| Oakton Community College |
| MAT 251 Calculus II |
| Section 001: TR 8:30 - 10:20 Room 3609 DP Campus |
| Paul Boisvert Professor of Mathematics |
| Spring Semester 2012 |
| Return to Homepage |
| Date | Page. | Problems | Read Section: |
| Tue 1/17 |
No Problems for HW BUT: REVIEW all old Precalculus Material, and Calculus Material: ESPECIALLY Precalculus Properties of Trig Functions, Inverse Trig fcns, e^x , ln(x), AND: All Derivative Rules from Calculus I (Power Rule, e^x, ln x, trig, Quotient Rule, Product Rule, Chain Rule, Inverse Trig fcn.s) |
4.7 | |
| Thu 1/19 |
p. 290 | 13-18 ALL ,
19-25 odd, 26 27-30 ALL (You may use Ratio of the Highest Powers instead of L'Hopital when appropriate) 31-33 ALL , 35 (Hint, LCD) Answers (if you don't have these answers in the back of your book): 13) -1 14) 1 15) 12/5 16) 6pi 17) 4 18) 4/7 19) 9/16 21) 4 23) -1/2 25) cos x Note, this is a trick question in two ways: first, the "variable" in the limit is h, not x, so you need to take d/dh derivatives, treating x as a constant. So, in LH, d/dh [sin x] = 0, since if h is the variable, x and sin x are constants. Secondly, the given problem is the DEFINITION of the derivative of sin(x), so naturally the answer has to be the actual derivative of sin x, which is indeed cos x. But it is strange to use the shortcut rule derivative of sin x to calculate a limit that is the definition of the derivative of sin x in the first place. 26) 1/4 27) 1/2 28) 4 / pi Note, you would normally be likely (and allowed) to use Ratio of Highest Powers on this one, not LH, since LH takes 3 repetitions to work. 29) Answer of 0 in Book is INcorrect--function is negative in the numerator's square root as x gets large, so it is undefined, and therefore cannot have any limit at all. Problem probably has a typo, and maybe was intended to have numerator of sqrt(8 + 4x^2), but still would not be a good problem to use L'Hopital on, as the first derivative step doesn't give an answer, and then the further derivatives get very ugly... But if that was the typo, it would indeed then have answer 0, though that would not be clear from L'Hopital directly. The correct method on the corrected problem would be to factor out 4x^2 from inside the root, then apply L'H to the part of the problem without the remaining root, giving 0. And the remaining root also obviously goes directly to 0, so 0(0) = 0 for the final limit. Obviously, such a problem will NOT be on the test... :) 30) Answer: 0 Note: at first, numerator tan(pi/2) has NO LIMIT, since it goes to +oo on the left of pi/2, but to -oo on the right of pi/2. However, because denominator is ALSO infinite, we CANNOT say right away that the whole fraction has no limit--instead, we are uncertain, and should use LH. What we are uncertain about is whether the bottom infinity may not be "larger" than the top--if it is, it will dominate BOTH the +oo AND the -oo in the top, and send the overall fraction to +0 and -0, but that would mean there IS a limit of 0 (since +0 and -0 are the SAME thing, unlike +00 and -oo.) And, in fact, this is precisely what happens! LH will show this by giving the answer of 0. 31) 1 32) Answer: 2 / pi NOTE: If you move (1-x) to the bottom, making it 1/(1-x) , and leave tan( ) in the top, you have 00 / 00 , and LH leads to another oo/oo, with the Deriv.'s getting worse. If you do it a second time, you still get oo / oo, with even worse deriv.'s. So STOP, try the other way! So correct way is move tan( ) to the bottom, making it 1/ tan( ) , or cot ( ). Leave 1-x in the top. LH now works out fine, Answer 2 / pi 33) 1 35) Lcd = x sin(x). Fraction becomes [x cos x - sin x ] / x sin x . Use LH once (note Product rule needed on x cox(x) ) , get [-x sin x ] / [x cos x + sin x] , but that is still 0/0 . Use LH second time, (again, with Product rule), this time get 0/2 = Answer = 0 |
4.7 |
| Tue 1/24 |
p. 290 | 39-43
ALL, 45, 47, 72, 73 Exponential LH problems: Follow these
steps: Name limit L, take ln of both sides, should
get 0(oo), flip one function to denom (not ln, if possible), proceed...
