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.
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Start the problem as you would LRAM, then find the midpoint of those numbers you use, starting from a and plug into the formula.
ReplyDeleteMRAM: All you have to do is plug into the formula
ReplyDeletedelta 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.
again i had no computer last night so hope this counts.....
ReplyDeleteMRAM-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) +…]