DylsexicChciken

The problem was that the limits and radius are the same for two different graphs. That didn't make sense considering they are two different graphs. It turns I used a graph that was a special case where the limits stay the same when converting between variables. But the question arises, why can't we use the same logic that studiot pointed out above this post and integrate using both variables for solids of revolution?

So limits of integration with bounds (0,0) to (1,1) for [latex] y=x^n [/latex] is a special case where the limits of integration are equal for both variables since [latex] 1^n=1 [/latex]. I guess that's why I was confused. Wow, I've outdid myself this time. But what about solid of revolution? We can't use both variables for solid of revolution?

The problem is the value of the radius. Surface areas can be calculated with either y variable or x variable. Solid revolutions can only be done with variable whose axis is perpendicular to the radius. I understand the integral for solid, but not surface area. e.g. from the example above for [latex]y=x^2[/latex], we can use as radius [latex]x[/latex] or [latex] \sqrt{y} [/latex]. If we use [latex]x[/latex] as the radius, that's when it confuses me. I understand why you would use [latex] \sqrt{y} [/latex] as the radius. This is because for each values of y, there is a unique value of x. But when you use x, you're assuming the radius is no longer unique, and hence the same radius x can be used for a graph of [latex]y=x^{10}[/latex], for example in the bounds of (0,0) and (1,1) of both graphs. But clearly the shape of the two graphs are different, then how can we claim they have the same radius x, where x takes values in [0,1]? We can use x as the radius because surface area integral works for both variables as can be verified. If it's still difficult to understand the question, I can scan a page from my book. I understand that the formula works, but I can't get an intuitive grasp of why it works when you integrate with a variable whose axis is parallel to a cross-sectional radius of the surface. This implies that the radius is the same in the bounds [0,1] for two different graphs. This is because we assume x takes values uniformly in [0,1], right? As you can see, the x taken as the radius implies the two graphs have the same cross-sectional radius, e.g. each discrete rubber band has an average radius that eventually adds up to the same number for two different graphs: [latex] \int_0^{1} 2 \pi x \sqrt{1 +4x^2} \,dx [/latex] [latex] \int_0^{1} 2 \pi x \sqrt{1 +100x^{18}} \,dx [/latex] But the formula is correct!

[latex] \int 2 \pi x \sqrt{1+\big(\dfrac{dy}{dx}\big)^2} \,dx [/latex] My textbook says, for rotation about y-axis, the formula above is used.

I understand that you use values of y to find values of x when a solid is formed from revolution around the y axis. This is because for each values of y, the resulting values of x from the equation [latex] y=x^2 [/latex] is unique. In other words, the radius of a discrete cross section, perpendicular to the y-axis, for [latex] y=x^2 [/latex] is unique from the cross section radius found in revolution of [latex] y=x^{10} [/latex], for any boundaries, which I chose (0,0) and (1,1) for simplicity. However, for me this intuition is thrown out the window when it comes to surface areas. Finding the surface area, you need the circumference of an approximating section of a cone and the (slanted) height, which is given by the formula [latex] 2\pi xh [/latex](the integration formula is similar), where x is the radius and h is the height. This is what is referred to as a "band", like a number of discrete rubber bands wrapped around the revolved graph, and the approximation of surface area gets better as there are more and thinner rubber bands. So the problem, if you have followed through above, is that x, directly, is the radius instead of having to use y to find x. The formula for the integral of revolution of surface area assumes that the sum of radius of cross sections or rubber bands, perpendicular to y-axis, of [latex] y=x^2 [/latex] and [latex] y=x^{10} [/latex] are equal and no longer unique for two different graphs.