At the end of the process you get ln(L) = Number, then
exponentiate both sides to get L = e^(Number) Answers: ODD: 39) 1 41) 1 43) e 45) 1 47) e 73) RECALL: Derivative of Log base-B of x is 1 / [x ln(B) ] You should memorize this deriv. Note the ln(B) is in the denom. with the x. So, Limit is oo / oo, so use LH, get 1 / [x ln(2) ] / ( 1 / [x ln(3)] ) , or, inverting, and cancelling the x's, Answer is ln(3) / ln(2) EVEN: 40) e^12. Hint: when you take ln of both sides and simplify, r.h.s. is (3/x) ln(1 + 4x), which is best thought of as 3ln(1 + 4x) / x . Use L'H on 3ln(1 + 4x) in the top, and "x" in the bottom, and it works right out. Exponentiate answer, get L = e^12. 42) 1 Hint: after taking ln and simplifying, for tan(theta) ln ( sin[theta] ), leave ln ( sin[ ] ) in the top, and move tan to bottom as 1/tan = ? Then L'H works straightforwardly, gives you "0", exponentiate to get L = e^0 = 1. Note: As you see in #40 and 42, when, in a 0 times oo situation, you have ln( anything) as one of the two functions, it is virtually certain that that is the one you should leave in the top, and move the other function to the denom. (and flip it). There are very few exceptions, and NONE that would show up on a test! The other function would have to be very hard and weird to make it worthwhile flipping the ln ( ) into the bottom, since deriv. of 1 / ln ( ) is already a complicated chain rule deriv., and whatever is inside the ( ) would require yet another chain rule step, making it even worse! 72) Trick question, 0^(oo) is NOT uncertain. LH not needed. Small numbers near 0 (i.e., between 0 and 1) get even smaller as you raise them to higher and higher powers, so 0^(oo) is definitely going towards a Limit of 0. BUT, of course, in this problem it is really 0^(-oo), which is 1/ [ 0^oo ], which is 1 / 0 = oo. So answer is L = oo. 44) [NOT assigned, optional, only do it if you want the challenge!] Tricky...couple hints...good luck! In first (simplified) step, leave ln(1 + 1/x) in top, move ln (x) to bottom as 1 / ln(x). Notice that the above Note: does not help, since BOTH functions are ln's! So we move the easier one down... Apply L'H, then simplify by cancelling x's and distributing where possible. You should end up with [ ln(x)]^2 / (x + 1) . Now use L'H TWICE more on that (remember the chain rule!), and you'll get... 0. Exponentiate, get e^0 = 1. |
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| p. 364 | 17-24 ALL, 36-42 ALL, 53-58 ALL, 26, 27, 44, 60, 65 45-50 All--Sorry, SKIP these for now, we did NOT yet cover integral of sin^2 and cos^2 in class. Will reassign next time. (Formulas for the two integrals and answers for the problems are already posted in the HW section below for next time, for those who wish to do them now. But most of you should wait until I cover the issue in class next time. I've included odd answers here, listed first, as well as even, listed later. The odd ones are also on the answer sheet, of course. Answers: ODD: 17) (1/100) (x^2 - 1)^100 + C 19) (-1/3) sqrt[ 1 - 4x^3 ] + C 21) (1/11) (x^2 + x)^11 + C 23) (1/28) (x^4 + 16)^7 + C 27) (1/30) (x^6 - 3x^2 )^5 + C 37) 1/3 39) (e^9 - 1)/3 41) sqrt(2) - 1 53) (1/10) tan (10x) + C 55) (1/2) tan^2 (x) + C 57) (1/7) sec^7 (x) + C 65) Integral (after u-subst.) = (-1/2) cos [ x^2] evaluated from 0 to sqrt(pi), = +1/2 - -1/2 = 1. EVEN: 18) (1/2) e^[x^2] + C 20) (1/5) [sqrt(x) + 1] + C 22) (1/10) ln |10x - 3| + C 24) (1/11) sin^11(theta) + C 26) (-1/10) cos (x^10) + C 36) 4/5 38) sqrt(2) - 1 40) 2 42) 1/2 54) (1/6) sin^6 (x) + (3/4) sin^4 (x) - (1/2) sin^2 (x) + C 56) (1/9) [ x^(3/2) + 8 ]^6 + C 58) Let u = e^(2x) + 1, du/2 = e^(2x) dx, integrate (1/2) 1/u du, get ... = (1/2) ln ( e^[2x] + 1) + C NOTE: here, original integral requires absolute value signs in ln|u|, but since e^[2x] + 1 is always positive, you can drop the absolute value signs in the final answer. 60) u = ln x , du = (1/x) dx, etc... Final Answer: 2 ************************************************************************************ EXTRA OPTIONAL Practice Problems, if you want them: (NOT assigned, do only if you want more practice.) 59) 1/3 61) (3/4) (4 - 3^[2/3] ) 63) 32/3 52) (1/4) sec(4w) + C |
5.5 | |
| Thu 1/26 |
p. 364 | 45-50
ALL FACT: Here are the two integrals needed, which depend on trig identities (u-substitution will NOT work): Integral of cos^2 (u) du = ... = u/2 + sin(2u) /4 + C AND Integral of sin^2 (u) du = ... = u/2 - sin(2u) /4 + C Answers: 45) pi 47) using "t" for theta: t/2 - (1/4) sin ( [6t + pi]/3 ) + C 49) pi/4 46) x/2 - (1/4)sin(2x) + C 48) I'll use "t" for theta. Let u = 8t, du/8 = dt, integrate (1/8) cos^2 (u) du , and get ... (1/8) [ u/2 + sin (2u)/4 ] , then replace u with 8t and simplify, then substitute in t = pi/4 , and t = 0 , to evaluate, and get ... Final Answer: pi/8. 50) Let u = x^2, du = 2xdx, integrate (1/2) cos^2 (u) du , proceed similarly to #48, replace u with x^2, and get ... Final Answer: (1/4)x^2 + (1/8) sin(2x^2) + C |
