| Oakton Community College |
| MAT 250 Calculus 1 |
| Section 004: TR 11am - 1:20 pm Room 2129 DP Campus |
| Paul Boisvert Professor of Mathematics |
| FALL Semester 2017 |
| Return to Homepage |
| Date | Page. | Problems | Read Sections |
| Tue 8/22 |
p. 1157 |
1, 2, 15, 16 Board Problems: A) Solve and write answer as an interval: (3x - 8)/5 - (x + 2)/15 < 2 + x/6 B) Simplify: sqrt(x^5) [cube root of (x^7) ] / [4th root of (x^15) ] Write answer as a root. C) Simplify using Laws of Exponents: [3 x^(-4) y^2]^(-3) [2 x^(-5) y]^2 / [6 x^(-2) y^3 ] ^ 3 Even Answers: 2) { x : x < 2 } This is read as: the set of all x such that x is less than 2. The colon is "such that". However, I don't require the { : } part, so you could just write: x < 2 16) -5 + (1/5) = -24/5 Board Problems: A) x < 112/11 which is, as an interval: (-oo , 112/11) B) x^[ 5/2 + 7/3 - 15/4 ] = x^(13/12) or, as a root, 12th root of (x^13) C) Partial steps: ... = [ (1/27) x^12 y^(-6) ] [4 x (-10) y ^2 ] / [216 x^(-6) y^9 ] = ... 4 /[27(216)] x^(12 - 10 - -6) y^(-6 + 2 - 9) = reduce fraction, and I will (optionally) move y to denom: = x^8 / (1458 y^13) |
Appendix A |
| Thu 8/24 |
p. 1157 |
17, 18, 19-23 ALL, 25, 37 ALSO, do ALL problems on the handout sheet given out in class today. Check answers on the back of the sheet. NOTE: To add like roots, add coefficients. Example: Add 5sqrt(7) + 8sqrt(7) = 13sqrt(7) The numbers inside the roots must be the same. If at first they are not the same, simplify the roots, and see if then they are the same. Even Answers: 18) FOIL out top and add like terms, get Top = 2ah + h^2. Then to divide, factor out gcf of h in top, and cancel with h in bottom. Final answer: 2a + h 20) Note: LCD = one of each group factor from denom's = (x+3)(x - 3) Then multiply each fraction by missing factor top & bottom, and add. Final answer: -12/[(x+3)(x - 3)] 22) Factor out GCF of x, then diff. of sq's. Final answer: x = 0, x = 3, x = -3 |
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| Tue 8/29 |
p. 1158 |
38-44 ALL Answers: 38) y-axis, i.e. , x = 0 40) Horizontal , so: y = 3 42) 2x - 5 = 0 is missing y, so it is vertical. Parallel line to it is also vertical. Vertical line through (-6 , 0) is: x = -6. 44) Line through (-9, 2) and (3, -5) has slope (-5 - 2) / (3 - -9) = -7/12. Perp. slope would be +12/7. NOTE: the rest of this problem is more involved than would be on a test--it asks you to find a "bisector", rather than just telling you the point it wants you to use. On a test I would just tell you the point to use. Other than this extra issue, the problem is simple and straightforward. Let's proceed: "Bisector"means it goes through the point halfway between our two points, which is the midpoint, which is ( average x-value, average y-value). To average numbers, add them, and divide by 2. So, Average x = (-9 + 3) / 2 = -3 Average y = (2 + -5) / 2 = -3/2 So our line goes through (-3, -3/2) , the bisector, with slope 12/7. Point-Slope Equation: y + 3/2 = (12/7) ( x + 3) Stop, do not simplify. |
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| p.10 |
13, 14, 16, 17,
18,20 State Domain only 25-35 ALL 42-45 ALL For these, do NOT state what the Domain is, don't worry about it. Just find the composition Board Problems: A) Find the Domain for y = 5 / (x^7 - 24x^5) B) Find the Domain for y = 5 - sqrt(4 - 7x) C) Find the Domain for y = cube root(x + 5) - x^2 + 2. This could also be written as (x + 5)^(1/3) - x^2 + 2 Answers: 14) Note, since this is a function of y, for this function, "y" plays the role of x. So I'm going to state Domain without using "x" or "y": Domain = All Real Numbers EXCEPT 3 and -2. 16) Domain: w < 2 , or (-oo, 2] 18) Domain: x > -5 , or [-5, oo) 20) For Domain, we worry "where is denom. = 0"? We try to find any "bad" points by setting 1 + t^2 = 0 and separating. But this gives t = sqrt(-1) which is imaginary--not real! So there are NO (real) solutions to our effort to set denom. = 0, and thus there are NO "bad" points where we should worry--all points are good. So Domain = ALL Real Numbers (they are all good points.) This can be written most easily as the interval (-oo, oo) 26) p^4 - 4 28) 1/(y^4 - 3) 30) w^6 - 4 32) 4 + h 34) 1 / (x^2 - 7)^3 42) x^2 - 4 44) | 1 / [ (x - 2)^2] - 4 | Board Problems A) Domain is: {x | x does NOT equal 0, and x does NOT equal + 2sqrt(6) } Note sqrt(24) has been simplified, and as always when you introduce a new square root to a problem, needs the plus/minus sign. B) Domain is x < 4/7, or, as an interval, (-oo, 4/7] C) Since there are no square roots and no denom.'s, there are no problems, so Domain is All Real Numbers, or (-oo , oo) |
1.1 |
|
| Thu 8/31 |
p. 22 |
25, 26, 27 Graph
these piecewise functions. Show 2 or 3 points for each part in
a table. Show all ghost points. Answers: 26) Top line: Ghost point (open circle) at (1,2), real point at (0, -1) , connect line going to left. Bottom line: Real point at (1,2). Since this is the same as the ghost point, you now draw it as a solid dot there after all. Best to show it as a big open circle with a smaller but still sizeable solid dot inside. [This sometimes happen, that a ghost point for one part is then inhabited by a real point from the other. Like a real person living in a haunted house with the ghost... :) ] And bottom line also has a real point at (2, 3) Connect up those points heading to the right... |
1.2 |
| |
p. 35 |
22, 24,
