Oakton Community College
MAT 250   Calculus I
Section 005:   TR  3 - 5:15 pm   Room 2806   DP Campus
Paul Boisvert     Professor of Mathematics 
FALL  Semester  2009
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Homework Assignments
Note:  Below, exponents are written with the wedge ^ notation.  So "x raised to the third power" would be "x^3". 
Fractions are written with slashes and parentheses:  A fraction with a + b in the top and c - d in the bottom is written as  ( a + b )  /  ( c - d ) .
Square Roots are written as "sqt(x)", and higher roots as fractional exponents, i.e.,  cube root of x would be  "x^(1/3)"
Date Page.  Problems Read Sections
Tue
8/25		 No Homework.    Read sections B.1, B.2 in the book.  Also, locate and review everything you covered in PreCalculus--old HW, Reviews, Tests, notes, etc.  You will need to know all that stuff as you do Calculus problems--I will review
some of it quickly, especially Trig, but you may need to do much more review on your own.                                                                                                                                                                                                         .		 B.1, B.2
Thu
8/27		 p. AP-29  4-6 ALL   Also, write the answers you get as INTERVALS.  That is, if your answer is x < 3,  write it as  (-oo , 3)
12, 13    Also, write the answers you get as INTERVALS.

Answers:  ODD:  5)  (-oo, -1/3]       13)  [0, 10]
                   EVEN:  4)  (-oo, 5/4 ]    6)  (-oo, -6/7 )     12)  (1 , 11/3 )		 B.1
p. AP-38 1,
5-17 ALL    ,  19, 20,
21-23 ALL,   #3,
27-29 ALL, 37-40 ALL
Odd Answers will be given out in class.
EVEN ANSWERS:  
6)   slope = 0 (Horizontal line)      Perpendicular slope would be for a vertical line, which is undefined slope  (or + oo )    Or, since perp. lines have opposite-signed reciprocal slopes, perp. slope
      to "0" is  "-1/0", but  1/0 is considered to be infinity.  The + or - sign is because we don't really know whether 0 itself is pos. or neg. (since it is neither.)
8)    Vertical line:   x = 0  ,  i.e. the y-axis       Horizontal line:  y = -sqrt(2)
10)  y - 4  =  (-1/5)(x - 3       12)  y = -9       14)  y + 1 = (-2/5) (x - 5)
16)   y-int.  (0,3)     x-int. (4,0)
22)  (x + 1)^2 + (y - 5)^2  =  10   Using sqrt(10) = approx. 3.2, we get it going thru (2.2, 5) , (-4.2, 5) , (-1 , 8.2) , (-1, 1.8) 
        y-intercept is where x = 0, substitute and solve (0 + 1)^2 + (y - 5)^2  =  10  to get :
        (y - 5)^2 = 9, take plus or minus sqrt of both sides, get  y - 5 = +3  or y - 5 = -3 , so y = 8 or y = 2
        So y-intercepts are (0, 8) and (0, 2)
        For x-intercepts, we can see from the graph that the circle never goes low enough to cross the x-axis, so there are 
             no x-intercepts. 
        Or, for x-intercepts, you set y = 0 and follow a similar procedure to what we did for y-int.'s , to get :
          (x + 1)^2 + 25 = 10, solving to get  (x + 1)^2  = -15, which is impossible, 
          since a square cannot = a negative number.  So this also shows that there are no x-intercepts.