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| p. 388 | 9-23 odd, all dx 24 -- dy , 26 -- dy, two integrals , 30 -- dy and dx , 31 -- dy only 33 -- dy and dx , 37 -- dx, 38 -- dy Answers: 24) Integrate e^y dy from y = 0 to y = 2, get... e^2 - 1 26) First, y = ln x ^ 2 means ln( x^2) means simplify bringing out the 2nd power to y = 2 ln(x). Solve for x, get x = e^(y/2) Also, y = ln x becomes, when solved for x, x = e^y For horizontal dA, the upper part of the Area goes from vertical line x = e^2 on Right to the Curve x = e^(y/2) on the Left. For the lower part of the Area, it goes from the Curve x = e^y on the Right to the Curve x = e^(y/2) on the Left. These Areas switch at the intersection point of the line and the lower curve, i.e. where x = e^2 is plugged into y = ln(x) = ln(e^2) = 2 Lower limit of lower Area is obviously y = 0, upper limit is where x = e^2 is plugged into y = 2 ln(x) = 2 ln(e^2) = 4 So: Area = (Upper part) Integral of [ e^2 - e^ (y/2) ] dy from 2 to 4 PLUS (Lower part) Integral of [ e^y - e^ (y/2) ] dy from 0 to 2. = etc... Note, integral of e^2 dy = (e^2)(y), with y not in exponent ... . = ... = evaluating: 4e^2 - 2e^2 - [2e^2 - 2e ] PLUS e^2 - 2e - [1 - 2] = simplifying = e^2 + 1 30) Integral of [x^1/2 - x^3] dx from x = 0 to x = 1 OR Integral of [ y^1/3 - y^2 ] dy from y = 0 to y = 1 BOTH = 5/12 38) Solve y = x/3 for x = 3y, integrate horizontal dA as Right x minus Left x. Intersection where y^2 - 4 = 3y, set = 0, factor, get y = -1 or 4. So, Integral of [ 3y - (y^2 - 4) ] dy from -1 to 4 = ... evaluating = 24 - 64/3 + 16 - [ 3/2 + 1/3 - 4 ] = 125/6 |
6.2 | |
| Tue 1/31 |
p. 400 | 5-21 odd 18, 22 Disks, around x-axis V = Integral of pi [ f(x) ]^2 dx 23-29 odd, 24 Washers, around x-axis V = Integral of pi [ (f[x])^2 - (g[x])^2 ] dx with f(x) = Outer function and g(x) = Innter function 31-35 odd, 34,36, disks and/or washers, you decide, around y-axis Same as above with functions of y, and Integral w.r.t. dy 42,44, 47 Various, you decide Answers: 18) (1/4) (pi)^2 22) pi 24) 16384 pi / 3 Note: to find intersection here, use calculator, OR: set functions =, then set equation = 0, get 16x^(1/4) - 2x = 0 . Now factor out 2x, and notice that to factor out x = x^1 from x^(1/4) you must divide x^(1/4) by x^1, which means subtract exponents and get x^(1/4 - 1) = x^(-3/4). So factorization is 2x( 8x^[-3/4] - 1 ) = 0. Set each factor = 0, and you get x = 0 for the first easy part, and then: For the other solution you must solve 8x^[-3/4] - 1 = 0. Separate, divide by 8, get x^[-3/4] = 1/8, then raise each side to the reciprocal power of -4/3. So x = (1/8)^(-4/3) = (8/1)^(+4/3) = (cube root of 8)^4 = 2^4 = 16. So intersection points are at x = 0 and x = 16 for your volume. 34) Washers Integrate: pi { 4^2 - (y^2)^2 } dy from y = 0 to 2 . get... 128pi/5 36) Washers: Integrate: pi { [ (1/2)sqrt(2)] ^2 - (sin y)^2 } dy from y = 0 to pi/4. Note, after integrating, simplify answer before evaluating!! Final answer: pi/4 42) Disks: After simplifying inside the integral before integrating, you have: Integrate: pi { (ln x)^2 / x } dx from 1 to 2. Use u = ln x, du = dx/x, integrate u^2 du, and you get, eventually: Evaluate (pi /3) [ (ln x)^3 ] from 1 to 2, ... Final Answer: (pi/3) [ (ln 2)^3 ] . 44) Disks: After simplifying inside the integral before integrating, you have: Integrate: pi { tan x } dx from 0 to pi/4 . Write tan x = sin x / cos x, Use an obvious u du substitution, and you get, eventually: Evaluate - pi [ln|cos x| ] from 0 to pi/4, which gives you First Final Answer: - pi [ln { (1/2)sqrt(2) } ] . But the ln simplifies by the laws of logarithms to - (1/2) ln(2). So Second Final Answer: +(pi/2) ln (2). First answer would be ok on test, but everyone in Calc. 2 should be able to simplify to the second answer if necessary. |
6.3 |
| Thu 2/2 |
p. 411 | 5-15 odd, 8, 10, 16 All by Shells (only) 27, 29, 31, 32--For these, set it up first by shells, and then by a second method (either disks or washers, whichever is appropriate.) Then, integrate and find the Volume by the EASIEST of the methods. If you're not sure which is easiest, try shells first. OPTIONAL: Then, if you want the practice, also find the volume by the other method, if possible. If this other method is shells, it's a good idea to try this optional step. If the other method is disks or washers, less need to do the optional step. Either way, you only need to do it if you want to. Answers: 8) 128 pi / 5 10) Hint: when ready to integrate, let u = 4 - 2x^2, du = -4x dx, so dx = -du/4, etc. Final answer: 8pi/3 16) Integrate: 2 pi { y (sqrt[-y^2 / 