Happy holidays, everyone! Hope you're having a great start to break! Today, I am going to be elaborating more on another important concept in calculus known as the disk method.
**NOTE: the shape extends from 0 to 1 (along x-axis) so the shape ends at 1** In order to find the volume of the shaded region, we have to subtract the volume of the cone shape from the volume of the dome shape. As per the solid of revolution formula, we know that V = π∫(b to a) [R(x)]^2 dx.
π∫(1 to 0)(x^1/2)^2 dx - π∫(1 to 0)(x)^2 dx π∫(1 to 0)x dx - π∫(1 to 0) x^2 dx = π[1/2 x^2] - π[1/3 x^3] = (substitue 1 and 0) 1/2 π - 1/3 π (3-2 π)/6 = 1/6π So the volume of the purple area is 1/6 π!! With this method, you can practically find the volume of any distinct 3 dimensional shape! Very interesting, right? Good luck and happy holidays,
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source: http://indusladies.com/community/threads/dot-kolams.77039/page-2 Can we use integrals to figure out the area of the insides of such complicated designs and shapes? Well, to introduce you to these designs, called 'kolams', a typical design drawn during the season of lights and happiness, I have posted a picture above for you to get the idea. So, lets say we had a simple kolam to deal with at first, such as this one: source: http://www.a1tamilnadu.com/singlePost.asp?cid=1199#.Wgj22bA-dPM Lets say the problem was to find the area inside of this complex shape. I want to try to use integrals to find the area of this shape. So, therefore, lets use a small, tiny arbitrary curve from this whole design. In other words, take a magnifying glass and find a curvature in which we can find a small approximation of the area inside that small curve. So, lets pick a small curve from this kolam and draw a x-y axis behind it to integrate it and approximate the area. This curve looks a lot like a natural log curve. We can use the left rectangular approximation method to find the approximate area. if we integrate ln(x) from x = 1 to x = 4, we get 2.5451.... ∫ [at interval 1,4] ln(x) d(x) = 2.5451.. However, lets try using rectangles to approximate the areas. So, for the first rectangle (from x=1 to x=2), we can solve the area, A, by getting the y value at x = 1 and multiplying it by one to get the area of that particular rectangle. Therefore, the area of the first rectangle would be ln(1) which is the y value, times 1, which yields 0. The next rectangle (from x=2 to x=3) is ln(2) times again 1 (which represents the width) which yields 0.6. The next rectangle (from x=3 to 4) is ln(3) times 1 = 1.09. Therefore, the total area which encompasses the area under the curve is: 1.09 + 0.6 + 0 = 1.7917, much lower than the actual area calculated above (2.5451). So, how do we solve for the area inside of these designs? It may seem a tedious task, but to be exact and accurate we must calculate the areas underneath the smaller curves of the design to eventually find the total area the design covers. In order to do this, we will have to put the design on to a grid and label the coordinates.
Good luck!!
I probably needed to repeat that. I know, though it sounds ridiculous, the sum of all natural numbers leading up to, I suppose infinity is proven to be -1/12 by many of the greatest mathematicians of all time. How about we become a mathematician and prove it once again? Instead of using complicated formulas and other ways of solving this puzzle, we can use a simpler way to prove it. I also wanted to talk about many other ways of solving this and mathematician's efforts to proving this as well. (especially from my favorite and highly regarded mathematician, Srinivasan Ramanujan). I saw this phenomenal fact in a youtube video on proving this (https://www.youtube.com/watch?v=w-I6XTVZXww), and wanted to share it with you all. Let's prove this infinite series in a simple way: Step 1. Write the first sum.First, we are going to write three different sums, labelled as S1, S2, and S3. The last sum, S3 should be the infinite series we are trying to prove (aka: 1 + 2 + 3 + 4 + 5......) The first sum is going to be: S1: 1-1+1-1+1-1+1-1+1....There are two possible answers for this infinite sum: If we end after a subtraction, the answer will be 0. If we end after an addition, the answer if 1. So, to compute the sum of the series, we would take the average of the two values ~ (1+0)/2 = 1/2. Step 2. Write the second sum twice.Write the next sum, which is slightly different, involving both addition and subtraction. We are going to duplicate this sum and add it to each other (2S2) to make the work easier. Also, shift the second S2 to one digit right so that the first number of the second sum is below the second number of the first sum. S2: 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8.....+ S2: 1 - 2 + 3 - 4 + 5 - 6 + 7 - 8.....2S2= 1 - 1 + 1 - 1 + 1 - 1 + 1 - 1 + 1...When you add both the sums digit by digit, you get the original sum we started with, which is S1. Therefore, we can conclude that 2S2 = S1 = 1/2., which means that S2= 1/4. Step 3. Write the original sum we are solving for and subtract S2 from it.S3: 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8.....- [1 - 2 + 3 - 4 + 5 - 6 + 7 - 8.....]0 + 4 + 0 + 8 + 12 + 0 + 16 + 0...Now, we are going to factor this new sum by 4, therefore, we obtain...you got it...the original sum times 4. 4 ( 1 + 2 + 3 + 4 + 5.....) = 4S3S3 - S2 = 4S3 S3 - 1/4 = 4S3 -1/4 = 3S3 -1/12 = S3 And...there is your proof that the sum of the infinite series is -1/12. It is absolutely surprising, but actually proven to be true. I know..weird. But I like it. Good luck,