27 Find inverses, use switch and solve method. 70) For each exponential function, see what the point on it is at either (1, ?) or (-1, ?) from the graph. That will tell you what the base is, and its shape will tell you whether it is B^x or B^(-x). Then you can match them up with their equations. Board problems: in each one, use switch and solve method to find the inverse function. A) f(x) = 7 - cube root(2x + 5) B) f(x) = 2 + 4(x + 7)^5 C) f(x) = (2x - 8) / (5x + 6) Hint, cross multiply here. Answers: 22) f-inverse(x) = 4(x - 1) or 4x - 4 24) f-inverse(x) = cuberoot of (x/3) 70) In order of equations as listed, answers are D , C, A, B A) f-inverse (x) = [ (-x + 7)^3 - 5 ] / 2 B) f-inverse(x) = - 7 + 5th root [ (x - 2) / 4 ] C) f-inverse(x) = (-6x - 8)/(5x - 2) OR, switching signs, (6x + 8)/(-5x + 2) Either answer is ok. |
1.3 |
| Tue 9/5 |
p. 35 |
47-52 ALL, 53-55 ALL, 45, 46 Board Problems: A) Simplify ln [ 12 x^5 (cube root of y) / (ez^3) ] B) Condense back into one big log: 4 ln x - (1/3) ln y + ln (e^3) Answers: 46) y = e^3 = 20.086 48) Assume here that logs have base "b": Simplify to 2 log x = 2(.36) = 0.72 50) Simplify to log [ (xy)^ 1/2 ] - log (z) = (1/2) log (xy) - log z = (1/2) [ log x + log y ] - log z = (.5) [.36 + .56] - .83 = .46 - .83 = -0.37 52) Note, log(b) of b^2 = 2, as the log cancels its own base, leaving the exponent 2. Answer: 2 + (5/2) log x - (1/2) log y = 2 + (2.5)(.36) - (.5)(.56) = 2 + .90- .28 = 2.62 54) Apply ln to both sides, get ln(2^x) = ln (55), or, simplifying the left side, x ln(2) = ln 55. Change ln's to decimals: .693 x = 4.007, or, dividing, x = 5.783 Board Problem A): ln 12 + 5 ln x + (1/3) ln(y) - 1 - 3 ln z Note, no need to add ln 12 and -1, unless problem asks for decimal answer. B) ln [ (x^4) (e^3) / cube root of y ] |
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| Handout
Sheet: Do I, II, and 0) parts e) through h) only Check answers on back of sheet. |
1.4 |
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| Thu 9/7 |
Handout
sheet: do all problems up through #6. Also do #8. (Do NOT do #7.) |
1.4 |
|
| p. 47 |
84-87 ALL NOTE: for
these, if they tell you, say, that sin theta = 7/9, that means
opp/hyp = 7/9, so draw a triangle with opp side 7 and hyp = 9. Then find the third side by the pythagorean theorem, and you will know all 3 sides, so you can write all 6 trig fcns from opp, adj, and hyp. BUT also the problem will tell you what Quadrant the angle is in, so use that to fill in the correct plus or minus signs on your answers for each fcn. Answer for 84 shows this in detail, if you get stuck. 15-28 ALL for these, list any reference angles you use. 47-54 ALL ANSWERS: 84) I will use "t" for "theta". Ignore the minus sign until later. Since sin = opp/hyp, = 4/5, opp must be 4 and hyp must be 5. Draw triangle with opp =4, hyp = 5, find 3rd side byPyth.., get adj = 3. Since t is between pi and 3pi/2, it is Quad. III. So using signs for Quad. III, we get: cos t = -3/5, sec t = -5/3, tan t = 4/3, cot t = 3/4, csc t = -5/4 . 86) Use same approach as for 84), hyp must = 5, adj. = 3. Then get... opp = 4, and we are in Quad. IV, so cos t = 3/5, sin t = -4/5, csc t = -5/4, tan t = -4/3, cot t =-3/4. 16) Ref. angle for 2pi/3 is pi/3, answer is = +sqrt(3) / 2 (Positive because 2pi/3 is in II) 18) Find smaller coterminal angle of 7pi/4. Ref. angle is pi/4. tan 7pi/4 = -tan pi/4 = -1. (Negative because 7pi/4 is in IV.) 20) Answer is 1/ cos (7pi/6) , which has ref. angle pi/6, = 1 / [-sqrt(3) / 2] = -2/sqrt(3) (Negative because 7pi/6 is in III.) 22) Find smaller coterminal angle of 4pi/3, ref. angle is pi/3, answer is -sqrt(3) / 2. (Negative because 4pi/3 is in III.) 24) -pi/2 is coterminal with 3pi/2, from p. 40 use y-value which is -1. 26) 3pi is coterminal with pi, from p. 40 use y/x = 0. 28) Use x/y from p. 40 which is -1/0 which is undefined or infinity. 48) Since cos (pi) = -1, therefore, cos-inverse of -1 = pi. 50) We know cos(pi/4) = sqrt(2) / 2. Two angles have their cos = -sqrt(2) / 2, namely those in II and III with pi/4 as a reference angle. BUT cos-inverse is supposed to come out in II for a negative value, never in III. So we need an angle in II with a pi/4 reference angle, which is pi - pi/4 = 3pi/4. Therefore, cos-inverse [-sqrt(2) / 2 ] = 3pi/4 52) if cos-inverse (2) = an angle t, then cos(t) = 2. But this is impossible, as cos and sin are always between -1 and 1. So cos-inverse (2) = undefined. 54) Since sin (3 pi/2) = -1, therefore, sin-inverse of -1 = 3pi/2. |
1.4 |
|
| Tue 9/12 |
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. 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 9/14 |
TEST 1 Covers
all material on which HW was assigned above, including the
Trig we covered in sec. 1.4 and on the Handout Sheet.Closed Book, No Notes.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 . Calculators must be stand-alone, NOT connected to Internet or outside via phone or computer. No electronics running during test, except stand-alone calculators. |
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| Tue 9/19 |
HW (and Review assignment) DUE. | ||
| p. 65 |
7,
9 2-sided limits 23, 21, 22 One-sided and 2-sided limits Answers: 22) a) 3 b) 2 c) 3 d) No Limit, since b) and c) are two different targets (one-sided limits) from the left and right sides. e) 2 f) 3 g) 2 h) 3 i) 3 [Note this limit = 3 since both LHL and RHL are the same y target value (3) from either side] |