In revolution of solids: What variable you integrate with respect to, i.e. the independent variable, matters. But in revolution of surfaces, the variable you integrate with respect to doesn't matter. Why is that? E.g., you have a segment of a graph bounded by [latex] y=x^2 [/latex] and [latex] y=1[/latex]. You rotate this about the y axis. To find the area of the solid, you integrate: [latex] \int_0^{1} \pi y\,dy[/latex] Where you have to use the variable y because the radius is in terms of x, and at different values of y, the value of x stays near different values longer because the graph is curved. i.e., the values of x, the radius, would be different from the above for a graph bounded by [latex] y=x^{10} [/latex] and [latex] y=1[/latex]. And it is precisely for this reason, I believe, that we use the variable y to find x. But when it comes to revolution of surface areas: Example: You revolve the graph of [latex] y=x^2 [/latex] from (0,0) to (1,1) around the y-axis. In this case the radius is the value of x. But this is what I don't get. Shouldn't you use y to find x instead of being able to just use the value of x? Like the revolution of solids, the x value, or the radius, changes from [latex]y=0[/latex] to [latex]y=1[/latex], or the value of x stays near different values longer because the graph is curved. For reference the formula for the surface area of the graph above [latex] y=x^{2} [/latex] from (0,0) to (1,1) is: [latex] \int_0^{1} 2 \pi x \sqrt{1 +4x^2} \,dx[/latex], where x in circumference [latex]2\pi x[/latex] is the radius and the square root is derived from the distance formula for the (slant) length(height) of approximating "bands" around the revolved surface. Consider revolving [latex] y=x^{10}[/latex] around the y-axis and formulating the integral again, with limits of integration as (0,0) to (1,1), same as before. The radius in this case is also x integrated from from (0,0) to (1,1)! Likewise, the formula for [latex] y=x^{10} [/latex] from (0,0) to (1,1): [latex] \int_0^{1} 2 \pi x \sqrt{1 +100x^{18}} \,dx[/latex] So how can it be? The radius is the same for both graphs? So how can we say that the radius is x in both cases of surface of revolution? This would imply that the radius of an approximating circle takes a value of radius uniformly between the infinitesimal values between 0 and 1 when it does not, or that the radius from 0 to 1 on the two graphs have the same radius at different intervals of the graph. To clarify my problem, consider the two graphs again: Graph of [latex] y=x^2[/latex] Graph of [latex] y=x^{10}[/latex] If we rotate each around the y-axis. Then we fit rectangular parallelepipeds of variable lengths(length being parallel to the major axis) but same width and height, with major axis parallel to the x-axis, into the center of the two revolved graphs such that the ends of the rectangle touches the boundaries of the revolved surface area, then the sum of the lengths of the rectangular parallelepiped that were fitted would be different for the two graphs, wouldn't it? Let [latex]t_i[/latex] denote the length of the ith rectangle fit into [latex] y=x^2[/latex] and [latex]k_i[/latex] denote the length of the ith rectangle fit into [latex] y=x^{10}[/latex], then: [latex] \sum_{i=1}^{n}t_i \neq \sum_{i=1}^{n}k_i. [/latex] Then how can we say the radius of the two graphs for the integral for finding their surface areas are both x, integrated from (0,0) to (1,1), when the sum of the radius or diameter are different?

It was a homework question on Webassign. The book for the Webassign didn't cover boundaries of Chebyshev's bounds, but it still gave us a question on boundaries of Chebyshev's bounds and whether to include the end points. Yes, the number [0, 1] example cleared things up(it helped when I drew a number line and marked arbitrary values within [latex][ 0, \dfrac{1}{k^2} ][/latex] and then the remaining would always be greater than or equal to [latex]1 - \dfrac{1}{k^2} [/latex]. Thanks for the help.

The book is giving me a conflicting definition of Chebyshev's rule with what John has posted. This is derived from the probability form of Chebyshev's rule: [latex]P(|X-\mu| \geq k\sigma) \leq \frac{1}{k^2}[/latex] Is this converse true? [latex]P(|X-\mu| < k\sigma) < 1 - \frac{1}{k^2}[/latex] This is what I translate from my book(Peck Introduction to Statistics & Data Analysis 4th edition, page 201). You can find similar statements around the web, such as here: http://www.csus.edu/indiv/s/seria/LectureNotes/Chebyshev.htm I edited the percentage into probability: " Consider any number k, where k is greater than or equal to 1. Then the (probability) of observations that are within k standard deviations of the mean is at least: [latex](1-\Big(\dfrac{1}{k^2}\Big))[/latex] " Therefore, translating definition from the book into probability form: [latex]P(|X-\mu| \leq k\sigma) \geq 1 - \frac{1}{k^2}[/latex] Did I do something wrong in translating from the sentence to the formula, or is something wrong here? The two formulas are contradicting: [latex]P(|X-\mu| < k\sigma) < 1 - \frac{1}{k^2}[/latex] [latex]P(|X-\mu| \leq k\sigma) \geq 1 - \frac{1}{k^2}[/latex]

So you're asking what percentage of the entire data is located at an instantaneous point on the graph? Wouldn't that be 0.00000..., which can be approximated by 0? So even if the equation/inequality includes the boundaries, we don't include it anyway?

So: [math]100(1/k^2)>[/math] (percent amount of data outside k standard deviations from the mean) [math]100(1-\Big(\dfrac{1}{k^2}\Big))\leq[/math] (percent amount of data inside k standard deviations from the mean) The question asks for the boundaries of where at least 89% of the data are inside some k standard deviations from the mean. Since k=3.015, so at least 89% of data are inside 3.015 standard deviations from the mean. At least means the data have to be greater than or equal to the boundaries, but the answer doesn't include the boundaries. So I am still confused. I got everything correct on this homework except the bracket signs(I chose the include brackets, or the sharp brackets: "[x, y]").

The question: " Mean=38.68, Standard deviation=11.42 Inside what interval is it guaranteed that at least 89% of the concentration observations will lie? " I solved for k in Chebyshev's rule equation: [math]100(1-\Big(\dfrac{1}{k^2}\Big))=89[/math] I get k= 3.015... So at least 89% of all data are within 3.015... standard deviations from the mean. The bounds are 4.247 to 73.113, rounded to 3 decimal places. The answer bound is (4.247, 73.113) and not [4.247, 73.113]. Why does it not include the boundaries?

