Instructor Hentzel Office Phone: 515-294-8141 E-mail: hentzel@iastate.edu Math Department Fax: 515-294-5454 http://www.math.iastate.edu/hentzel/class.166.ICN Textbook: Calculus Early Vectors Preliminary Edition By Stewart. Wed, Jan 15 6.1 Summation p368: 2,5,10,12,15,18,21,25,(28),(30),38, 41,43,44,(48) Main Idea: We want to add up the area of little rectangular slices. Key Words: Summation Notation, Sum i, Sum i^2, Sum i^3, Sum i^4. Induction Goal: We want to simplify and evaluate the sums. Actually, the sums will be areas of little rectangles. ------------------------------------------------------------------ Previous Assignment Page 353 Problem 28 f'(x) = 3x^(-2) Initial conditions: f(1) = f(-1) = 0 x^(-1) f(x) = 3 ------- + C -1 f(x) = -3/x + C There are two pieces to the curve. One had domain (-Infinity,0), the other has domain (0,Infinity). The function is not defined at x=0. On (-Infinity,0) On (0,Infinity) f(x) = -3/x + C f(x) = -3/x + C f(-1) = 3+C = 0 so C = -3. f(1) = -3/1 + C = 0 so C = 3. f(x) = -3/x -3. f(x) = -3/x + 3. Check f(-1) = 0 f(1) = 0 f'(x) = 3x^(-2) f'(x) = 3x^(-2) ------------------------------------------------------------------- Page 353 Problem 66 Two balls are thrown upward from the edge of the cliff in Example 8. The first is thrown with a speed of 48 ft/sec and the second is thrown one second later with a speed of 24 ft/sec. Do the balls ever pass each other. first case second case a(t) = -32 a(t) = -32 v(t) = -32 t + 48 v(t) = -32t + C s(t) = -16 t^2 + 48 t + 432 v(1) = -32 + C = 24 C = 56 v(t) = -32 t + 56 s(t) = -16 t^2 + 56 t + C s(1) = -16 + 56 + C = 432 C = 392. s(t) = -16 t^2 + 56 t + 392 check check s(0) = 432 s(1) = -16 + 56 + 392 = 432 v(t) = -32 t + 48 v(t) = -32 t + 56 v(0) = 48 v(1) = 24 a(t) = -32 a(t) = -32 Solving: -16 t^2 + 48 t + 432 = -16 t^2 + 56 t + 392 48 t = 56 t -40 0 = 8t - 40 t = 5. s(5) = 272 s(t) = 272 v(t) = -32 t + 48 v(t) = -32 t + 56 v(5) = -112 v(5) = -104 ----------------------------------------------------------------------- The two projectiles pass when both are going down. The second projectile is going slower. Page 353 Problem 74 a(t) = (0,2) v(0) = (1,-1) r(0) = (0,0). a(t) = (0,2) v(t) = (0,2t) + C v(0) = (0,0) + C = (1,-1) v(t) = (1,2t-1) s(t) = (t,t^2-t) + C s(t) = (0,0)+C = (0,0) s(t) = (t,t^2-t). <========= answer. check. s(t) = (0,0) v(t) = (1,2t-1) v(0) = (1,-1) a(t) = (0,2). ======================================================================= New Material: Theory of integration: Find the area under y = f[x] from x=a to x=b. | . . | . ' ' | . ' | '. | . ' |Find this area |. | | | | | | -------------------+--------------------------------- | a b | The idea is to divide the area into small rectangles. | | _ . .__ | . _'_| | | ' | . ' | | | |___'. | . ' | | | | |. | | | | | | | | | | | | -------------------+--------------------------------- | a b | If we keep the tops of the rectangles under the curve, then the area of the rectangles is less than the area under the curve. No matter how small we make the width of the rectangles, then the area of the rectangles will be less than the area under the curve. | | _______ | ____| .