2.2 |
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| p. 76 |
25, 27, 39, 41, 45, 40, 42,
51 If you get 0/0, factor and cancel.
If you do NOT get 0/0 at first, do not factor, you are done. Answers: 40) 4 42) -5 Note, (t-2) and (2-t) are the same factors with signs switched, so they cancel to a (-1). That then multiplies the remaining factor as it is evaluated. |
2.3 |
|
| x |
p. 86 | 17, 19, 20, 21, 23 Here, if you get nonzero/0, answer is +/-
infinity, make a table of one x on left and right of limit, see if +
or -. If they disagree, write "No Limit". Answer: 20) a) -oo b) +oo c) No Limit |
2.4 |
| Thu 9/21 |
Trig Handout Sheet:
Do 7, 9, 12, 13 |
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| P. 47 |
92, 93, 95, 88,
89 for these, show period, 4 marks labeled, top and
bottom values, 2nd period, tails, and relabel y-axis for vertical
shift. Answers: 88) New Period is pi, the first 4 marks are pi/4, pi/2, 3pi/4, and pi. Amplitude is from 2 to -2. Sine function starts at 0, goes up to +2 first. Second period goes out to 2pi, or backwards to -pi. Tail goes up at right-hand side, down at left-hand side. No vertical shift. 92) Same as #88) but Amplitude goes from 3 to -3. |
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| p. 96 |
9-23 odd, 10, 12,
20, 22, 25-33 odd. For fractional functions, use
R.H.P. in the fractional portion. Answers: 10) plug in oo, get 5 + 1/oo + 10/oo = 5 + 0 + 0 = 5 12) RHP is 4x^2 / x^2 cancels to 4, so limit is 4. Notice you do not end up plugging in oo at all here, since RHP has you wait until you have looked at the powers, and the x's canceled before it was time to plug in oo's for them in the end. 20) RHP Note: no denom., so denom. = 1, as the one is the "highest power"in the bottom, but can be dropped. In future, no need to show a "1". So RHP is 3x^7 / 1 = 3x^7, plug in -oo and get 3(-oo)^7 = 3(-oo) = -oo. Note odd power keeps -oo negative. 22) 2x^(-8) = 2/ (x^8), plug in -oo and get 2 / (-oo)^8 = 2 / +oo = +0, so function approaches its limit of 0 from above. |
2.5 |
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| Tue 9/26 |
p. 133 |
16, 18, 19, 21 Use definition (2) in the book, which is the only one I showed you in class. Do NOT use definition (1), even if the problem says to. For the above, find f-prime of x, evaluate at the given value of x, then write the point-slope equation of the tan. line. Simplify that equation only if book wants you to. 37, 39 Here, the book says "evaluate at x = a", then in each problem give you "a = number". This is just a fancy way of saying, "evaluate at x = that number". Otherwise, do these two problems in a similar fashion to the ones above them. Answers: 16) Deriv. is f ' (x) = 6x - 4, so slope at x = 1 is 2, point slope equation of the tangent line is y + 1 = 2(x - 1). 18) Deriv. is f '(x) = -4x, so slope at x = 0 is 0, point slope equation is y - 8 = 0(x - 0). When there are zero values involved in the Point-Slope equation, it is then so easy that you should simplify to y = mx + b form, which gives us final answer of y = 8 for the equation of the tangent line. Note that the equation y = 8 is the equation of a horizontal line, so we have a horizontal tangent line, which is a very important concept in later chapters of the book |
3.1 |
| Thu 9/28 |
p. 150 |
7-23 odd, 14,
18, 24, 49 27, 29, 30 Use precalc. to simplify mult. or div. here first, so you get add/subt. of powers. Then use calculus power rule. 35, 37, 38 Tan lines [note, ones above are generally NOT entire tan lines, just slopes or deriv.s.] 44-48 ALL, SKIP 47. These are Higher Order Deriv.s, meaning 2nd deriv, 3rd deriv, etc. Answers: 14) dg/dw = 10w^11 18) f-prime (s) = 1/( 8 sqrt(s) ) 24) ds/dt = 2 / sqrt(t) - t^3 + 1 30) Simplify before taking deriv. by short division to: y = 3s^2 - 2s + 3 Then dy/ds = 6s - 2 38) dy/dx = (1/4) e^x - 1 . Evaluate at x = 0, and get slope = 1/4 - 1 = -3/4. Point slope is then y - 1/4 = (-3/4)(x - 0) Which simplifies to y = (-3/4) x + 1/4. 44) f-prime (x) = 9x^2 + 10x + 6 2nd deriv. = f-double prime (x) = 18x + 10 46) f-prime (x) = 6x + 5e^x 2nd deriv. = f-double prime (x) = 6 + 5e^x 48) f-prime (x) = 10e^x 2nd deriv. = f-double prime (x) = 10e^x |