2 + 25 ] } dy from y = 0 to sqrt(50) ... Same u-substitution process as in #10... Final Answer: 500pi/3 32) NOTE: My book says the hyperbola is y = 6/(x + 3), but at least one other student's book says it is 2/(x + 3), which of course changes the problem slightly. Don't know which version most students have, but do it for 6/(x + 3), so you can check to see if answer agrees with my solution below. Note: GRAPH requires some carefull plotting or zooming in to see that the curve and line form a bounded area only for NEGATIVE x-values. On that area, Line y = 2 - x is above (and to right) of other curve (Hyperbola) y = 6/(x + 3). Find intersections by setting equal, multiply both sides by LCD of (x + 3), simplify to... 0 = x^2 + x, factor, so intersections are at x = -1 and x = 0. WASHERS: Integrate: pi { [ 2 - x ] ^2 - 36/(x + 3)^2 } dy Hint, for each part of the integral, use a trivial u-substitution (watch minus signs, though). Eventually, evaluate pi { -(1/3) (2 - x)^3 + 36/(x + 3) ] from -1 to 0 , get... Final Answer: pi/3 SHELLS: Solve for x: Line: x = 2- y Hyperbola: multiply by LCD, then solve.... get: x = 6/y - 3 Integrate: 2 pi y {2 - y - [6/y - 3] } dy from y = 2 to y = 3 , simplify, distribute before integrating, get: Integrate: 2 pi {5y - y^2 - 6 } dy from 2 to 3, proceed easily... get: Final Answer: pi/3 Hooray, answers match! :) |
6.4 |
| Tue 2/7 |
REVIEW: Before coming to class (i.e., over the weekend) REVIEW all Homework problems. Any HW questions you have had trouble with you should be ready to ask me for help with during the review session. Bring your Book to class. ALSO, there will be an IN-CLASS Review Assignment, which will count as a regular HW assignment, and be handed in with your other HW after the Test. |
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| Thu 2/9 |
TEST 1 Covers all material on which HW was asssigned above. Covers up through Section 6.4 Closed Book, No Notes. Make sure you have your special trig function values memorized (as well as everything else you need to know.) BE ON TIME! Time is valuable, and while I try to make it so most students can finish on time or early, this will NOT be likely if you arrive late! BRING SOME PAPER of your own. This can be paper you will rip out of a notebook (as long as it has nothing written on it.) Or any other kind of blank paper you wish. Bring several sheets. BRING your OWN Calculator. NO SHARING of calculators during the test |
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| Tue 2/14 |
All HW and Review Assigned before Test 1 is now Due | ||
| p. 418 | 3- 9 odd, 8 These are dx 21-24 ALL These are dy But actually SKIP # 22--But read Answer for it below. Answers: 22) SKIP, because it is actually the same problem as EXAMPLE 6 at the top of page 418, but with a minus sign change in the middle. That sign eventually gets squared if you read EXAMPLE 6, so it is irrelevant. This would NOT be a typical Test problem, too difficult... NO NEED to read through Example 6 unless you are really interested in it... The fact that y = ln [ x + sqrt(x^2 - 1) ] can be solved for x is quite a lucky coincidence, and is not typical of such a complicated function. Plus, then you need the fancy shortcut anyway! 8) Take dy/dx = (1/2) sqrt(x) - 1 / [2 sqrt(x) ] , square it by FOIL, and get (dy/dx)^2 = x/4 - 1/4 - 1/4 + 1/(4x). Middle terms add to -1/2, so add 1 to the middle term, and it changes from -1/2 to + 1/2. That is the signal that the shortcut will work. So LET'S USE the SHORTCUT!! So I can skip to integrating dy/dx but with the middle sign changed from - to +. IN FACT, let's use the double shortcut: the answer to the integral will just be the original function y = f(x) with its middle sign changed: So: Evaluate (1/3) x^(3/2) + sqrt(x) from x = 4 to 16, get.... 64/3 + 4 - [ 8/3 + 2 ] = ... Final Answer: 62/3 24) Take dx/dy, square it, and get (dx/dy)^2 = 8 e^(2sqrt(2) y) - 1/2 + (1/128) e^(-2 sqrt(2) y) Add 1 to the middle term, and it changes from -1/2 to + 1/2. That is the signal that the shortcut will work. So LET'S USE the SHORTCUT!! So I can skip to integrating dx/dy but with the middle sign changed. IN FACT, let's use the double shortcut: the answer to the integral will just be the original function x = f(y) with its middle sign changed: So: Evaluate 2e^( sqrt(2) y) - (1/16) e^(- sqrt(2) y ) from y = 0 to y = [ln 2 ] / sqrt(2) = etc.. Final Answer: 65/32 |
6.5 | |
| p. 428 | 12, 9-15 odd, Hint on 15, do 2 separate integrals on the two domains, then add. Answer: 12) (5/2)[ 1 - 1/(e^8) ] |
6.6 | |
| Thu 2/16 |
p. 428 | 18--random
force Note, this really shouldn't be in this section, as it