Remember these tips for SAT with calculator! These are important:
Since m is the product of all integers: m = 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 10 = 1 x 2 x 3 x (2 x 2) x 5 x (3 x 2) x 7 x (2 x 2 x 2) x (3 x 3) x (5 x 2) <-- there are 8 2's Therefore, the answer to this question is 8 = n. 4. If you are more confident with a sample, there must be a wider range of the numbers, which means a greater margin of error. 5. (IMPORTANT TIP FOR Exponential Growth questions): If they give you an exponential equation for a particular scenario and ask you to find out when it is doubled, do this: For ex. 18 x 16^0.0125t <-- what is the value of t when the equation is doubled? 18 x 2^4 x 0.0125t <- change the 16 to 2^4 18 x 2^0.05t <- simplified <- to find the value of t, simply make 0.05t equal to one so that the equation can double 0.05t = 1 t = 20 Good luck,
The Law of Sines and Cosines are fundamental to determine and derive the unknown variables of a triangle (which is not right angled) such as its angles, and sides with some information already given. We are going to prove this theorem right: The Law of SinesThe Law of Sines says that to find the value of an angle or side of a triangle (with either one given), we can calculate the other angle/side by using this formula: sin ß / B = sin α / A
The Law of CosinesThe Law of Cosines says that to find the value of any side with a given angle and other two sides, we use this formula: a^2 = b^2 + c^2 -2bc cos θ. Good luck!!
Lets breeze through some SAT Math tips which really helped me score in the high band for the SAT Math Section. I might be doing a series on this. Let's start with Math without Calculator: MATH WITHOUT CALCULATOR 1. Right Angle Trigonometry Problems ~ a lot of the SAT Math problems!!
2. Data Analysis (Standard Deviation)
EXTRA TIPS FOR WITHOUT CALC SECTION
Good luck,
Hello, Everyone!! Its been a while, but I have decided to start a series on Calculus topics, especially since I will be taking Cal BC this coming school year, I thought it would be cool to refresh these topics in my mind and give you some info on how Calculus can help you out. Really. I may not be in Calculus already, but I have been exposed to many calculus related problems during my ninth and tenth school years in Canada. Calculus is so amazing, and broad but at the same time, can be used to find volumes of solids of various different kinds. So, don't worry if you don't know the formula for finding the volume of a sphere during a test or quiz. Calculus has got you.
Lets do an example together. Let's find the volume of a sphere (the midpoint being 0) from the values of a and -a. lets first find the graphic formula for the area of the cross-sectional surface: x^2 + r^2 = a^2 <--- finding area through the circle formula - step 1 r = √(a^2 - x^2) <--- solving for r - step 2 S(x) = π(a^2 - x^2), -a ≤ x ≤ a <--- finding area of the circle, domain - step 3 V = ∫(from a to -a) [π(a^2 - x^2) dx] <--- set volume to definite integral - step 4 = π[ a^2x - 1/3x^3](from a to -a) <--- solve - step 5 = π[4/3a^3] = 4/3a^3π So, we finally derived the volume formula of a sphere, and since it is a generalized formula, a can be any value (radius). Good luck,
1. If the area of a sector of a circe is = 18(pi) cm sqaured and the radius is 6 cm, what is the measure of the central angle in degrees? FIrst off, lets look at what we have. We have:
Next, we are going to see what the area of a sector is equal to: Asector = (degrees)/360 * area of circle Now, substitute: 18(pi) = x/360 * 36(pi) 1/2 = x/360 x = 180 Good luck!!
Interestingly, I proved the value of pi today! My sister made apple pie (she is, no doubt, an extraordinary baker) today, and I decided, why not incorporate some math to it? So I decided to measure the circumference of the pie tray and then the diameter, to calculate the value of pi. So, I used my earphones as a referent. The total length of the circumference of the pie was equal to about 66 cm. I then measured the diameter of the pie, which turned out to be about 21 cm. When you divide the circumference of any circle by the diameter of the same circle, you get the value of pi! Obviously, due to human error, you can't receive the exact value but I got pretty close: 66cm (circumference measurement) / 21 cm (diameter measurement) = 3.142857... Ta da! So, I basically proved the value of pi which a delicious apple pie as a circle! I know we all know about this, but why can't we try crazy stuff like this! This is super cool! I am a total nerd, I know. Good luck,
Extraneous Solution? Yup, throughout high school math, you will be hearing this term for a loooong time. A super long time. When you are dealing with equations with square roots, you have to always check whether your answer is correct by substituting with the variable. If the solution does not support the equation it is labelled as an extraneous solution. But in this blog post, I will be explaining to you why this happens. Why do these solutions happen to exist? Real Mathematical Reasoning Extraneous Solutions occur because squaring both sides of a square root equation results in 2 solutions (the positive and negative number). Therefore, one of those numbers will be an extraneous solution, or an extra solution which does not fulfill the original equation. For example, x = 4 (original equation) <squaring both sides of the equation> x^2 = 16 x = -4, 4 <substituting x = -4 into the original equation> -4 is not equal to 4 Good luck,
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