Wow, 8 weeks into the school year. sounds like a long time. this week was just reviewing for our exam which is in two parts on wednesday and thursday of this week, wednesday we are taking the multiple choice part of the exam. and on thursday we are taking the free response section, which mrs. robinson helped me abbey stephanie and trina with today at her house :) i would tell you what's on it, but you should've came to the study group.
ok, ill stop being mean now, and lets talk about calculus. at the study group today, i learned a lot about the process of elimination when it comes to multiple choice questions. also, if there is nothing else you can do with a problem that gives you a function and asks you to find something on a certain interval, like find the absolute extrema on the interval [0,2pi], then you can plug it into your calculator if you absolutely cannot figure it out by hand. also, short cuts are your friend :) like using the table function for problems like these. all of this really helped on the chap. 3 multiple choice part of our study guide.
some things i understand now are how to find a tangent line. the rules of finding a limit approaching infinity. how to take simple derivatives. instantaneous and average speed. all of these things. also how to look a a graph of a function and be able to know what the graph of the derivative of that function will look like, thank you dylan becnel :) haha. i also understand rolle's theorem and mean value theorem very well. what should i explain today? i think i'll explain how to look at the graph of a function and be able to know what the graph of the derivative of that function will look like.
Example:
let's say you are given the graph of a horizontal line. well, you know that the function has to look like this f(x)=#. like f(x)=3. and the horizontal line would be at 3. well the derivative of that would be 0. so the graph of the derivative of that function would be nothing. (chap. 3 multiple choice number 4)
the same thing i still don't understand every week. OPTIMIZATION. i hate it. i did learn today though that whenever you optimize a rectangle you will always end up with a square, which means whenever you work it out your x & y should be equal to the same thing. (chap. 3 multiple choice number 3) the x & y both ended up being equal to 11/2.
even though i got b rob to help me today with it, i still cannot do it by myself. any tips? just for optimization in general? please & thank you!
good luck on all of your exams tomorrow everyone :)
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ok for that optimization question...the number four on the test we just got back.....
ReplyDeleteso we know that we have a perimeter of 24 right...well to make your secondary and primary equation you make two squares...we know that the area of a square is lenth times width! and that there is two lengths and two widths for each square so your seconday equation will be 2(2L+2w)...that's your perimater because you're adding them...and since we know that the sum of the two perimeters is 24 you set 2(2L+2w) equal to 0!
so now...dealing with optimization:
take your secondary equation and solve for...lets go with length
2[2(L)+2(w)] = 0
24=4(L)+4(w)
24-4(w)=4(L)
(L)=6-w
so now plug in L to the primary equation which is your area...so that's
A=(L)(w)
A=(6-w)(w)
A=6w-(w)^2
TAKE THE DERIVATIVE OF THAT:
A'=6-2w
-2w=-6
w=3
SO NOW TAKE WHAT YOU KNOW....W=3....and PLUG IT INTO THE SECONDARY FORMULA:
24=2(2(L)+2(w)
24=2(2(L)+2(3)
24=4L+12
12=4L
L=3
so knowing your W is 3 and your L is 3
[it's a square so they had to be the same]
you know that the width is three and the length is three!!
ellie, you just saved my life. haha
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