becomes infinite at one limit, and ."answer is "Infinite amount of work 19, 20, 21 -- Spring problems. NOTE: on #21, they tell you Work = 100, which means they must have already integrated an unknown "kx" and evaluated, and gotten 100. So you do that--integrate an unknown "kx", keeping "k" in the answer, evaluate, and then set the result = 100 and solve for k. Now that you know "k", repeat the problem using the known "k", and integrate to find the new value of work they are asking for. Answers: 18) Integrate 2/(x^2) to get -2/x evaluated from x = 0 to x = 3. BUT 2/0^2 = infinity even before integrating, and so does -2/0 after integrating, so answer is that the force was infinite at 0 in the first place, and an infinite amount of work is required to move it. SO: this problem really shouldn't be in this section, as integrating functions that are or become infinite has not yet been covered in the book (though it will be.) 20) Change 2 cm to 0.02 meters, and 4 cm to 0.04 meters. Use F = kx, F is told to you as 500kg(9.8m/sec^2) = 4900 Newtons, so solve 4900 = k(0.02) , or get k = 245,000. Then use that to integrate kx from 0 to 0.04, get W = 196 Joules. |
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| Tue 2/21 |
p. 458 | 7-35 odd, skip 23, 25 | 7.1 |
| Thu 2/23 |
p. 458 | 23, 25 Circular Integration by Parts | |
| p. 466 | 11, 12, 14, 15, 16 odd powers of sin, cos Answers: 12) First remove the issue of the 20x by letting u= 20x, du/20 = dx, and do (1/20) Integral of cos^3 (u) du Answer becomes ... (1/20) [ sin u - (1/3) sin^3 (u) ] = (1/20) [ sin (20x) - (1/3) sin^3 (20x) ] + C 14) (-1/6) cos^6 (x) + (1/8) cos^8 (x) + C 16) u = sin x, du = cos x, Integrate u^(-3/2) - u^(1/2) , get... eventually -2 sin^(-1/2) ( x) - (2/3) sin^(3/2) (x) + C Factoring might be good here, factor out -2 sin^(-1/2) (x) [ 1 + (1/3) sin^2 (x) ]. Then note sin^(-1/2) = sqrt(csc ), so best answer may be -2 sqrt [csc(x)] [ 1 + (1/3) sin^2 (x) ] |
7.2 | |
| Tue 2/28 |
p. 473 | 7-45 odd, SKIP 29 | 7.3 |
| Thu 2/30 |
p. 473 | 48-50 ALL These are completing the square, then using trig substitutions | |
| p. 466 | 20,
24, 30, 21, 27 More trig powers, this time of tan and sec (we covered
these ideas previously, but you did not have HW on them yet.) 10, 13 -- Even powers of sin and cos |
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| p. 1080 | NOTE: I have decided NOT to put these hyperbolic functions on Test 2--instead, they will be covered on Test 3. SO: IF you DON'T want to do this HW now, you can POSTPONE it until after Test 2. I will tell the class this on Tues, but if you read this before then, please let others in the class know. Here is the assignment, in case you do want to do it now--but it's up to you, and if you have a lot of other work, you should postpone it. 3-8 ALL, 10-12 ALL, 13-18 ALL, 20 A) only, and 21 ANSWERS ARE BELOW in the section where I remind you gain to do these, for Tuesday 3/20 |
Appendix C | |
| Tue 3/6 |
REVIEW: Before coming to class (i.e., over the weekend) REVIEW all Homework problems. Any HW questions you have had trouble with you should be ready to ask me for help with during the review session. Bring your Book to class. ALSO, there will be an IN-CLASS Review Assignment, which will be handed in with your other HW after the Test. |
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| Thu 3/8 |
TEST 2 Covers all material on which HW was asssigned above, since the first test. Covers
up through Section 7.3 Closed Book, BUT: You MAY USE ONE 8 1/2 by 11 NOTESHEET (both sides is fine.) BRING SOME PAPER of your own. This can be paper you will rip out of a notebook (as long as it has nothing written on it.) Or any other kind of blank paper you wish. Bring several sheets. BRING your OWN Calculator. NO SHARING of calculators during the test |
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| SPRING BREAK MAR 12 -18 College Closed | |||
| Tue 3/20 |
All HW (and Review) since Test 1 is Due | ||
| p. 1080 |
Don't
forget that the HW on sinh and cosh from Appendix C, page 1080 , was
assigned before Test 2, but not covered on Test 2. But it will be on Test 3! Review that HW above and make sure you have done it and asked Q's on it, if needed. Answers: 4) x cosh (sqrt [x^2 + 1] ) / sqrt [ x^2 + 1] 6) (1/2) cosh (x^2 + 4) + C 8) x sinh x - cosh x 10) x sech^2 (x) + tanh (x) 12) -3 tanh (3x) csch^2 (3x) or -3 sech(3x ) csch(3x) (same answer when simplified.) 