So I get that light is scattered off when the sun's rays shine at sharp angles toward the surface of the Earth, which results in more longer wavelengths reaching our eyes and thus making the sky look red or orange. But today there was a heavy, dark, rain cloud, but it wasn't raining where I stood and the cloud was few hundred miles away(when I checked a satellite map after I got home). In perspective, these clouds covered the eastern half of the sky I was seeing at around 4:00PM, and it looked almost like it was directly over of me. At the end of the eastern half of the sky, just above the horizon, are bands of reddish lights between the heavy, gray clouds covering the eastern horizon. Meanwhile the sky above me, behind the clouds is still bright blue daylight. So the sun hasn't set, and even if it did set, it would be reddish where the sun set and not the eastern end. Anyone know what this phenomenon is and what is happening? Is the light bouncing off the rain clouds, is that what's causing this? It looks more like the light is coming from behind the clouds, but that might be a mislead observation. X is approximately where I stood and the cloud is circled.

In solving an integral of sqrt(9-x2), we substitute x=3sin(t) to get an integral of 3cos(t). Why does this work? In the original function, x can take on any value(if you include imaginary numbers), but the new function is equivalent to the original one even though sin(t) takes values only between its amplitudes. P.S. I remember a guide to writing math in this forum. Anyone have the link?

Is the summation below true? [latex] \sum_{a}a * Pr[R=a] \leq \sum_{a\leq b}b * Pr[R=a]. [/latex] Where R is a random variable and Pr[R=a] means that the probability that the random variable is equal to some number 'a'. You can ignore that part and replace Pr[R=a] with x: [latex] \sum_{a}a * x \leq \sum_{a\leq b}b * x. [/latex] The first summation provides all 'a' values, so the summation is over larger amount of terms. The right hand side sums only those a<=b, so the right hand sums over less amount of terms. But at the same time the right hand is multiplied by b>=a. So it is not clear which one is bigger. I am not extremely familiar with Riemann sum, so hopefully someone here knows what I am talking about. My first intuition is that the above is only true if the upper limit is finite, or that the above is true regardless since you can always have infinite plus 1, i.e. you can always have a number greater than the upper limit for [latex] \sum_{a}a * x [/latex] even if the upper limit for [latex] \sum_{a}a * x [/latex] is infinite. This is the gist of my question. I am trying to prove Markov's Theorem for when Pr[R<=x].

Oh yea, I forgot to mention something that may be important. In [latex] \frac{d}{dx} \frac{dy}{dx} = \frac{\frac{d}{dt} \frac{dy}{dx}}{\frac{dx}{dt}} [/latex] The variable t is a parameter for both y and x. So x and y can are functions of t. Meanwhile, y is a function of x. I forgot to mention this as I thought it would not be important. My main trouble is with understanding how to "operate" in Leibniz notation since I have never learned any rules to manipulate the Leibniz notation. E.g., you learn properties of commutation and distribution and etc, which you can apply to arithmetic, but I have no idea how to start operating in Leibniz notation. For example, how to even begin expanding: [latex] \frac{d}{dx} \frac{dy}{dx}[/latex] Given that x and y can are functions of t, and y is a function of x. I would have never known the above can be expanded to the second derivative: [latex]\frac{\frac{d}{dt} \frac{dy}{dx}}{\frac{dx}{dt}} [/latex] If no one told me so. I understand how to get the first derivative, you have by the chain rule of differentiation: [latex] \frac{dy}{dt}=\frac{dy}{dx}*\frac{dx}{dt}[/latex] The book divides both sides of the equation by [latex]\frac{dx}{dt}[/latex] to get: [latex] \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} [/latex] But how would you start calculating the second derivative: [latex]\frac{d}{dx}\bigg(\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\bigg) [/latex] I have tried applying the quotient rule of differentiation, which I assume is the wrong approach since I keep getting 0 for the numerator, what do you do to the above to get: [latex]\frac{\frac{d}{dt} \frac{dy}{dx}}{\frac{dx}{dt}} [/latex]

Ah, so the author of my book multiplied the top and bottom by (1/dt)/(1/dt). The book didn't say you can "multiply" and the book just did this, which is really uninformative: [math]\frac{d}{dx} \frac{dy}{dx} = \frac{\frac{d}{dt} \frac{dy}{dx}}{\frac{dx}{dt}}[/math] Which skipped a lot of steps that would have helped me understand the notation better. Thanks for the help. Yea, I can see why. I already dislike this notation.

For the second derivative, why is (d/dx)(dy/dx)= [(d/dt)(dy/dx)]/(dx/dt)? I don't see the logic. Am I supposed to apply the quotient rule of differentiation on dy/dx? If I do that I get the second derivative is 0 because dy-dy=0. Am I allowed to use algebra on dy and dx?

Edit, first line should read: I am confused as to why merge sort for a list of size 2^k=n takes n steps.

I am confused as to why merge sort for a list of size 2^n takes n steps. If you have n=4 lists 3,9,8,10 and you sort from least to greatest: 3 9 8 10 (3,9) (8,10) 2 work to compare 3 with 9 and 8 with 10. (3,8,9,10) 3 work to compare 3 with 8, 8 with 9, and 9 with 10. That's a total of 5 work, not 4. Every definition out there says it takes 4 steps to merge when it obviously takes 5 steps in the example given. Why is that?