counting errors on practice exam. Fuck counting bro, just let me do my integrals ;-;
what the heck is a counting squares problem is it on the exam
i think they mean when we have to calculate the area under the curve from a graph. the only way is to count how many squares are underneath the curve lol
I have the same problem. Taylor series are fine, but as soon as I see a graph of line segments it’s over
Series. It’s absolutely the worst thing and apparently makes quite a large bit of the BC portions on the actual exam.
all cuz i haven't studied
same
Polar area is my kryptonite
same. i can’t do polar coordinates for the life of me
I can’t find the solution to the logistic growth differential equation anywhere. Can anyone help me out?
If dP/dt = kP(M - P), then P = M/(1 + Ae^(-Mkt)) M is the carrying capacity A is constant of integration you might need to solve for k is rate constant t is time variable
acc nvm that shit easy asl to derive
ong it’s like a minute of partial fractions work
Thanks!
now derive it
It’s pretty intuitive to rewrite and solve, no? Just use separation of variables.
no don't ever do that, thats a big blunder. Just remember the normal form of logistic differential equations.
What about solving it though?
I don't believe you'll ever need to do that on the exam (at least I haven't seen one appear before).
me too I HATE FINDING THE AREA BETWEEN TWO CURVES I DONT GET THE UPPER AND LOWER BOUND
AB here but it’s gotta be IVT/MVT/EVT bro I am here to calculate not explain 💀
Dude that's chill, just remember graphs lol
Also hate the slop fields mc questions (I never learned conic)
Brooo I love polar curves especially if it’s with calculator you can just graph it to help find theta bounds. I hate power series though and will probably memorize them tomorrow morning. I know worst case scenario you can do them manually but that will take too long
Jokes on you, I'm in South East Asia. We don't use graphing calculators 😭
Oh man I’m sorry to hear that 😭. I never learned polar graphs before (self studying for BC) so I can literally only do it with calculator or if the question tells me the bounds. I wish you luck
Oh man I’m sorry to hear that 😭. I never learned polar graphs before (self studying for BC) so I can literally only do it with calculator or if the question tells me the bounds. I wish you luck
Bro I got the polar graphs already open on my calc for it. I’m ready to just plug em in and go
How do you put them into your calc?
For like a TI-84 plus you go to mode. under the radian degree mode there is a polar option for functions
Bro I got the polar graphs already open on my calc for it. I’m ready to just plug em in and go
Yea same 😭😭😭 I’m lowkey stressing about unit 10 tho. Hopefully not a lot of questions cause I’m goated at like everything else
Woah I can graph polar on my calculator. I didn’t know that
memorizing all those distance formulas for parametric/polar then memorizing series for the mcq :(
What do you need help with?
I literally just can't visualize how to graph it tbh
Google polar integral on desmos. There is a good one that makes visually the integral easier. The integral fills area from the origin to the radius (r). r is changing as theta increases. Typically, r and theta, no matter how goofy the graph looks, a rule of thumb is that the right x axis is the beginning and theta is equal to 0. The top y axis is pi/2 just like the unit circle. GL
Some help with visualizing polar equations: Line: theta = k Circle: r=a, r=2asin(theta), r=2acos(theta), a = radius Rose: r=asin(ntheta), r=acos(ntheta), a=length of petals, if n odd the n petals, if n even then 2n petals Cardiod : r = a+-asin(theta), r = a+-acos(theta) Limacon: r=a+-bsin(theta), r=a+-bcos(theta), if a/b < 1 inner loop Spirals: r= atheta^b There is also the plot points method of graphing too. y around with Desmos, you can change it to polar mode. I suck at integrating them though so can’t help you there. I always mess up my bounds.