|. |____ | . | ' | | | ' | | . ' | | | | '. | . ' | | | | |. | | | | | | | | | | | | -------------------+--------------------------------- | a b | We can do the same process putting the tops of the rectangles above the curve. Then the area of the rectangles will be more than the area under the curve. So the main idea, which should be obvious is that Any Lower Sum <= the area under the curve <= Any Upper Sum. Now we can approximate the area of elementary curves. Find the area under y = x^2 from x=0 to x = 1. | | hn | . | . | | . | | h3. | | | | | .'| | | h2 .' | | | h1 . | | | ----------------------+-----'---|----|------------------- | 1/n 2/n 3/n ... Adding the areas which are below the curve we have ho/n + h1/n + h2/n + ... + hn-1/n Adding the areas which are above the curve we have h1/n + h2/n + h3/n + ... + hn/n Since we are working with the function y=x^2, the heights hi are given by hi = (i/n)^2. Lower Sum = [(0/n)^2 +(1/n)^2 +(2/n)^2 +(3/n)^2 + ... +((n-1)/n )^2]/n Upper Sum = [(1/n)^2 +(2/n)^2 +(3/n)^2 +(4/n)^2 + ... +(n/n)^2 ]/n Now we do some algebraic manipulation to get Lower Sum = (1/n)^3 ( 0^2 + 1^2 + 2^2 + 3^2 + ... + (n-1)^2 ) Upper Sum = (1/n)^3 ( 1^2 + 2^2 + 3^2 + 4^2 + ... + n^2). Fortunately, there is a formula for the sum of the squares of the integers. n(n+1)(2n+1) 1^2 + 2^2 + 3^2 + ... + n^2 = --------------- 6 (n-1)(n)(2n-1) (n-1)(2n-1) 2n^2 -3n +1 3n-1 Lower Sum = ------------ = ----------- = ------------- = 1/3 - ----- 6 n^3 6n^2 6n^2 6n^2 n(n+1)(2n+1) (n+1)(2n+1) 2n^2 +3n+1 3n+1 Upper Sum = ----------- = ----------- = -------------- = 1/3 + ---- 6 n^3 6n^2 6n^2 6n^2 The Lower Sum is a trifle less than 1/3. The upper sum is a trifle more than 1/3. Lower Sum < 1/3 < Upper Sum Since the Lower Sums approach 1/3 from below, the area has to be at least 1/3. Since the Upper Sums approach 1/3 from above, the area has to be less than 1/3. Therefore, the area can only be 1/3. If we wish to nail down our assumptions: If there is such a thing as "area" which behaves as area should. That is: A larger region cannot be inside a smaller region. Then the area under y=x^2 from x=0 to x=1 can only be 1/3. ------------------------------------------------------------------ But maybe, just maybe, for the sake or argument, the concept of area does not really exist. Mathematicians have concerned themselves with the question and have come up with the following observations. If one has that the lower and upper sums converge from above and below to the same value, Then that value works as the area. It is a reasonable value to assign to the area. Finally, for a continuous function, the upper and lower sums will always converge to a common value, so the area under continuous functions exists. Useful formulas. n SUM 1 = n i=1 n n(n+1) SUM i = ------- i=1 2 n n(n+1)(2n+1) SUM i^2 = ------------ i=1 6 2 n |n(n+1)| SUM i^3 = |------| i=1 | 2 | ------------------------------------------------------------------- Example 6 Page 368. n Evaluate SUM i(4i^2-3) i=1 n n n n SUM i(4i^2 -3) = SUM (4i^3-3i) = 4SUM i^3 -3 SUM i i=1 i=1 i=1 i=1 n^2 (n+1)^2 n(n+1) = 4 ------------- - 3----------- 4 2 2 n (1 + n) (-3 + 2 n + 2 n ) = --------------------------- 2 ----------------------------------------------------------------- n Page 369 Problem 32 Evaluate SUM (3+2i)^2 i=1 2 n (47 + 24 n + 4 n ) ans = -------------------- 3 ------------------------------------------------------------------ n n(n+1)(2n+1) Prove that SUM i^2 = -------------- i=1 6