14) (-1/2) sech (2x) 16) Not obvious or simple, but there is a (somewhat tricky) way to integrate this: When you see the Integral of sech^5 x csch x, you wish you had tanh x (instead of csch x), to match with a sech^2 (x) as u and du. Temporararily, I'll switch to regular trig functions, then back to hyperbolic later. Since you see csc x, which is 1/sin x, you think: "Easy to make it into cot x, then change that to tan x." Make it cot x easily by multiplying by cos x in the top (and bottom, to balance it) of integral, getting Integral of ( 1/ cos (x)) sec^5 (x) [ cos x / sin x ], or Integral of (sec x) sec^5 (x) [ cot x ]. Then write cot x as 1/ tan x, and also combine the sec x with the other 5 powers of sec x, giving us Integral of sec^6 (x) / [ tan x ]. Then change back to hyperbolic version: So we get: Integrate: sech^6 (x) / tanh x dx , and write the top as sech^6 (x) = sech^4(x) sech^2 (x) Now u = tanh, du = sech^2, and the remaining sech^4 (x) in the top is [sech^2 (x) ]^2 , then use cosh^2 - sinh^2 = 1 in its alternate form (divide by cosh^2) of: 1 - tanh^2 = sech^2. Replacing the [ sech^2 ]^2 by that, and with u = tanh, we get : Integrate: (1 - u^2)^2 (1/u) du Square out the top by Foil, divide each term by the denom. of u, and Integrate: 1/u - 2u + u^3 du Easy integrals, replace u's, get: ln |tanh x| - tanh^2 (x) + (1/4) tanh^4 (x) + C. Note: The same process would integrate other power combinations of sec x and csc x, though odd/even powers might be an issue... 18) 1/coth = tanh, so Integrate tanh x dx , write as sinh/cosh = du/u, get ln|u| = ln|cosh x| evaluated from ln 2 to ln 3. Use definition of cosh as [e^x + e^ (-x)] / 2 , and simplify via e^ln 3 = 3, and e^ -ln(3) = 1/ e^ ln(3) = 1/3 , etc., and we get: ln [ (3 + 1/3) / 2 ] - ln [ (2 + 1/2) / 2 ] = ln (5/3) - ln (5/4) = ln [ 5/3 divided by 5/4 ] = ln (4/3). 20) a. Area = 2 Integral of ( 2 - cosh x ) dx from 0 to Q, where Q is the right-hand x-value of the intersection, and I've doubled the Area to pick up the stuff to the left of the y-axis (by symmetry.) To find Q at the intersection of cosh x and 2, write cosh x = 2, i.e., [e^x + e^(-x)] / 2 = 2 . Multiply by 2e^x on both sides, and get (e^x)^2 + 1 = 4e^x . Call e^x by the name "w", and we have w^2 - 4w + 1 = 0, solve by Q.F. for w = 2 + sqrt(3) , and since e^x = w, solve for "x" by taking ln of both sides. So x = Q = ln (2 + sqrt(3) ) for the right-hand intersection value. Integral is [2x - sinh x] evaluated from 0 to ln (2 + sqrt(3) ), giving: [2 ln (2 + sqrt(3) ) - sinh [ln (2 + sqrt(3) ] ] - (0 - 0) Plug in ln (2 + sqrt(3) to the definition of sinh using e^x and e^-x, and you get, after some simplifying with conjugates of roots, sqrt(3) !! So final Area is 2 [ 2 ln (2 + sqrt(3) ) - sqrt(3) ] , having doubled the result. |
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| p. 510 | 21, 23, 24, 25 Volumes--do these problems after you do the ones below. Use normal formulas for disks, washers, etc. from Volumes in Ch. 6 5-17 odd These are Situation 1, where the limits of integration are infinite. 27-35 odd These are Situation 2, where the function becomes infinite--set denom. = 0 to find out where (or otherwise figure out where the function is infinite--note that ln x is -oo at x = 0, etc... ) This value of x = c where the function becomes infinite is the vertical asymptote. Unless c occurs as one of the limits of integration, you will then have to do two separate integrals, one stopping at x= c, the other starting from x = c. Remember that if either one of the two separate integrals is infinite, the whole thing Diverges. But if the first part is infinite, do both anyway, in case you made a mistake in the first one! Answers: 24) Integrate by DISKS, use dy, need radius = x , solve for x = y^(-1/3) - 1 Integrate pi [ Integral of [ y^(-1/3) - 1 ]^2 dy ] from y = 0 to y = 1 . Foil out the [ ]^2, integrate by power rule, Get: pi [3y^(1/3) - 3y^(2/3) + y ] from 0 to 1 = ... = pi. OR SHELLS: get 2pi Integral of x (x + 1)^(-3) dx from 0 to oo, which uses the "switch", let u = x + 1, x = u - 1, So get: 2pi Integral of (u - 1) u^(-3) du = ... 2pi [ -1/(x + 1) + 1/ [ 2(x + 1)^2 ] from 0 to oo = ... = pi. Same answer, good sign! |
7.7 | |
Thu 3/22 |
No Homework | ||
| Tue 3/27 |
p. 534 | 2, 9-12, ALL, 17-22 ALL, a) and c) only 23-29 odd |
8.1 |
| Thu 3/29 |
p. 546 | 9-19 odd, 27-33 odd Sequences, use various limit processes, including "Growth Rates" | 8.2 |
| p. 553 | 19-31 odd Geometric Series, find a and r and use formula (if |r| < 1) Sum = a / (1 - r) | 8.3 | |
| Tue 4/3 |
p.553 | #49-->#47 That is, the expression in #49 is really equal to #47, so do #47 (as a telescoping series.) 55, 56, 58 Answers: 56) As k grows, for each partial sum Sk, all terms cancel except sin(0) which is 0, AND sin [(k + 1)/(2k + 1) ] pi. As k-->oo, (k + 1)/(2k + 1) goes to 1/2, so remaining term is approximately sin [ (1/2) pi ] , which is 1, which is the final Sum: Series adds to 1. 58) As k grows, for each partial sum Sk, all terms cancel except - tan-inverse 1 and tan-inverse (k + 1) As k-->oo, tan-inverse (k + 1) goes to tan-inverse (oo), which is pi/2, since tan(pi/2) = oo. Since - tan-inverse 1 = - pi/4, Final Sum is -pi/4 + pi/2 = pi/4. . |
