Sunday, October 18, 2009

post 9

This week in calculus we took our exam. We took the multiple choice portion on wednesday and the free response portion on thursday. We had six packets that we worked on for two weeks to help us review all the way up until the exam. it had chapters 1 through 3. It went over stuff from finding points of discontinuity of the graph of a limit approaching a number, all the way up to optimization. On Friday, we started learning again. We learned implicit derivatives. They are very easy, in fact, the only thing is you MUST know how to simplify correctly. It is just like finding a regular derivative, except instead of only having an x-value in the function, you have both an x-value and y-value. you find implicit derivatives by identifying whether or not it is an implicit der by seeing if it has a x and y value. you take the der of both sides. everytime you take the der of a y value you put dy over dx. then you solve for dy over dx

im still not completely comfortable with optimization because im not really good at finding the different variables. holla at me if you wanan help
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1 comment:

  1. LE2010October 21, 2009 at 4:42 PM

    for optimization..

    lets say that the promblem says:
    solve for the total length and width of two squares that are not connected which have a perimeter of 24!

    so 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!!

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