3.3 |
| Tue 10/3 |
p. 160 |
9, 10, 11, 13, 14, 20, 21, 23, 26,
27, 33, 35, 37, 38, 39 Here, use Product Rule and Quotient Rule. Simplify enough to get the Book's answer. Answers: 10) dg/dw = (e^w )(10 w + 3) + (5w^2 + 3w + 1)(e^w) = Factor: = e^w (10w + 3 + 5w^2 + 3w + 1) = Simplify group factor = (e^w )(5w^2 + 13w + 4) = dg/dw 14) s = (4e^t) (sqrt(t) ) Let first fcn. = 4e^t, and second = sqrt (t). Use prod. rule, get: ds/dt = 4e^t (1 / [2sqrt(t) ] ) + sqrt(t) [ 4e^t ] = Factor, simplify: ds/dt = 4e^t [ 1/ (2 sqrt(t) ) + sqrt (t) ] 20) Q. R. : f-prime (x) = [ (x - 2) (3x^2 - 8x + 1) - (x^3 - 4x^2 + x) ( 1) ] / [x - 2]^2 We will not simplify. 26) Q. R. : dh/dw = [ (w^2 + 1)(2w) - (w^2 - 1) ( 2w) ] / (w^2 + 1)^2 Simplify top, being sure to distribute minus sign: TOP = 2w^3 + 2w - 2w^3 + 2w = 4w. So final answer is dh/dw = 4w / (w^2 + 1)^2 38) y= 4p^(-3), so dy/dp = -12p^(-4) , or dy/dp = -12/ p^4 |
3.4 |
| x |
p. 169 |
17-27 odd, 22, 24, 32-38 ALL [SKIP 35] 43, 45, 46 Answers: 22) Prod. rule: e^6x (cos x) + sin x (6e^6x) NOTE: the extra 6 in last factor is from a rule we haven't covered in class yet, BUT which we WILL cover on Thursday. So it will be explained soon, don't worry about it yet. Then Factor: Final answer is e^6x [ cos x + 6 sin x] 24) Q.R.: [ (1 + sin x) (-cos x) - (1 - sin x)(cos x) ] / [1 + sin x]^2. Simplify top, distributing, esp. minus sign, and find that -sin x cos x and + sin x cos x CANCEL. Final answer: -2 cos x / [1 + sin x]^2 32) sec^2 (x) - csc^2 (x) 34) Use prod. rule with sec x = first fcn., tan x = 2nd fcn. Get sec x [sec^2 (x) ] + tan x [sec x tan x]. Answer must collect powers: sec^3 (x) + sec x tan^2 (x) Factoring out sec x here is also possible, but not very helpful... 36) Use Q. R.: ( [1 + tan w] (sec^2 (w) - tan w [ sec^2 (w) ] ) / [1 + tan w ]^2 Distribute and simplify the top, and the tan (sec^2 ) terms will cancel, leaving: sec^2 (w) / [1 + tan w ]^2 38) Q. R., so somewhat similar to both #34 and 36. Simplify top as in 36, collecting powers as in 34. Final answer: [ sec^2 (t) + sec^3 (t) - sec(t) tan^2 (t) ] / [1 + sec t]^2 Again, sec t could be factored out of top, but not very helpful... |
3.5 |
| Thu 10/5 |
p. 191 |
7-33
odd These are two-level chain rules 41-44 ALL These are two-level chain rules 57-61 odd These mix chain rule with Prod. Rule, Q.R. All the problems above are similar to what you might be tested on on Test 2. 45-49 Odd, 50 Three-level chain rules. BUT: this topic will NOT be on Test 2. I will wait until Test 3 to test you on three-level chain rules. ONLY two-level chain rules will be on Test 2. Answers: 42) 8(cos x + 2 sin x )^7 [ -sin x + 2 cos x ] 44) 4( 1 - e^x)^3 [ - e^x]. Must move sign to front, and should move e^x also, so final answer: -4e^x (1 - e^x)^3 50) Think y = [cos( )]^4, with 3 levels: [ ] ^4, then cos( ), then innermost 7x^3 dy/dx = 4[ ]^3 times chain rule (-sin ( ) ) times chain rule 21x^2 Then simplify...and write as cos^3 ( ), not [cos ( ) ]^3 Final answer: -84x^2 [cos^3 (7x^3) sin (7x^3) ] |
3.7 |
| Tue 10/10 |
REVIEW:
Before
coming to class (i.e., over the weekend) REVIEW all Homework
problems since the last test. Any HW questions you have had trouble with you should be ready to ask me for help with during the review session. 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 10/12 |
|
TEST
2 Covers all
material on which HW was assigned above since Test 1 EXCEPT:
Three-level chain rules will NOT be on this test.