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| p. 567 | 11, 13, 14, 15-21 odd, 23-29 odd, 24, 28, 31, 33 Answers: 14) Separate into two sums, factor 2 out of first sum: 2 [ SUM of (1/6)^k ] - [ SUM of (3/6)^k ] First SUM is geometric series with a = 1, r = 1/6, sum = 1 / (1 - 1/6) = 6/5, multiply by the 2, get First SUM = 12/5. Second SUM is geometric series with a = 1, r = 1/2, sum = 1 / (1 - 1/2) , get 2nd SUM = 2 Total is thus 12/5 - 2 = 2/5 24) Integrate [ x / sqrt(x^2 + 4) ] dx = u-subst. = sqrt(k^2 + 4) evalutated from 1 to oo, get sqrt (oo^2 + 4) - sqrt 5 = oo, Diverges. So the Series Diverges since the Integral Diverges. Note, you could have seen this result sooner by using the Divergence Test on the Individual Terms: Lim k--> oo of k / (k^2 + 4)^(1/2 ) = RHP = Lim k--> oo of k / (k^2)^ (1/2) [dropping the "4"] = k / k = 1 = NON-Zero, so series DIVERGES. 28) Integrate 1 / x [ ln x ]^2 dx , use u-subst.., get -1/ (ln x) evaluated from 2 to oo = -1 / ln(oo) - - 1/ ln(2) = 0 + 1/ ln(2). Since the Integral Converges to 1/ ln(2), the Series ALSO Converges (but to a different sum than 1/ln(2). The actual sum of the Series is difficult and is beyond the level of Calculus 2. |
8.4 | |
| Thu 4/5 |
p. 599 | 7,
9, 12 Do parts b) and c) only Here the
"quadratic" approximation means the Taylor Series up to the c2x^2 term
only. See hint below for #9. 13-19 odd Again, here do the "quadratic" approximation, meaning the Taylor Series up to the c2x^2 term only. 33, 35 Here, do one more term than in the previous problems, up to the "4th" term, which means c3x^3 (counting c0 as the "first" term.) Answers: 12) Even though this only asked for terms up to c2, I will give one more term as well, up to c3. You did NOT have to do that here, though. Notice that the 3rd deriv. is already quite complicated, so further ones are very difficult. We will see an easier way to find this series in a later section. f(0) = tan-inverse (0) = 0 = c-zero. (after dividing by 0!, but that equals 1, and 0/1 = 0) f '(x) = 1 / (1 + x^2) | x = 0 = 1 , then divide by 1!, and get 1 = c1 f '' (x) = -2x / (1 + x^2 ) ^2 | x = 0 = 0 = c2 (after dividing by 2!, but answer is still 0) f '''(x) = [ - 2 ( 1 + x^2 )^2 - (-2x) ( something ) ] / [ 1 + x^2 ]^4 | x = 0 = ... -2 , then divide by 3!, get c3 = 1/3 Note we needed the Quotient Rule here. So, Taylor series for tan-inverse x = 0 + 1 x + 0 x^2 - (1/3) x^3 + ... = x - (x^3) / 3 + ... Book answer stopped with the x, though. Book's approximation of tan-inverse (0.1) is: x evaluated at x = 0.1 = 0.1 More accurate answer using the extra term of -(x^3) / 3 is: x - (x^3) / 3 evaluated at x = 0.1 = 0.1 - (0.1^3) / 3 = 0.1 - 0.000333 = 0.0996666 approx. equals tan-inverse (0.1) Calculator answer is tan-inverse(0.1) = .0996687 Note that the answers are already very close!, even after only two (non-zero) terms! 9) Hint: your series should eventually turn out to be: 1 - x + x^2 Then, to find approx. for 1/ (1.05), set it equal to 1/ (1+ x), which means the denom's are equal: (1.05) = (1 + x), then solve for x : you get x = 0.05 That is what you need to substitute into your series, the value x = 0.05, then proceed to calculate... |
9.1 |
| Tue 4/10 |
REVIEW: Before coming to class (i.e., over the weekend) REVIEW all Homework problems. Any HW questions you have had trouble with you should be ready to ask me for help with during the review session. Bring your Book to class. ALSO, there will be an IN-CLASS Review Assignment, which will be handed in with your other HW after the Test. |
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| Thu 4/12 |
TEST 3 Covers all material on which HW was asssigned above, since the second test. Covers
up through Section 9.1 Closed Book, BUT: You MAY USE ONE 8 1/2 by 11 NOTESHEET (both sides is fine.) BRING SOME PAPER of your own. This can be paper you will rip out of a notebook (as long as it has nothing written on it.) Or any other kind of blank paper you wish. Bring several sheets. BRING your OWN Calculator. NO SHARING of calculators during the test |
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| Tue 4/17 |
p. 621 | 9-15 odd part a) only Maclaurin Series at x = 0 17-21 odd, Taylor Series at various a not equal to 0 29-33 odd Binomial Series |
9.3 |
| Thu 4/19 |
p. 629 | 7, 9, 13 21, 23, 25, 26 31-37 odd Answers: 26) f(x) = sqrt(1 + x) = (1 + x)^(1/2) = Binomial Series with p = 1/2 = = 1 + (1/2)x - (1/8)x^2 + (1/16)x^3 - (5/128)/x^4 + ... df/dx = take term by term derivatives in series = 1/2 - (1/4) x + (3/16)x^2 -(5/32) x^3 + ... df/dx must be the derivative of (1 + x)^(1/2) , i.e., it must be the series for g(x) = f-prime of (x) = (1/2)(1 + x)^(-1/2) . Sure enough, if you find the Binomial series for (1/2)(1 + x)^(-1/2) [using the Binomial series with p = -1/2, and then multiplying by (1/2)] you will get the same series that we got for df/dx. |
