1/20/09 We started class today by looking at the integral from 0-1 x^2 dx compared to the integral from 0-1 t^2 dx We decided that it would be the same numerical answer because it doesn't matter what variable is used, the equation and dx is the same, so it would result in the same numerical answer next we looked at accumulating area function for Ax=the integral from 1-x 6dt. We graphed this function as t equalling the horizontal axis, y the verticle, and the equation at y=6. Next, we made a table of value for x vs. Ax. When x=0, Ax=-6 x=1 Ax=0 x=2 Ax=6 x=3 Ax=12 This was to show that it doesn't matter what variable is used in the equation, but in this exercise it was importaint to differentiate between x and t
Next, Mr. Hansen passed out the worksheet Accumulating Area Function (Indefinite integrals)and we looked at the similarities and differences between the graphs and values of the intergral graph vs the graph of the function. If A=the integral from a to x f(t)dx is an anti-derivitive of f. Or f is the derivitive of A.
On Wednesday, January 21, we went over what the integral of f(t)dt as a approaches x equal in terms of f(x) and a shortcut to be able to evaluate limits using this rule. First we needed to review that A’(x) was really equal to the limit as h approaches zero of A(x+h) –A(x)/h. So then we had to go over and clarify what we had discovered yesterday. First we decided that A(x) equals the integral of f(t)dt as a approaches x is actually a function of x, meaning that f(t) does not change, the x value does. Then we finally decided that yesterday we discovered that A is an anti-derivative of f or A’=f (either way is correct to look at it). So then we used a graph to discover why A’ must equal f. Finding the graphical representations of every part of the limit as h approaches zero of A(x+h)-A(x)/h we found that it was approaching f(x) meaning that A’(x) must equal f(x). A’=f is the Fundamental Theorem of Calculus or FTC because it ties together the two most important concepts in calculus this year: derivatives and integrals. So the FTC shortcut is to find the anti-derivative of f(t) (without worrying about a possible constant c) and use that for both endpoints subtracting the ending point and starting point as you plug them into the anti-derivative equation. So for example in the integral of 3t^2dt as 2 approaches 5 we find that the anti-derivative equal x^3 so then we take that and create the equation 5^3-2^3 must equal the integral of 3t^2dt as 2 approaches 5. So a practice problem we went over in class was the integral of x^2dx as 0 approaches one, so you can try that on your own (answer = 1/3).
Today in calculus in class our new material was to learn to rotate a shape around a line other than a coordinate axis. So taking this example of three points connected by two perpendicular lines with point R at (6,x), Point P at (x,y) and Point Q at (10,y).We had to review some things to help with learning how to do our goal in class. So in terms of x, PQ = 10-x. And In terms of y, PR= 6-y. So next we actually went to rotate a graph around a separate line. So taking the graph y=4√x (which you can put on your graph screen) we rotated that around the line x=9 to produce a “pitcher’s mound.” Remember when you draw in the boxes of your individual pieces you always want to put them perpendicular to the line you rotated so for this graph these will be horizontal boxes making your thickness dy. Then finding the radius of a disc to estimate the area would make the radius 9-x. That makes the equation for the volume of the disc, V= π (9-x)^2 dy. But you can not have one term in terms of x with a dy so we find x in terms of y to substitute making our final expression, the integral from 0 to 12 of π(9- y^2/16)^2 dy. Basically the process is the similar to rotating an equation around one of the axis but with some minor differences.
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1/20/09
We started class today by looking at the integral from 0-1 x^2 dx compared to the integral from 0-1 t^2 dx
We decided that it would be the same numerical answer because it doesn't matter what variable is used, the equation and dx is the same, so it would result in the same numerical answer
next we looked at accumulating area function for Ax=the integral from 1-x 6dt. We graphed this function as t equalling the horizontal axis, y the verticle, and the equation at y=6. Next, we made a table of value for x vs. Ax.
When x=0, Ax=-6
x=1 Ax=0
x=2 Ax=6
x=3 Ax=12
This was to show that it doesn't matter what variable is used in the equation, but in this exercise it was importaint to differentiate between x and t
Next, Mr. Hansen passed out the worksheet Accumulating Area Function (Indefinite integrals)and we looked at the similarities and differences between the graphs and values of the intergral graph vs the graph of the function. If A=the integral from a to x f(t)dx is an anti-derivitive of f. Or f is the derivitive of A.
On Wednesday, January 21, we went over what the integral of f(t)dt as a approaches x equal in terms of f(x) and a shortcut to be able to evaluate limits using this rule. First we needed to review that A’(x) was really equal to the limit as h approaches zero of A(x+h) –A(x)/h. So then we had to go over and clarify what we had discovered yesterday. First we decided that A(x) equals the integral of f(t)dt as a approaches x is actually a function of x, meaning that f(t) does not change, the x value does. Then we finally decided that yesterday we discovered that A is an anti-derivative of f or A’=f (either way is correct to look at it). So then we used a graph to discover why A’ must equal f. Finding the graphical representations of every part of the limit as h approaches zero of A(x+h)-A(x)/h we found that it was approaching f(x) meaning that A’(x) must equal f(x). A’=f is the Fundamental Theorem of Calculus or FTC because it ties together the two most important concepts in calculus this year: derivatives and integrals. So the FTC shortcut is to find the anti-derivative of f(t) (without worrying about a possible constant c) and use that for both endpoints subtracting the ending point and starting point as you plug them into the anti-derivative equation. So for example in the integral of 3t^2dt as 2 approaches 5 we find that the anti-derivative equal x^3 so then we take that and create the equation 5^3-2^3 must equal the integral of 3t^2dt as 2 approaches 5. So a practice problem we went over in class was the integral of x^2dx as 0 approaches one, so you can try that on your own (answer = 1/3).
Today in calculus in class our new material was to learn to rotate a shape around a line other than a coordinate axis. So taking this example of three points connected by two perpendicular lines with point R at (6,x), Point P at (x,y) and Point Q at (10,y).We had to review some things to help with learning how to do our goal in class. So in terms of x, PQ = 10-x. And In terms of y, PR= 6-y. So next we actually went to rotate a graph around a separate line. So taking the graph y=4√x (which you can put on your graph screen) we rotated that around the line x=9 to produce a “pitcher’s mound.” Remember when you draw in the boxes of your individual pieces you always want to put them perpendicular to the line you rotated so for this graph these will be horizontal boxes making your thickness dy. Then finding the radius of a disc to estimate the area would make the radius 9-x. That makes the equation for the volume of the disc, V= π (9-x)^2 dy. But you can not have one term in terms of x with a dy so we find x in terms of y to substitute making our final expression, the integral from 0 to 12 of π(9- y^2/16)^2 dy. Basically the process is the similar to rotating an equation around one of the axis but with some minor differences.
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