|
|
| Tue 10/17 |
HW and Review since Test 1 is due. | ||
| p. 181 |
17
parts a) to e) Hint for d) set s(t)
= 0, solve for t by factoring or Quad. Formula. Note this
part d) is NOT a calculus issue, just algebra. 18 parts a) to e) 11-15 odd, parts c) and d) only. Board Problem A) An object undergoes Simply Harmonic Motion with position function s(t) = -2 cos (3t) + 6 Find position s(t), velocity v(t), and acceleration a(t) for t = pi/12 Answers: 18a) v(t) = ds/dt = -12t + 64 b) At highest point, horiz. tan line, so slope = 0 = deriv. = v(t) So set v = 0, solve for t. -12t + 64 = 0, solve, get t = 16/3 = 5.33 sec's. So reaches highest point at t = 5.33 sec's. c) Plug in time found in b) to positions function s(t), that is, find s(5.33), get 362.67 =max height. d) set s = 0 to find out when its position hits the ground (height = 0). So solve -6t^2 + 64t + 192 = 0 Use Q.F., or could use graph. calc, get: t = 13.108 sec. ( Also t = -2.44, before t = 0). e) Use time from d), put it into velocity function v(t). So find v(13.108) = -93.296 ft/sec Board Problem A) : s = -sqrt(2) + 6 v = 3 sqrt(2) a = 9sqrt(2) |
3.6 |
|
| Thu 10/19 |
p. 200 |
5-21 odd, 14, 16, 18, 24. 25, 27, 29 Tan lines Answers: 14) dy/dx = y e^(xy) / [2 - xe^(xy)] 16) dy/dx = 1 / [ 1/2sqrt(y) - 2 ] or, multiplying by the Total Lcd of 2sqrt(y) top and bottom, you get: 2sqrt(y) / [1 - 4sqrt(y) ] 18) dy/dx = [ 1/(y - 1) ] / [ 1 + (x + 1) / (y - 1)^2 ] 24) dy/dx = [1 / 2sqrt(x + y^2) ] / [cos y - y/sqrt(x + y^2) ] or, multiplying by Total Lcd of 2sqrt(x + y^2), you get: = 1 / [2 cos(y) sqrt(x + y^2) - 2y ] |
3.8 |
| p. 211 |
9-21
odd |
3.9 |
|
| Tue 10/24 |
p. 211 |
CORRECTION From what I said in
class: do NOT do 64, 65. Instead do 77, 79, 81 Here, simplify function first using precalculus laws of logs, then take deriv. after that. So here is the CORRECT Assignment: 23-29 odd, 55-59 odd, 56, 58 Note: in 58, ignore the absolute value signs. 77, 79, 81 Answers: 56) dy/dx = 1 / [ x ln(10) ] 58) dy/dx = ( 1/ [ (tan x) ln 8 ] ) times chain rule deriv. of tan x,which is sec^2(x) . Recall chain rule deriv. always goes in top. Final answer: sec^2 (x) / [ tan (x) ln 8 ] |
|
| x |
p. 221 |
7, 8, 10, 13-29 odd, Tan lines: 31, 33, 34 DO These, I'm not sure if I assigned them in class, but regardless, definitely do them! Answers: 8) sin-inverse (x) + [ x / sqrt(1 - x^2) ] 10) 1 / [ x sqrt( 1 - [ln x]^2 ) ] 34) f-prime(x) = 1 / [ (e^x) sqrt( [e^x]^2 - 1) times chain rule deriv. of inside e^x, which is another e^x, and which goes in the top. That top e^x then cancels the bottom e^x outside the root. (note that the absolute value signs are not needed, since e^x is always pos.) Final answer for f-prime(x) is therefore: 1 / sqrt( [e^x]^2 - 1) Now evaluate at x = ln 2, note that e^(ln 2) = 2, and get... : 1/ sqrt(3) = slope. so Tan Line is y - pi/3 = 1/ sqrt(3) [ x - ln 2] |
3.10 |
| Thu 10/26 |
No Class, I was out sick. |
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| Tue 10/31 |
p. 243 |
23-27
odd Here, just find critical
points, where dy/dx = 0. Do NOT check endpoints or
find Absolute Min or Max. 37, 49, 50, 38, 47 For these, find Critical Points, ALSO check endpoints, make a table of x and y-values, and say what the ABS. MIN and ABS. MAX are. These involve the full procedure as we did it in class. Answers: 38) Crit. pts: dy/dx = (4/3) (x + 1)^(1/3) = 0. Divide by 4/3, raise to 3rd power, and you get x + 1 = 0, or x = -1 = Crit. pt. Also check endpoints of x = -9, x = 7. This gives the following points: (-1, 0) (-9, 16) and (7, 16). So: Abs. Min at (-1, 0), and, tied for the highest y-value of 16: Two Abs. Max's: (-9, 16) AND (7, 16). 50) Crit. pts: dy/dx = 12x^5 - 60x^3 + 48x = 0 = Factor: 12x(x^4 - 5x^2 + 4) = 12x ( x^2 - 1) ( x^2 - 4) = 12x (x + 1)(x - 1) (x - 2) (x + 2) = 0, for five Crit. Pts: x = -2. -1. 0, 1, 2. Since endpoints are x = 2 and -2, which are ALREADY crit. pts, no need to list endpoints separately, they are already included. This gives us: (-2, -16) (-1, 11) (0, 0) (1, 11) and (2, -16) So two Abs. Max's at (-1, 11) and (1,11), and two Abs. Min's at (-2, -16) and (2, -16) |
4.1 |
| p. 289 |
41, 43, 47, 48, 49,
50 For these, just find differential dy = f-prime(x)
dx. Be sure to show the dx on the right hand side. 13-19 odd Here, find dy using the numbers given. First, evaluate f-prime at x = a, then multiply that by dx (which is new x-value minus "a".) The result will be dy. Then use: New y-value = Old y-value + dy. Old y-value is f(a), dy is from previous step. Decimals ok, though if book avoids decimals try it that way too. But on test decimals will be ok. Answers: 48) dy = [ 1/ sqrt(1 - x^2) ] dx 50) dy = -1/(1 - x) dx, OR (switching signs) dy = 1/ (x - 1) dx |
