Please describe how you would approach a known cross section problem that you haven’t done before (like when the sections are trapezoids for examples). What is your approach if you can’t cut paper to help you visualize the solid that will be created? How did this activity help you understand finding volumes with known cross sections? Why might this be useful in the real world?
This week we learned the disc method and the washer method. Grasping the concepts of these methods is pretty difficult. It's hard to picture rotating a solid around an axis in your head. The disc method is used for solids. When rotating a function around an axis, I'd first graph the function, then draw in a representative rectangle, then label the rectangle accordingly. Next I would write my integral. For discs, the volume would be v=pi * the integral from a to b of f(x)^2 dx. Then I'd plug that into NINT on my calculator to get the final answer.
An example of a disc problem would be f(x)=(sinx)^.5 , from 0 to pi , around the x-axis. The graph of this function looks kind of like a semi-circle from 0 to pi. The representative rectangle would be vertical because the integral is in terms of x. the width of the rectangle would be dx while the heigh would be f(x) which was given f(x)=(sinx)^.5. My integral would be pi * the integral from 0 to pi of ((sinx)^.5)^2 dx. After integrating this, my final answer would be 2 pi. Washers are basically discs with a cylindrical hole in the middle. The volume for a washer would be V=pi * the integral from a to b of R(x)^2 - r(x)^2 dx. An example of a washer problem would be y=x^.5 , y=x^2 , around the x-axis. The representative rectangle would be vertical because the integral will be in terms of x. The width of the rectangle is dx, the big radius of the function is x^.5, and the small radius of the function is x^2. The integral would be V=pi * the integral from 0 to 1 of ((x^.5)^2) - (x^2)^2 dx. My final answer would be 3pi/10. This week we covered separation of variables. The idea of this concept was to basically get the y's on one side and the x's on one side, and then anti-deriv both sides. An example of this would be dy/dx=(x^2)/(1-y^2) find x such that y(0)=1. First I'd multiply both sides by (1-y^2) and dx to get them out of the denomonater and on the other side. My equation would be (1-y^2)dy=(x^2)dx. Then I'd anti-deriv both sides to get y-(y^3/3)=(x^3/3)+c. Next, I'd plug in the point given (0,1) to get (1)-((1)^3/3)=(0^3/3)+c. Then I'd solve for c to get 2/3. Finally, I would rewrite the anti-derived equation plus the c we just found to get y-(y^3/3)=(x^3/3)+2/3. You could rewrite this to make it look better by multiplying it all by 3 to cancel out the 3 as a denomonater. My final answer would be 3y-y^3=x^3+2.
The second part of section 6.4 was on exponential growth and decay. We were given the following equations: dy/dt=k*y y=yo*e^kt y=yo*e^-kt These types of problems are pretty easy too. You'd use the first one if they gae you k, t, and y. You'd use the 2nd one for exponential growth, and the last one for exponential decay. An example using the first equatioin would be if you were given that k=1.5 when t=0 and y=100. First you would plug in 1.5 for k to get dy/dt=1.5y. Then you would use separation of variables to get the y's on one side and the x's on the other, so you'd get (1/y)dy=1.5dt. Then take the anti-derivative to get ln(y)=1.5t+c. Then plug in the given t and y values to get ln(100)=1.5(0)+c. Solve for c to get that c=ln(100). Then rewrite the anti-derived equation with the c we solved for to get ln(y)=1.5t+ln(100). You could leave the answer as that, or you could rewrite it to look nice to be y=100e^1.5t. This week we worked on u substitution, only this time it was with anti-differentiation. It works the same way a normal u sub would work, except you need to anti-deriv the function too. An example of this would be the integral of the function cot(7x)dx. First, I changed it cot to cos(7x)/sin(7x). Next, I chose a u. The u I chose was sin(7x). You would follow the same steps as u sub by taking the derivative and then solving for du to get du=7cos(7x)dx. However, we want to get cos(7x) to cancel out instead of 7cos(7x). In order to get cos(7x) I'd divide by 7 so (1/7)du=cos(7x)dx. Then you would go back to integral, and replace cos(7x)dx with u. So you'd get the integral of the function (1/7)(1/u)du. Anti-deriv this and you get (1/7)ln(u)+c. Plug back in sin(7x) for you to get the final answer (1/7)(sin7x)+c.
Another section we covered this week focused on slope fields. We did an activity using dot paper that helped in understanding the concept. Basically you plug a point into the equation to get the slope at that point. Then you would go on the dot paper and make a small line that had the slope that it equaled. For example, if the equation was dy/dx=x+y and the point you were evaluating was (-1,-1) you'd plug the point into the equation so you would get -1+-1 and get -2. You'd go to the point (-1,-1) and draw a small line with the slope of -2. If you do this for all the points you'll be able to see the shape of the graph of x+y. While learning about the Fundamental Thereom of Calculus, I found that deductive learning was more helpful that inductive learning. I prefer deductive learning because I feel like it's more effective for me to hear a lecture, take notes, then do examples. When the process is reversed, I don't understand the material as well. For me, doing examples and then trying to understand the concept doesn't do much because I don't know what I'm doing during the examples. It's like trying to build a machine without the instructions. However if I'm given the instructions first, I kind of know the general idea of what I'm attempting to do. Deductive learning can be extremely boring at times, but in the end I have solid notes and examples that I can use if I get stuck on the homework.
I believe the Fundamental Thereom of Calculus is important because, just like any other fundamental idea, it's the basic building block of what's to come. This is the central idea of which we base more complicated concepts on. In order to get into more complicated problems, it's essential that the Fundamental Thereom of Calculus is understood. The first part of the thereom shows the relationship between the derivative and the integral. It says that the derivative of f(t) on the integral from a-x = f(x). So basically the orginal function equals the derivative. The second part of the thereom shows us how intergrals and anti-derivatives are related. It says that integral f(t) on the interval a-b = the anti-derivative of b - the anti-derivative of a. An example of this would be the integral of sinxdx on the interval 0 to pi. First you'd anti-derive it so you'd get -cosx. Next, you would subtract the anti-derivative of a from the anti-derivative of b so it would be -cos(pi) - -cos(0) which would essentially be 1 + 1 which would equal 2 as your final answer. |