Today in class we talked about the Reverse Chain Rule/ U-substitution/ Simple Substitution (pick your favorite name for it). Pretty much, when we need to solve for the anti-derivative of a composite equation [such as f(g(x)) ], we use this technique to solve antiderive accurately, taking into account the inside function as well as the outside function.
This process basically consists of two parts:
1) Find u (or the inner function). Once we have established what the inner function is, we can rewrite the original integral in terms of "u" and make our lives easier.
2) Find a relation between dx and du. Since our integral is in terms of dx when we first start off, it is useful to make it in terms of du instead. How do we do that? you may ask. Well, du is pretty much equal to the derivative of the "u" equation that we found above. Once we figure out that, we set the du/dx = (the derivative of du). then we solve for dx.
We plug in this value for the dx in our integral. Pull out any coefficients to the front and find the anti-derivative of the "u" term. Don't forget to multiply the coefficients in the end!!!
Then, once that is done with, you plug in your inner equation function for the "u" value and simplify if needed. ------- Sometimes, you get a function where you have two choices for "u".
Such as the function:
[tan x]^3[sec x}^2
Here: one of the options works better than the other, and it is up for you to decide which choice is better. (in this case, it is easier to set [tan x] as "u" because you end up with all [sec x] in the end)
Today in class, we explored what happens when we make a table of values for the function A(x) equals the integral from 1 to x of 1/x. We determined the domain and range of A(x) by looking at the table and graph of 1/x and found that the graph of A(x) would never have a slope of zero. We also found that A(c) + A(k) = A(ck), k(A(c)) = A(c^k), and A(c) – A(k) = A(c/k). By putting together all of this information, we determined that the function A(x) is actually the naturally occurring logarithmic function or natural log, denoted by lnx. This was further proven by the fact that A(e) = 1. We finished the period by doing a worksheet to review logarithmic functions.
Date: 3/9/2009 7:20:02 PM Monday. Today in class we talked about e and the means by which “e” was calculated. We were given a handout in order to explore the methods of discovery. Some really important things to remember are located at the top of the first page of the “e” finding packet. Quote: If a function is continuous the limit of the function = the function of the limit = a value f(c) After the exploration we discovered that the definition of e is: e= lim(x-0) of (1 + x)^ (1/x) or e = lim (n-infinity) (1+ 1/n)^n we also figured out that if x were to be placed in stead of the numerator in the equation above then its limit would be e^x We then took all of these concepts and applied them to interests. Continuous interest is calculated by Money = initial e^R%T
Today, March 10, we first looked at the homework. Then, we began to look at "inverse" function, which as Euler pointed out, could be better described as reverse functions, or undoing functions. Our prime example of inverse functions is f(x)=e^x, whose inverse is f^(-1)(x)=ln(x) (To calculate the inverse function, take the original and switch the x's and y's). e^(lnx)=x because the effects of the two functions cancel out, similar to x^3 and x^(1/3).
We then began to work on differentiating e^x packet. Here, we applied the rules of differentiation on this versatile function and applied the rules of exponents to simplify equations. Parts of the packet are included in homework.
Today in class, we pretty much discussed slope fields. Slope fields are just visual representations on what slopes of individual points on a particular graph would be. Basically, to make a slope field, you need to find the slope of an individual point. Since you would be given the dx/dy equation, all you have to do is go around the coordinate graph, pick points, plug the numbers in for the variables in the dx/dy equation, and then draw a little tick mark resembling what a possible tangent line would be at that particular point.
Through this process, you can easily guess what certain graphs would look like. If you are asked to graph a function that goes through a certain (x,y) point, all you have to do is guess what the function looks like at that point in relation to the slope tick mark and draw the rest of the graph in accordance with the other slopies (or tick marks of slopes).
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Today in class we talked about the Reverse Chain Rule/ U-substitution/ Simple Substitution (pick your favorite name for it). Pretty much, when we need to solve for the anti-derivative of a composite equation [such as f(g(x)) ], we use this technique to solve antiderive accurately, taking into account the inside function as well as the outside function.
This process basically consists of two parts:
1) Find u (or the inner function). Once we have established what the inner function is, we can rewrite the original integral in terms of "u" and make our lives easier.
2) Find a relation between dx and du. Since our integral is in terms of dx when we first start off, it is useful to make it in terms of du instead. How do we do that? you may ask. Well, du is pretty much equal to the derivative of the "u" equation that we found above. Once we figure out that, we set the du/dx = (the derivative of du). then we solve for dx.
We plug in this value for the dx in our integral. Pull out any coefficients to the front and find the anti-derivative of the "u" term. Don't forget to multiply the coefficients in the end!!!
Then, once that is done with, you plug in your inner equation function for the "u" value and simplify if needed.
-------
Sometimes, you get a function where you have two choices for "u".
Such as the function:
[tan x]^3[sec x}^2
Here: one of the options works better than the other, and it is up for you to decide which choice is better. (in this case, it is easier to set [tan x] as "u" because you end up with all [sec x] in the end)
Today in class, we explored what happens when we make a table of values for the function A(x) equals the integral from 1 to x of 1/x. We determined the domain and range of A(x) by looking at the table and graph of 1/x and found that the graph of A(x) would never have a slope of zero. We also found that A(c) + A(k) = A(ck), k(A(c)) = A(c^k), and A(c) – A(k) = A(c/k). By putting together all of this information, we determined that the function A(x) is actually the naturally occurring logarithmic function or natural log, denoted by lnx. This was further proven by the fact that A(e) = 1. We finished the period by doing a worksheet to review logarithmic functions.
Date: 3/9/2009 7:20:02 PM
Monday. Today in class we talked about e and the means by which “e” was calculated. We were given a handout in order to explore the methods of discovery.
Some really important things to remember are located at the top of the first page of the “e” finding packet.
Quote:
If a function is continuous the limit of the function = the function of the limit = a value f(c)
After the exploration we discovered that the definition of e is:
e= lim(x-0) of (1 + x)^ (1/x)
or
e = lim (n-infinity) (1+ 1/n)^n
we also figured out that if x were to be placed in stead of the numerator in the equation above then its limit would be e^x
We then took all of these concepts and applied them to interests.
Continuous interest is calculated by Money = initial e^R%T
Today, March 10, we first looked at the homework. Then, we began to look at "inverse" function, which as Euler pointed out, could be better described as reverse functions, or undoing functions. Our prime example of inverse functions is f(x)=e^x, whose inverse is f^(-1)(x)=ln(x) (To calculate the inverse function, take the original and switch the x's and y's). e^(lnx)=x because the effects of the two functions cancel out, similar to x^3 and x^(1/3).
We then began to work on differentiating e^x packet. Here, we applied the rules of differentiation on this versatile function and applied the rules of exponents to simplify equations. Parts of the packet are included in homework.
Today in class, we pretty much discussed slope fields. Slope fields are just visual representations on what slopes of individual points on a particular graph would be. Basically, to make a slope field, you need to find the slope of an individual point. Since you would be given the dx/dy equation, all you have to do is go around the coordinate graph, pick points, plug the numbers in for the variables in the dx/dy equation, and then draw a little tick mark resembling what a possible tangent line would be at that particular point.
Through this process, you can easily guess what certain graphs would look like. If you are asked to graph a function that goes through a certain (x,y) point, all you have to do is guess what the function looks like at that point in relation to the slope tick mark and draw the rest of the graph in accordance with the other slopies (or tick marks of slopes).
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