4.5 |
|
| Thu 11/2 |
p. 257 |
71-77 odd, 72,
81 For these, do only steps 3 and 4 of the big
"graphing" process. BUT, Do NOT Graph! That is do step 3), which is find Crit. Pts. by setting dy/dx = 0. Then do step 4), which is evaluate 2nd Deriv. at crit. pts, see if results are pos. or neg., and use that info to say if the Crit. Pt. is a Max or a Min. Then STOP, do NOT do step 5) (infl. pts), do not do steps 1 and 2 (intercepts), do NOT Graph. Answer: 72) Crit pts: (0,0) This has pos. 2nd deriv, so is a MIN (4, 32) This has neg. 2nd deriv, so is a MAX |
4.2 |
| p. 268 |
9-13
ALL, 51, 33 [ACTUALLY,
make 33 OPTIONAL, as it has an issue that would not occur
on a test. See
Answer for 33) Below. For each graph make a table of x, y, dy/dx, d2y/dx2, and TYPE of point. Find all intercepts, crit. pts, and inflection points. Use 2nd deriv-test to classify crit. pts as max or min, and ALSO use 2nd deriv = 0 to find inflection points. Then draw graph by hand. Of course, you may also check on graph. calc. at end of problem. Answers: 33) OPTIONAL: Here, you may assume that when x = 0, x ln(x) also = 0. That is, 0 [ ln(0) ] = 0. This may seem obvious, but it is not obvious--it is actually a calc 2 issue, since ln(0) is actually infinity, and (0)( infinity ) creates issues, which makes the answer not so obvious. But it does = 0. The rest of 33 is not too bad...so do it if you are going to Calc 2, otherwise optional. See answers in book. 10) y-int at (0,0) which is also an x-int. Other x-int's at (sqrt(3), 0) and ( -sqrt(3) , 0) Use decimals to graph.. Crit. pts where dy/dx = 0 are at x = 1 and x = -1. Find y-values for those, get (1, 2) and (-1, -2). Evaluate 2nd deriv. = -6x at crit. pts. At x = 1, 2nd deriv. = neg, MAX. At x = -1, 2nd deriv. = pos.,MIN. So (1, 2) is a local MAX, and (-1, -2) is a local MIN. Find infl. pts. by setting 2nd deriv. = 0 = -6x, solve, get x = 0. Since dy/dx at x = 0 was NOT zero, since we didn't find it when we looked for crit. pts, dy/dx is therefore nonzero. So that is an infl. pt., i.e., (0,0) is also an infl. pt, in addition to being an intercept. Draw graph, check on calc. 12) y-int at (0,0) which is also an x-int. Other x-int's at (sqrt(3/2), 0) and ( -sqrt(3/2) , 0) Use decimals to graph.. Crit. pts where dy/dx = ... = Factor: 12x^3(x - 1)(x + 1) = 0 are at x = 0, x = 1 and x = -1. Find y-values for those, get (1, -1) and (-1, -1), as well as the (0,0) you already had. Evaluate 2nd deriv. = 60x^4 - 36x^2 at crit. pts. At x = 1, and also at x = -1 , 2nd deriv. = pos., so MIN's. At x = 0, 2nd deriv. = 0, which is the UNCERTAIN situation, (and which would not happen on a test.). See LAST step at end of problem. So (1, -1) and (-1, -1) are local MIN's, and (0,0) we will have to wait to study further below. Find infl. pts. by setting 2nd deriv. = 0 = , factor, get 12x^2 (5x^2 - 3) = 0, solve, get x = 0 (still uncertain) and x = +sqrt(3/5) and x = -sqrt(3/5). Again, use decimals to graph. Since since we didn't find them when we looked for crit. pts, dy/dx is therefore nonzero. So for the sqrt's, dy/dx was NOT zero, and they are infl. pts. You find y-values for them... LAST, what about the uncertain (0,0) , where dy/dx and d2y/dx2 are BOTH zero? This would not be on the test, but once you graph the other points, see if you can use common sense to figure out what it is. Note, if it was an inflection point, think about how the graph would connect the dots--would it create another max, which we didn't find, and another x-int., which we didn't find? Yes it would, so that can't be right. Same problems if it was a min, no good. So it must be a MAX at (0,0), and if so, the graph should now make perfect sense, and agree with graph. calc. OR you could use the little table we did in class, of a point on the left and right of x = 0, check dy/dx, see how pos. or neg. tan lines show you what type of point it is. Again, this issue will not occur on the test. |
4.3 |
|
| Tue 11/7 |
REVIEW:
Before
coming to class (i.e., over the weekend) REVIEW all Homework
problems since the last test. Any HW questions you have had trouble with you should be ready to ask me for help with during the review session. 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. |
||
| Thu 11/9 |
|
TEST
3 Covers all
material on which HW was assigned above since Test 2, up to
and including sec 4.5, 4.2, and 4.3 NOTE:
Three-level chain rules WILL be on this test (as well as
2-level ones, which should be routine by now.)