28)  y-int:  (0,0)     x-int's:  (0,0)  and   (4,0)    Vertex: (2,4)   Points Downward
38)  Intersections where x + 1 = x^2   , set = 0, Quad. Form. gives... x = 1.62, x = -0.62
        Substitute to get  y = 2.62   and y  = .38     Final Answers:  (1.62, 2.62) and (-0.62, 0.38)
40)  Since both equations can be solved for y^2, set those two equal, cancel the 1's on both sides,
        you get:  x^2 = (x - 1)^2.  Expand and solve, etc...   Answers:  (0.5, 0.87) and (0.5, -0.87)		 B.2
    Read B. 3, and Look at Table on page AP-43 and start memorizing all radians to degrees measures for special angles.  B.3 
 Tue
9/1		 p. AP-46  1, 2, 3,
5--By Hand, special angles, not using Calculator
6, 7, 8
Board Problems:   For each problem, find 3 other angles with the same reference angle -- as close to 0 as possible.
  A)  3pi/4     B)  -pi/6     C )  505 degrees    D)  234 degrees
ANSWERS:
Board Problems:   A)  pi/4, -pi/4, and (-3pi/4 or 5pi/4)    B)  pi/6, 5pi/6,  (7pi/6 or -5pi/6)    
C)  All angles in this answer are in degrees.  Coterminal is 505-360 =  145, which has ref angle  180 - 145 = 35.   So:  { 35, -35, and (215 or -145) }
D)  All angles in this answer are in degrees:     234 - 180 =  54,   so  { 54, 126, and -54 }
 Book Problems:   2)  Angle = 5pi/4 rad, or 225 Degrees                       6)  cos x  = -4/5  ,  tan x  =  -3/4                     8)  sin x  =  -1 / sqrt(5)  ,  cos x  =  -2/ sqrt(5)		 B.3 
Thu
9/3








































   Homework Sheet given out in class.  Answers below, and will be handed out in class..
    NOTE:  The answers below were ALSO given out in HARD-COPY form in Class on Thursday Sep. 10th.  So you don't need to print them out unless you feel like it.
    NOTE:  #5 Below has an INCORRECT ANSWER on the Hard-Copy given out in Class.  CORRECTED answer is given below.

ANSWERS:   P. AP-46 #10)   Period =  2,   Key points at 1/2, 1, and 3/2 .  Second Period out to 4.   Amplitude = 1, curve starts at top (0,1) and heads downward. Both tails head downward from top.
Board Problems:  1)  Before Vertical Shift:   Period =  2,   Key points at 1/2, 1, and 3/2 .  Second Period out to 4.   Amplitude = 3,  curve starts at (0, 0 ) and heads downward .
                                 After Vertical  Shift of +5:   makes  new top and bottom values 8 and 2, with center y-values at 5.

2)  Before Vertical Shift:   Period =  pi/2,   Key points at pi/8, pi/4, 3pi/8 .  Second Period out to pi.   Amplitude = 5,  curve starts at Top at (0, 5) and heads downward .
                                 After Vertical  Shift of -3:   makes  new top and bottom values 2 and -8, with center y-values at -3.

3)   Before Vertical Shift:   Period =  4pi,   Key points at pi, 2pi, 3pi .  Second Period out to 8pi.   Amplitude = 7,  curve starts at (0, 0) and heads upward.
                                 After Vertical  Shift of -4:   makes  new top and bottom values 3 and -11, with center y-values at -4

