merf's 1st post

This week in calculus, the learning began with average speed, which is basically a slope. To find the average speed you use the formula m=(y2-y1)/(x2-x1) , but first you have to find your y's and x's. For example, an anvil falls off the roof onto Ryan G's head, what is its average speed during the first two seconds if y=16t^2 describes the fall. Alright, to begin, the points you have to use are 0 and 2 because the length is the first two seconds, which would be the starting point (0) and the other point (2). So you plug both of those into the formula given in the problem, and so that's how you find your y's. So if you plug your x's and y's into the formula m= (y2-y1)/(x2-x1), you will get m=64/2, which is 32. so your final answer is 32 length per second. For Instantaneous speed you do the same thing, except you use the formula lim h->0 (f(x+h)-f(x))/(h). The example problem given to use for that one is extremely long, so I'm not even going to go there... Another thing I learned and know how to do pretty well (other than taking derivatives) is using the quotent rule, the formula is (vu^1 x uv^1)/v^2, but the way I learned it is, you take the derivative of the top, multiply it by the bottom, minus defivative of the bottom, times the top. For example, let's say you were given the problem y=(4x^(3/2))/(x), first you have to take the bottom, and multiply it by the derivative of the top. So the begginning of the problem would like like, x(6x^(1/2)). After that, you must subtract the derivative of the bottom, multiplyed by the top. So the second part of the problem would look like, -1(4x^(3/2)). So the full problem would now look like x(6x^(1/2))-1(4x^(3/2)). After you multiply all that out, you get (6x^(3/2))-4x^(3/2)), after you further simplify that, the final answer should be 2x^(3/2). THE END. ( I think this is like 330 something words)

mher out...