Closed Book, No Notes. 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! |
|
| Tue 11/14 |
HW, Review due |
||
| |
p. 327 |
11-22 ALL, 23-33 odd, 30, 34,
37-45 odd, 38, 44, 46, 47, 48 NOTE: Actually, SKIP #21, it involves a trig-inverse integral, which we did not cover yet. I shouldn't have assigned it yet! Answers: For 11-22, I will write the antiderivative or Integral as I (x) [or I (variable) if it is a different variable than x]. This is to remind you that it is a new function, whose derivative is the original function. However, when asked for an antiderivative, or asked to find an integral, or to "integrate", you do not need to give the result a name like I(x) if you don't want to--you can just write down the answer. [There is no "natural" name for it--it would be up to you if you wanted to name it.] In each case, the derivative of the answer, that is, the deriv. of the new function I( ), should equal the original problem. It is a good idea to verify and check this occasionally, especially if you are not very confident of your derivative rules. And of course each answer will have the "+ C" on the end. 12) I(x) = x^11 + C [To check, see if you agree that the deriv. of I(x) = the original g(x) in the problem.] 14) I(x) = -4 [sin (4x) / 4] = simplifying = -sin(4x) + C Note: here the inside 4 plays the role of "a" in the various formulas. 16) I(s) = -cot(s) + C [To check, see if you agree that the deriv. of I(s) = the original Q(s) in the problem.] 18) I(z) = z^(-6) or 1/z^6 + C 20) I(y) = ln y + C 22) I(t) = (pi)t + C 30) (9/2)x^4/3 + C 34) Integrate r^(2/5) dr , and get (5/7)r^7/5 + C 38) [-cos(4t)]/4 + 4cos(t/4) + C 44) Note, the variable is the greek letter "phi" (lower case, not capital). Answer - (1/3) csc (3 phi) + C 46) -(1/6) cot (6x) + C 48) (1/2)e^(2t) + (4/3)t^(3/2) + C |
4.9 |
| Thu 11/16 |
p. 327 |
DO #21, the one I told you to
Skip up in the row above. But do it now, it's a trig-inverse
integral. ALSO do #56 ALSO Do these IVP problems: 67-75 odd, 83-87 odd Answer: 56) 2x^11 - 2e^(12x) + C |
|
| |
p. 374 |
29-50 ALL, [but SKIP 30, 38,
42, 43. 46] NOTE:
SKIPPED
ones are Optional, do
them if you want extra practice, answers
are also below. Answers: 30) Integral = x^3 + x^2 evaluated from 0 to 2 = 12 32) Integral = -2 sin x evaluated from 0 to pi/4 = -sqrt(2) 34) Separate fraction into two parts and simplify: 2/t + t^(1/2) / t. Second term simplifies to t^(-1/2). So Integrate 2/t, get 2 ln(t). Integrate t^(-1/2), get 2 sqrt (t). Evaluate 2ln(t) + 2 sqrt(t) from 4 to 9. Factor out the 2, get : 2[ln(9) + sqrt (9) - ( ln(4) + sqrt(4) ) ] = 2 [5 + ln(9) - ln(4)]. Good enough...and if 2 is not factored out, OK too. 36) Integral = e^x evaluated from ln 8 to 0. Get 8 - 1 = 7 38) Multiply out the integral the long way, then Integrate: x^3 - 6x^2 + 8x and get: x^4 / 4 - 2x^3 +4x^2 evaluated from 0 to 4 = 64 - 128 + 64 - (0 - 0 + 0) = 0. 40) Integral = sin-inverse (x) from 0 to 1/2 = sin-inverse(1/2) -sin-inverse(0) = pi/6 - 0 = pi/6. 42) Integral is easy, answer is ... pi - 2 44) Similar to #42... lots of minus signs when evaluating... answer is 2 - pi 46) Separate fractions, simplify, Integrate x^(-2) - x^(-5/2) dx = -1/x + 2/3x^(-3/2) evaluated from 4 to 9 = ... = -1/9 + 2/81 + 1/4 - 1/12 = 13/162 48) 5(e^2 - 1) 50) Integrate, get -2cot(4x) evaluated from pi/16 to pi/8 = -2(0 - 1) = +2 |
5.3 |
| Tue 11/21 |
p. 391 |
17-27 ALL, 39-45 odd,
62-70 Even, 65, 67, 69 U-substitutions
, show u = , du =
, New u-integral, and u-answer, then
switch back to orig. variable. Hints
for 65, 67, 69 are below, also. Answers: 18) (1/2) e^(x^2) + C 20) (1/5) (sqrt(x) + 1)^5 + C 22) (1/10) ln |10x - 3 | + C 24) (1/11) sin^11(theta) + C 26) Note, problem shows sin x^10 without parentheses, which always MUST mean sin [ x^10], NOT (sin x)^10. If it meant (sin x)^10, it would have been written sin^10 (x), although the parentheses might have been skipped. Note that sin^10 (x) is what WAS intended in #24! Also note that on Tests I will write a problem like 26) as sin [ x^10], showing the brackets, so there will be no confusion. So, 26) uses u = x^10, du = etc, and is very different from 24). Final Answer to 26): (-1/10) cos (x^10) + C 62) u = 4w, du = 4 dw Final answer: (1/4) sec(4w) + C. NOTE: you could also simply do this by the shortcut rule for Integral of sec(ax) tan(ax) = (1/a) sec(ax). But the reason that shortcut rule works is because of the u-substitution! 64) u= sin x, du = cos x dx Final answer: (1/6)sin^6 (x) + (3/4)sin^4 (x) + (1/2)sin^2 (x) + C 65) Hint: u = cot x, end up integrating 1/u^3 du, basically... 66) u = x^3/2 + 8 du = (3/2)x^(1/2) dx = (3/2)sqrt(x) dx. Multiply both sides by 2/3 to remove the extra 3/2. So (2/3)du = sqrt(x) dx, which is what you see in the Integral, and your New u-integral is: Integrate u^5 (2/3) du Final answer: (1/9) (x^3/2 + 8)^6 + C 67) Write sec^8 (x) as 1 / cos^8 (x), then let u = cos, and the rest goes similar to 65)... 68) denom. = u = [ e^(2x) + 1 ] du = 2e^(2x) dx In the top of the integral you see e^2x dx, so the extra 2 must be removed. So du/2 = e^2x dx in the top, and your New u-integral is: Integrate 1/(u) (du/2) This creates an " ln |u |" result. Final answer: (1/2) ln | e^(2x) + 1 | , but since e^(2x) and 1 are always positive, so is their sum, so the abs. value can be dropped. So you get: (1/2) ln ( e^(2x) + 1 ) + C |