4)  Before Vertical Shift:   Period =  1,   Key points at 1/4, 1/2, 3/4 .  Second Period out to 2.   Amplitude = 4,  curve starts at Bottom at (0, -4) and heads upward.
                                 After Vertical  Shift of +6:   makes  new top and bottom values 10 and 2, with center y-values at 6.
5) INCORRECT:  sin (23.7) = coterm.  = sin( 4.86 rad.) in IV.   Ref angle alpha = 1.5676       So   +  sin(1.5676) =   -.9999  (neg. because in IV.)
     CORRECTED:  sin (23.7) = coterm.  = sin( 4.86 rad.) in IV.   Ref angle alpha = 6.28 - 4.86  =  1.42      So   +  sin(1.42) =  -.9887  (neg. because in IV.)
6)  All in DEGREES:  cos(-740)  = coterm.   =  cos (-20)  in IV     Ref. angle alpha = +20,    So  + cos(20) =  +.9397   (pos. because in IV)
7)  Tan (31pi/6) = coterm. = tan (7pi/6) in III.    Ref angle alpha = pi/6        So  +  tan (pi/6)  =  + 1/sqrt(3)  (pos. because in III)
8) Cot(-15pi/4) = coterm. = cot (pi/4) in I, so alpha is same, all positive, answer is cot(pi/4)  =  1.
9)  Sec (-9) = coterm. = sec(3.56) in III.    Ref angle alpha = 0.42     So +  1/cos(.42)  =  - 1.095  (neg. because in III)
10)  csc(8) = coterm. = 1.72 in II.    Ref angle alpha = 1.4216.    So +  1/sin(1.4216)  =  + 1.011  (pos. because in II.)
11) All in DEGREES:    sec(3783) = coterm. subtract 10(360)  = sec 183 in III.    Ref angle alpha = 3      So +  1/cos(3) =  -1.0014  (neg. because in III)
12) I will use "x" for "theta" here.   Sec x = 9/4 = hyp/adj, so hyp = 9, adj = 4, find... opp = sqrt(65).    Draw picture with angle "x"  (NOT "2x") of this right triangle.
       Sin (2x)  = 2 sin x  cos x   =  From above, or from picture:    =  2 (sqrt(65) /9) (4/9)  =  8sqrt(65) / 81
       Cos(2x)  =  (cos x)^2   -  (sin x)^2   =  (4/9)^2  - (sqrt(65)/9 )^2   =  -49/81.       So angle "2x" is in II (Why?)
       Tan (2x) =  sin(2x) / cos(2x)   = ...  =  -8sqrt(65) /49
13)   As in #12, with less detail:   hyp = sqrt 146,     cos(2x)  =  -96/146  =  -48/73          sin(2x)  =  reduced = 55/73
         Csc(2x) = 1/sin(2x)  =  73/55
14)   cos (pi/12) =  cos ( half of pi/6)  =  sqrt (  [1 + cos pi/6 ] /  2 )  =  sqrt [ ( 1 +  sqrt(3) ) / (2) ]   =  simplify by mult. top and denom inside root by LCD of 2  =  sqrt(2 + sqrt(3) ) / 2
         sin (pi/12) =     etc...                                                                  =   sqrt (  1 -  sqrt(3) )  /  sqrt (2)     =  etc.  =   sqrt(2 - sqrt(3) ) / 2
        cot (pi/12)  =   cos ( )  / sin ( )   = ...   sqrt [  (2 + sqrt 3) / (2 - sqrt 3) ]
15)  Similarly to #14:    sin ( half of pi/4)  =  ... = sqrt (2 - sqrt (2) ) / 2                 cos (half of pi/4)  =  sqrt (2 + sqrt (2) ) / 2
        tan (pi/8)  =  ...  =  sqrt [ (2 - sqrt(2) ) / (2 + sqrt(2) ) ]
16.  Sin x = 4/7    so First angle x1 = sin-inverse (4/7)  =  .6082 rad        Sine is also pos. in II       So find angle in II with ref angle of .6082
                                 2nd angle  x2  in II =  pi - .6082  =  2.5344
          Answer:  { .6082, 2.5344, + 2kpi }
17.     x1 = cos-inverse(-3/8)  =  1.9552 in II.     Cosine is also negative in III, so find angle in III with same ref. angle as 1.9552.
           Ref angle for 1.9552 is  pi - 1.9552  =  1.1864          Angle in III with ref. angle of 1.1864 is  pi + 1.1864  =   4.328 = x2
           Answer:  { 1.9552 ,  4.328  +  2k pi }
18.    x1 = tan-inverse (5)  =  1.373           Instead of finding a second angle and then adding 2kpi's, for Tangent we can simply add 1pi to the first angle, which automatically finds the second angle.  
         Then adding more 1pi's to both of those gives us coterminal angles.  But since all this simply involves adding 1kpi's to the first angle, that's the shortcut answer:
         Answer:  [1.373 +  k pi }
19.  csc x  =  -4    Take reciprocal of both sides:   sin x   =  -1/4     Now solve sin x = -1/4 like in #16.   Etc...    x1 = -.2527,        x2 in III =  pi + .2527 = 3.394
        Answer:  {  -.2527 ,  3.394 + 2k pi }
20.  Sec x = 2   All Answers here in DEGREES:   Take reciprocal of both sides:   cos  x   = 1/2    Now solve cos x = 1/2 using Special Angles (so why not Degrees!)
                          If cos x = 1/2,  x = 60 Deg.     [Or cos-inverse(1/2) = 60 deg.]    Cosine is also pos. in IV, using ref. angle of 60 deg., easy way is to use  - 60 deg.  in IV.
        Answer:  { 60 deg.,  -60 deg.   + 2k pi }