9.4 |
| Tue 4/24 |
p. 642 | 45-49 odd Parametric Equations dy/dx = (dy/dt) / (dx/dt) | 10.1 |
| p. 653 | 9, 11, 13 15-25 odd |
10.2 | |
| Thu 4/26 |
p. 653 | Graphs: 27, 29, 37, 39, 43 For each graph show by hand values in a table for r and theta for all special angles in I (pi/6, pi/4. pi/3), and selected angles in other quadrants if needed. Also be sure to show pi/2, pi, 3pi/2, and 2pi, of course. You may then of course check it on your graphing calc. | |
| p. 662 | 5, 7, 11 Derivatives:
Use dy/dx = fraction with: top = d/d(theta) [
( r ) sin(theta) ] , and denom. =
d/d(theta) [ ( r ) cos(theta) ] In these, r is of course replaced by f(theta). Simplify before taking deriv., if possible, to avoid product rule, if possible. 21-25 ALL: Areas, using Integral of (1/2) r^2 d(theta), with r again replaced by f(theta) (and then squared...) Answers: 22) Graph is a cardioid (upside-down heart) starting at [in Cartesian (x, y) coordinates] (4, 0), going to (0,8), back down to (-4, 0), then spiralling through Quad. III in to (0,0), then spiralling back out through Quad. IV back to (4, 0). This starting point has been reached again as theta comes to 2pi, so it repeats after that. Now I will use "t" here for "theta": Area = Integral of [ (1/2) (4 + 4 sin t )^2 ] dt from t = 0 to t = 2pi. Foil out the ( )^2, multiply by 1/2, and factor out 8 from the new terms, and you get: 8 ( Integral of [ 1 + 2 sin t + sin^2(t) ] dt ) Integrals are easy, add like terms when finished of t + t/2 = 3t/2, and we are ready to evaluate the following: 8 [ 3t/2 - 2 cos(t) - (1/4) sin(2t) ] from t = 0 to t = 2pi. Answer: 24pi. 24) Graph carefully, using multiples of pi/4, you get that from theta = 0 to theta = pi/2 the curve has already completed one "loop" in Quad. I. Note carefully that for next angles from pi/2 to pi, as angles go through Quad. II, radiuses are NEGative, going backwards through points in Quad. IV to make a second loop there. As theta goes from pi to 3pi/2 in Quad. III, third loop is also formed in Quad. III by POSitive radiuses. And as theta goes from 3pi/2 to 2pi in Quad. IV, the radiuses are again NEGative, so point are backwards into Quad. II, forming the 4th and final loop in Quad. II. Then it repeats after that. Since first loop is in Quad. I from theta = 0 to pi/2, we can simply find that area, then multiply by 4 to find total area. Now I will use "t" for theta: Area = 4 ( Integral of [ (1/2) (3 sin 2t )^2 ] dt ) from t = 0 to t = pi/2. This gives: 2(9) ( Integral of [ sin^2(2t) ] dt ) Be careful, Integral needs a u = 2t substitution! du/2 = dt, and proceed... Eventually, after doing u-integral and re-substituting in the "t"'s, we get: 18 [ t/2 - (1/8) sin(4t) ] evaluated from 0 to pi/2 . Answer: 9pi/2 . |
10.3 | |
| Tue 5/1 |
REVIEW: Bring all HW questions you have had trouble with to class for help. Bring your Book to class. ALSO, there will be an IN-CLASS Review Assignment, which will count as a regular HW assignment, and be handed in with your other HW after the Test. |
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| Thu 5/3 |
TEST 4 Covers all material on which HW was assigned above. Covers all HW assigned up through Section 10.3 You may use ONE 8 1/2 x 11 NOTESEHEET. (or smaller) Otherwise, closed book, no other notes BRING SOME PAPER of your own. BRING your OWN Calculator. NO SHARING of calculators during the test IMPORTANT NOTE: If you miss Test 4, you can NOT take a makeup in the normal way. You have only two options: 1. Take a zero for your grade, and make that the one that you raise to the (Final Exam minus 10%). OR: 2. Take an Incomplete in the class, and make up Test 4 very soon after Finals week. YOU MUST STILL TAKE THE FINAL ON TIME, though! If you choose this second option, there will still be the usual late penalty on Test 4. Once you take it, I will grade it and then change your Incomplete to the new grade. So, if you miss Test 4, talk to me about these options very soon afterwards! But, I urge you NOT TO MISS TEST 4 ! |
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| Tue 5/8 |
Return Test 4 REVIEW for Final Exam [you should already have reviewed all old HW and old Tests and old Reviews before you come into class this day.] Bring your BOOK to class for the review. Bring a GRAPHING CALCULATOR, too |
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| Thu 5/10 |
Last HW Assignments Due.
FINAL EXAM: Bring
Paper, 2 Pencils or Pens, and a GRAPHING Calculator of your Own! YOU MAY USE ONE 8 1/2 BY 11 NOTESHEET FOR THIS TEST. (OR SMALLER.) Starts on Time. If you are late, you get no extra
time.
If you miss the Final Exam, you will get a ZERO for it, which will
destroy your grade.
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