Ok, let me just say that I was somewhat scared when I walked in on Monday not because I didn't think I could do the work, but jsut because it was CALCULUS. However, I stressed myself out over nothing as usual.
To me, calculating the average speed was the easiest thing I've learned so far because it's just basic algebra with the slope formula. First, you identify the time you start and the time you end [t1, t2]. Each t value is the equivalent of a x value in the basic slope formula (y2 - y1)/(x2 - x1). Then you plug each t value in as your x in the equation given to describe the fall. The outputs are your y values. After plugging everything into your equation for slope, you get your average speed.
B-Rob stressed to us that we had to remember the Long way to find instantaneous speed (rate of change, definition of derivative), and I feel confident that I can do this becaus plugging into a formula is my thing. I like things thatare set in stone, so there's no question in my mind that I'll have a problem with this. You just take the limit as h goest to 0 of the function plus h minues the function all over h. So, this is it:
lim f(x + h) - f(x)
h-0 h
Then you just manipulate algebra to get the instantaneous speed or the derivative.
I really have a true understanding of everything, even after thirty-six formulas for derivatives were thrown on the board. For instance, the product rule states that when multiplying two things, you take one original times the derivative of the other and add that to the other switched up (this is better: (uv1 * vu1). I also understood how the quotient rule works and among other things (except, I keep forgetting the bottom of the quotient rule--my problem!).
The only things I'm unsure about do not actually involve doing the steps and formulas, just how to end the problem.
1. Like, after I take a derivative using the quotient rule and get a fraction, do I leave the bottom squared or foil it out?
2. Also, if I have a fractional exponent in the denominator, do I have to change it to a root and multiply by the conjugate to get rid of the radical on the bottom?
3. On the Section 2.3 sheet # 35, I didn't know what to do with three, so if anyone could help me, that would be great!
4. On the big packet (one with like everything), I didn't know how to do the ones where it started at a certain height. Could anyone explain to me an example (like #34?)? Thanks!
So overall, I think I had a good first week in Calculus. I learned a lot and am so psyched about the possiblity of getting college credit by taking the AP exam.
--Malerie
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Very nice Malerie
ReplyDeleteAlrighty now. I feel accomplished, I'm actually going to attempt to answer Malerie's problem other than the other way around.
ReplyDelete1. You leave it squared because it's in simplest terms this way.
2. Also, if I have a fractional exponent in the denominator, do I have to change it to a root and multiply by the conjugate to get rid of the radical on the bottom? No. If it's a simple problem such as this one: 5/x^2 you follow the steps below.
1) Bring the 5 down and make the exponent negative: 5x^(-2)
2) Multiply the constant by the exponent and minus 1 from the exponent: -10x^(-3)
3) Take the 10 and put it back on top (where the 5 was) and take the negative off of the exponent.
Whatever you do, DON'T mess with radicals. Change them into fractions; they're much easier.
3. I had a question on #35 too, but this is what I did. (3x^3+4x)(x-5)(x+1) [[original]]
1) I took the product rule for (3x^+4x)(x-5) and got 12x^3-45x^2+8x-20
2) I then took the derivative of that and got (36x^2-90x+8)
3) Then I multiplied that and (x+1) using the product rule and got 48x^3 - 99x^2 -74x-12
4. I'm pretty sure I did this wrong, but I just ignored that part, thinking that they gave you too much information to confuse you. I didn't remember doing anything like that in class.
thanks ashley!
ReplyDeletewow, this is a change! joking!:)
haha, thankks, way to make me feel better about myself
ReplyDeleteDid it help you any? Maybe? Even a little?
I can help with the whole question if you have to change the bottom to a radical and then multiply by the conjugate. The answer is no with a simple problem like 8/x^3. You simply move the bottom to the top, making the exponent negative. You then take the derivative, which would be -24x^(-4). But after doing so, you must remember to bring everything back down so there's no negative exponents. So you will just keep the -24 on top and the x^4 on bottom as your answer.
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