Alright so calculus is not all that I thought it would be. Its not that hard, if you can grasp the basic concepts. It uses your basic algebra skills and techniques that you picked up in algebraI and algebraII. Finding derivitives is not that hard, and quotient and product rule is pretty basic also. I find these so basic because it reliys on those algebra skills. The only thing that I need with these concepts is more work and play so that I can fully understand them and all the minor details that they involve, which I'm sure we will all get. For instance, I can find the simple derivitive of x^2 which = 2x. And I can find the derivitves of things such as (9x-2)/2x by using quotient rule. This concept is not hard. Something that is a little more confusing to me is x(1-(4x-2/2x)). This problem was something like one that was on the worksheet that we got on friday. I understand that it is product rule, but with that hole thing...(4x-2/2x) is messing with me. Its like quotient rule inside of product rule. lol ahhh!!! Like I have no clue where to start. Also I need more help on the shortcuts that we learned in class. I can pretty much get them besides like the ones that are like (2x-2/3). Instead of using qoutient rule cant i just make the top multiplied by the three raised to a negative 1 power (3^-1)???? idk i think that was one of the shortcuts.
Also I need more work with the shortcuts of finding average speed and instantanious speed. I get you find the derivative or something but idk where to start for the most part. Need work on those concepts. Im going to just through this out there too...I need to remember my junk from adv. math that I forgot.
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Ryan,
ReplyDeleteYou need to EXPLAIN something!!
so for Average Speed:
ReplyDeleteTake the problem they give you then take two numbers from the problem and make them a point. For example:
A water balloon is dropped and falls at y=4.9t^(2)m in (t) seconds. Find the average speed during the first 3 seconds.
so you know that your point will be (0,3) because it's the FIRST THREE SECONDS. so you plug in the 0 in order to get the answer for t1. [you should get 0] Then take the 3 from the point and plug it in, in order to get the answer for t2. [you should get 44.1]
now, the ANSWER you got AFTER you pluged it in goes AT THE TOP. it'll be t2 - t1
it should be 44.1 - 0
this will go over the point..remember instead of x and y it could be known as t1 and t2.
take t2 and subtract t1 from it.
it should be 3 - 0
now that you have both t2 and t1 for both things, the one that you got when we pluged it in...the one that ended up being 44.1-0...you put that at the top of the equation and then put the point which ended up being...3-0...as the denominator. giving you:
44.1 - 0
________
3 - 0
whis is 44.1 divided by 3 giving you
14.7 meters per second for the average velocity
[dont forget the meters per second!!]
now for instantaneous speed:
ReplyDeleteall you need to do is take the derivative of the equation they give you then plug in what they give you "t" is equal to.
for example: [number 29 on the HUGE packet]
Given the position equation: s(t)=t^(2)-3t-1, find the instantaneous velocity at t=3.
so all you need to do is take the derivative which is:
2t-3
so now just plug in 3 for all the "t"s. giving you:
2(3) - 3
so that's 6 - 3
which is 3!
THATS YOUR ANSWER!