I'm the type that can't let a problem go, and even though class is tomorrow I want to know right now why the **** the height of this particular rectangle is 1/4 and NOT what it CLEARLY should be in 1/2. Hell, you don't even have to know calculus to help me. You just have to have a pair of eyes capable of seeing 20/20. It's an Upper and Lower sum problem for the area under/above a function using circumscribed rectangles (those that go slightly above the function in their formation) and inscribed rectangles (those that go slightly below the function in their formation.) Now, I've separated the widths (or delta x's) into fourths between 0 and 1. You're supposed to pick the right endpoint of each interval all the way to 1 , starting from 0. so the right end point of 0 and 1/4 would be 1/4 and the right endpoint of the next interval between 1/4 and 2/4 would be 2/4 , so on and so forth. I'm trying to find the height of the right endpoint of x = 1/4 of the first interval. I'm finding the upper sum so you're supposed to use the circumscribed rectangle's height for the height. Simple enough right? The problem is I swear to God , up and down on a stack of 5000 bibles, that the ****ing height for x = 1/4, is ONE HALF. The book , calcchat.com, cramster.com swear that the height for the rectangle whose base is 1/4 .... is 1/4. I'm not blind am I ?
dude, relax. you are right, so I'm not sure what you've read elsewhere that is confusing you. the graph is of the formula: y = square root of x. Thus, when x = 1/4, y = 1/2. Clearly, we can all agree that the square root of 1/4 is 1/2.
You are right and it's good to question text books - they are definitely not infallible and I used to find about one error every hundred questions even several editions down the line. Only thing I could think of is that it wants you to average out the approximations of heights 0 and 1/2 = 1/4 to use in approximating your rectangles... but it doesn't explicitly say so so I think you're doing nothing wrong. Bring it up with the teacher.
yeah that's what you plug into the function after you get the height. for x = 1/4, y = 1/2... that's absolutely right here's how the book's site worked the problem out the general formula for the upper sum deal is where delta x would be 1/4 and Ci would be the height. so the answer for delta x = 1/4 should be .... sqrt (1/2) X (1/4) = .176
I'm taking Calculus online, I will pay someone to do all my homework and take the online quizzes and tests.
How is that going to remotely help you with your degree? You mine as well not even take the class if you're going to do that.
I just want to pass, this stuff in Cal is ridiculous. When in life will you ever use this stuff... unless you want to become a engineer.
Man, looking at this I realized that I could have done this in a heartbeat in high school, but after college now, I have no clue how it's done.
TMac640 - the website's answer is correct - maybe you mis-worded what you thought the answer should be in your original post? Remember that approximating areas upper and lower sums means we're adding up areas of rectangles. The points 1/4, 2/4, 3/4, 1 are the right-end points on the x-axis that we plug into the function to find out the corresponding y-values. These y-values in turn are the respective heights of each rectangle. The width for each of these rectangles is the same: 1/4 (which we got from dividing the length 0-1 into four parts). So for the first rectangle, you should be reading (I keep where we get the height of rectangle values on left side and width of rectangle values on right side so make sure you understand where each step is coming from): A = height*width = y* delta_x = f(1/4) * 1/4 (plugged in the point 1/4 into the function [our right-end point for the first approximation]) = sqrt(1/4) * 1/4 = 1/2 * 1/4 = 1/8 (which is what the site also has for the first rectangle but they combined it under the common denominator 8) I think this is where you might be confused. Ci are not heights. They are the points we pick on the right or left side of the rectangle and what we plug into the function f(Ci) that churns out y values that do correspond to heights. - To Kate81 I am a student who also works as a tutor/educator and I have to say people like you really should be ashamed. You are not only cheating what should be one of the last institutions where certain principles should be upheld, but you are also making honest hardworking people less likely to get ahead.
Everyone else may demonize you but I commend you are getting the job done at all costs. College degrees are stupid and a waste of time. You paid your money, you should get your piece of worthless paper. Kudos to you ma'am.
I don't know what to say to this other than this mentality is why America is losing jobs to lower priced yet better talent overseas. When you're a basketball athlete, you train your triceps, your form, your rhythm, your focus and patience until you can shoot at the highest % possible. Your brain is the same - it requires exercise. If you don't believe me, look up how neural pathways work and how they increase their interconnectedness and get thicker myelin sheaths. What you're missing is that you don't know how to perceive mathematics and what benefit you can get from it. I ask history or literature students the exact same question at the start of every single semester. Why do you think they force you to take math in your major? It's not so you can learn how to count or find areas. It's because it teaches you how to think, how to perceive and how to focus and solve problems. There is no field more dedicated to pattern matching than math. As for the commending the cheating aspect of it... great... +1 to you but -a lot more to all of society.
I think I see where you're coming from and I did indeed misinterpret Ci as the height. I'll look at this in a little while, I'm currently busy not understanding Java. Really really appreciate you helping out. I know these threads are subject to a few wisecracks and what not because of the nature of my username, but it's really helpful to have replies like yours. Thanks man.
The book is correct. the upper sum is the height times width of *each* rectangle. The height changes for each rectangle but the width remains the same. You seem to be making the height the same for all rectangles. delta x is the width = 1/4 n=number of rectangles j=the current rectangle in the sum F(Cj) = height for current rectangle. F(Cj) = sqrt (j/n) since j/n is the x-value (right-most position) on the graph for a given rectangle. so, since area is width x height width = 1/4 height = sqrt (j/n) where n=4, and j goes from 1 to 4. note: the second line in the books solution is just factoring out line #1.
College degrees are worthless. If colleges really wanted to prepare students, how about they teach something that will help them in the real world and not create a ridiculous amount of debt for people that are just trying to get started in life. The vast, vast, vast majority of people coming out of college are no better prepared than when they went in, they are just several years older and have paid a whole lot of money to fool themselves into thinking that they are.
note also that if you only what part of the area under a graph, it slightly complicates. Your "general formula" assumes x starts at point 0.