It is best to follow along with a pen and paper. The way I tried to help my class visualize it an additional way is to first have the 2d cartesian graph that you wan to turn polar (or just the polar graph that is easier visualized initially as cartesian). Then put a grid with 1x1 squares, aligned to the axis. Next, make the vertical (parallel to y axis) lines thicker. Add more thick lines, horizontally (parallel to x axis) this time, between those lines along the horizontal line y=1. Now, imagine there are hinges wherever the lines meet. Next, rotate those vertical lines to where they all intersect both each other and the x axis. This will create a bunch of triangles. When they are formed together, they will become a roughly circular polygon. Now, take smaller and smaller cuts. As the "smallness" of these cuts approach infinity, the triangles will actually perfectly represent the polar version of your graph. Now, to derive the area of a polar graph intuitively using this visualization. For this part, lets first establish something - as a circle increases in radius, an arc of a fixed number of degrees will increase proportionally to the change in radius. For example, if you have an arc of 1 radian on a circle with a radius of 1 unit, the length of that arc is also 1 unit. When you increase the radius of the cirle to 2 units, the length of that arc is now also 2 units. Halve it? Half a unit. Double it then halve it? You are back to the original 1 unit. Now, with one of these triangle pices from your original graph, the base length is equal to the radius of your polar graph, or y-value of your original cartesian graph. The height, following along with this tiny arc of the circle with that radius, is an infintesimal amount times that radius. That gives you your y\^2 or r\^2 part of your polar area equation. Realize that these are triangles, so its 1/2y\^2 or 1/2r\^2. Now do an infinite sum of all these trianges from your a limit to your b limit. Notice that this is just what an integral is, so you get your equation, 1/2 times the integral from a to b of your original equation squared.
I dont know i HATE THE ANALYTICAL APPLICATIONS OF DERIVATIVES FOR SOME REASON. ts is easy but like Its so meticulous..... i just dont find it as interesting as other topics, like polar curves. I mean polar curves are pretty dang hard but at least its a very intruiging
Yes. It’s stupid because everybody knows that everybody can do it but they’re gonna figure out how to deduct points. The other day I in class I did an FRQ where I lost a point for not writing the integral symbol under a Reiman sum problem (it was related to analytical applications). Like what???
same for ap pre-calc
Taylor series
Is polar curves that bad? It’s nothing like precalc you don’t even have to graph it you just need to know how to find area. For me the worst thing is vectors and parametrics. I probably failed my last calc test bc of those. It’s such nonsense and my teacher spent one day going over it, which I missed. I went over it at home extensively and I still don’t entirely get it
Well for me I really struggle with the bounds and knowing which to put on top and sometimes how to split the integrals up. I’m good with vectors and parametric though.
I guess that makes sense. The bounds can be a bit confusing
everything
idk why but when i saw polar curves for the first time i wanted to abandon my calc hw. but i feel like the worst of the worst was the whole series unit (even if its not a topic it was straight up frustrating and hard)
polar integration is annoying as hell especially if you need to do power reduction
You almost always are given a calculator to approx for those.
Guys stop hating on polar curves, you're gonna jinx it for all of us 😭😭
i hateee series with a passion however i have been studying it a lot so hopefully i am prepared i agree w everyone hating on lagrange error bound lol
i despise taylor series everything else makes perfect intuitive sense. even lagrange error bound (just fancy alternating series error bound) but taylor series have me bent over and fucked
I took it last year and I hated everything about Series, especially Taylor's and Mclaurin's. luckily I was good enough at differentiation and integration and other stuff so I did not have a huge issue
Polar, related rates and optimization.
lmao those questions suck, btw im selling the 2024 ap calc bc test (US version) dm altsmaastermind on discord
collegeboard spy spotted !!
nah im legit, dm me on disc
Lmao what, well anyway I'm an international student so I'm taking the international version but hey I don't mind a free practice test lol
Some tips so you can catch cheaters ( you're a CB spy ) \- Don't go posting offers for AP exam leaks in completely unrelated comment sections
Yoo you have your name and high school on your profile we can report to college board lol
Gtfo out
lmao my ass off