Monday, March 1, 2010

post 28

all we did this last week was take an AP test, and then make corrections on it. and this three day weekend thing screwed me up, so i'm a day late on my blog, but anyways, i'm going to do this one on implicit derivatives because that's a super easy topic. :)


Implicit derivatives involve both x's and y's, unlike normal derivatives.
1: So, first you have to take the derivative of whatever they give you as you normally would.
2: Whenever you take the derivative of y, you have to note it with dy/dx.
3: Solve for dy/dx
(if you want to find slope plug in an x and y value)

example: y^3+y^2-5y-x^2=-4
First you just take the derivative, but don't forget to not the derivatives of the y's! So you get: 3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0

Then you have to solve for dy/dx, so you get:
dy/dx(3y^2+2y-5)=2x which then is further solved for to get dy/dx=2x/(3y^2+2y-5)

and that's it for that problem, it's done.

Another example:
Okay, let's say you want to find the slope of 3(x^2+y^2)^2=100xy at the point (3,1)

First you take the derivative, which involves all kinds of product and exponent rule...
6(x^2+y^2)(2x+2y(dy/dx))=100(y+x(dy/dx))

then, you need to foil it n stuff, so you get:
12x^3+12x^2(dy/dx)+12xy^2+12y^2(dy/dx))=100y+100x(dy/dx)

then, as usual, you would have to solve for dy/dx:
dy/dx=(-12^3-12xy^2+100y)/(12x^2+12y-100x)

after you solved for dy/dx, you plug in your x and y value from the point given to get your answer, so I think the final answer would be: 1.84 (if I put it in my calculator right)

that is it for implicit derivatives, and they are really easy to identify, it is the exact same thing as a derivative pretty much, just with x's and y's. Just don't forget to plug in the point that some problems will give you at the end, I have forgotten to do it before.

One thing that i still screw up on is MRAM, i don't know what my problem is, even when i look at the formula i still can't solve the problem, so if someone could review that, it would be good.

3 comments:

  1. Start the problem as you would LRAM, then find the midpoint of those numbers you use, starting from a and plug into the formula.

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  2. MRAM: All you have to do is plug into the formula
    delta x [f(a+delta x) + ... + f(b)]
    Take the problem Find the area of f(x) = x-3 on the interval [0,4] with four sub intervals.
    Step 1: is to find delta x, which is found by b-a/n- 4-0/4 = 1
    Step 2: Find the midpoint between a and b, by adding delta x to a until you get to b.
    So your number will be: 0 1 2 3 4
    Find the midpoint between each which you do by adding the two then dividing by two.
    0+1 / 2 = 1/2
    1+2/2 = 3/2
    2+3/2 = 5/2
    3+4/2 = 7/2
    Step 3: Plug the midpoint values into the formula
    1 [f(1/2) + f(3/2) + f(5/2) + f(7/2) ]
    Step 4: Plug each into equation and simplify
    1 [(1/2 -3) + (3/2 - 3) + (5/2 -3) + (7/2 -3) ] = -4
    Something to remember: area cannot be negative, so in this case, the answer would be 4.

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  3. again i had no computer last night so hope this counts.....

    MRAM-middle hand approximation

    The trick to finding the middle sum is having the capability of finding all of the midpoints in the given interval.

    x[f(mid) +…]

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