5.5 |
| Tue 11/28 |
p. 374 |
None of these will use
u-substitutions. Some will have one area, some two. You
decide--GRAPH it in order to see what's happening. 55-60 ALL, Hint for 58): 2 Areas, and before integrating, FOIL out the function completely. Then you should have easy power rule integrals. 51 part ii) only 53 part ii) only 54 Answers: 54) Two areas. Function switches from above axis to below axis at x = pi/2. So first Area is: Integrate (6 cos x - 0 ) dx from x = -pi/2 to +pi/2, then ADD it to the 2nd Area: Integral of ( 0 - 6 cos x ) dx from pi/2 to pi. First Area is 6[1 - -1] = 12, and 2nd Area is -6 [0 - 1] = +6. So Final answer is: 12 + 6 = 18. 56) Graph it, note that function switches from below zero to above zero at x = 1. So do two integrals: Integrate 0 - (x^3 - 1) dx from -1 to 1, then ADD to it the Integral of (x^3 - 1) - 0 dx from 1 to 2. Final answer is 19/4. 58) Graph and find zeros, note that function has zeros at -1, 0, and 2. Area from -1 to 0 has function on top, axis on bottom, while area from 0 to 2 has those reversed. Multiply out function to get x^3 - x^2 - 2x So do 2 integrals: Integrate (x^3 - x^2 - 2x) - 0 dx from -1 to 0, then ADD to it the Integral of 0 - (x^3 - x^2 - 2x) dx from 0 to 2 Final answer is 37/12. 60) Only one area, and one integral needed. Integrate 0 - cos x dx from pi/2 to pi, get answer of +1. |
5.3 |
| p. 391 |
All of these should use u-substitutions.
79 80 -- for this one, you can switch "theta" to x, and do the problem with x's. 81 2 Areas 82 Answers: 80) I am switching to x for the variable here: Graph of y = sin x cos x (in RAD!!) shows there is only one area between 0 and pi/2, with function above x-axis the whole way. So Integrate (Top - Bottom) dx = Integral of (sin x cos x - [0] ) dx Let u = sin x, du = cos x. [Note, if you let u = cos x, du = -sin x, it will also work out.] Integrate u du , get (u^2)/2 = (1/2)sin^2 (x) evaluated from x = 0 to pi/2. That gives: (1/2) [1^2 - 0^2] = 1/2. 82) Graph, Area is all above x-axis, so Integrate (Top - Bottom) dx = Integral of [ x / sqrt(x^2 - 9) - 0] dx Let u = inside of sqrt, so u = x^2 - 9, du = 2x dx remove the extra constant 2, so du/2 = x dx, which is a match for what is outside the sqrt. Integrate 1/sqrt(u) du/2, use power rule on root after changing to fractional exponents, get u - answer is: sqrt(u). Change back to x's: sqrt(x^2 - 9) evaluated from x = 4 to x = 5 = ... = 4 - sqrt(7). |
5.5 |
|
| Thu 11/30 |
|
REVIEW for
TEST 4: Before coming to class (i.e., over the
weekend) REVIEW all Homework problems since the last test Any HW questions you have had trouble with you should be ready to ask me for help with during the review session. 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. |
|
| Tue 12/5 |
|
TEST 4 Covers
all material on which HW was assigned above. Up through
section 5.5 NOTE: DIFFERENT
MAKEUP POLICY for this test: Read
Carefully:
Note: you may use a notesheet for
the Final, but NOT for this Test 4!. |
|
| Thu 12/7 |
Return
Test 4 [Any Makeup for Test 4 must have
been started in the Testing Center by 11am, the time class
starts.] 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 a GRAPHING CALCULATOR, too |
||
| Thu 12/12 |
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
NOTE-SHEET FOR THIS TEST.
(OR SMALLER.) Starts on Time. If you are late, you get
no extra time. If you have an emergency, call me IMMEDIATELY at 847-635-1935 and leave a detailed message with the nature of the emergency, YOUR PHONE NUMBER, and when you can make up the test. Also Email me the same information. Emergency situations will still need to be made up soon, with a late penalty. ------------------------------------------------------------------------------- Final Grades will be available ONLINE through the MyOakton BANNER system shortly after they are due, which is Tues., Dec. 19. You can check starting either Wed. the 20th, or Thurs. the 21st, to see if the College has processed them. I am NOT allowed to give out grades by email or phone, which are not secure. If you wish to see your final exam, email me after Dec. 21st and set up a time to come in and look at it. |
||
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