Page 230:  2)   A)  -pi/4    B)  pi/3    C)   -pi/6                    4)  A)  pi/6    B)  -pi/4     C) pi/3
6)  A) 2pi/3            B)  pi/4           C)   5pi/6                        8)  A)  pi/4           B) 5pi/6            C)  pi/3
10)  A) -pi/4           B) pi/3            C)  -pi/6                       12)  A)  pi/4           B) 5pi/6             C)  pi/3
14)  tan (alpha)  =  4/3  =  opp/adj  , so hyp =sqrt( 3^2 + 4^2)  =  5.     So sin = 4/5,  cos = 3/5,   cot = 3/4,   sec = 5/3, csc = 5/4
16)  sec(alpha) = -sqrt(13) /2 =  hyp/adj.  [so alpha is in II, since sec-inverse comes out in II for negative values, just like cos-inverse].  
       Draw picture of triangle,  use Pyth. Thm.,  get  opp =  3.    So, using knowledge that we are in II for plus/minus signs:  
        sin = 3/sqrt(13)  ,    cos = -2/sqrt(13)  ,   tan = -3/2  ,   cot = -2/3,   csc = sqrt(13) / 3
82)  Give csc-inverse (1/2) a name of an angle: "x".   So csc-inverse (1/2) = x.   Therefore, csc (x) = 1/2.   Take Reciprocal:  sin(x) = 2/1  = 2.   But this is impossible:
        -1 < sin x < +1    That is, sin x can never be bigger than 1, so it certainly can't equal  2.    So original "x" is actually undefined!
          NOTE:   since sine and cosine are trapped between -1 and 1, i.e. their absolute value is always LESS than one, and since csc and sec are their reciprocals,
          and since the recip. of a number less than one is bigger than one, and vice versa, we get  that
                    the Absolute Value of sec and csc are always GREATER than 1.
84) Similarly to #82, you get that cot x = -1/2 ... but that certainly is possible!   Tan and Cot can range from 0 to infinity, and from 0 to negative infinity.  So any values for
       Tan-inverse and Cot-inverse are possible, and are never undefined.   [If you took reciprocals, get tan x = -2, calc. will tell you an answer for Tan-inverse (-2). ]		 B.3
Tue
9/8		  p. 45 1,  11-20 ALL
Answers:  12)   9^1/2 = sqrt(9) = 3     14)  3^(3/3)  =  3^1  =  3     16)  13^1  =  13
18)  sqrt(6)       20)  2/3		  1.5
   p. 59 15, 18, 19, 21, 22, 24, 25, 26, 27, 29, 31, 45     [30, 32 are OPTIONAL]
Answers:    NOTE:   ln (1/x) = ln 1 - ln x =  -ln(x)   [since ln(1) = 0] 
18)  f-inverse(x)  =  x^3/2            22)  f-inverse(x)   =   2x + 7          24)  f-inverse(x)   =   x^-1/3    or     1 / [cube root of (x) ]
26) a.   ln(1/5^3)  =  ln(1)  -  3ln(5)  ... Since ln(1) = 0 ,  Answer is -3ln(5)
       b.  9.8 = 98/10 = 49/5 = 7^2 / 5    So answer is ...  2ln(7) - ln 5
        c.  ln(7) + ln (7^ 1/2 )  = 1 ln(7) + (1/2) ln(7)  =  (3/2) ln(7)
        d.  1225 = 49 times 25 , so answer is ....  2ln(7) + 2ln(5)
        e.  .056 =  56/1000   = 7/125 ,  so answer is ... ln(7) - 3ln(5)
        f.   [ ln(5) + ln(7) - ln(7) ] / 2 ln(5)   =  ln(5) / 2 ln(5)   =  1/2.
30)  a.  x^2 + y^2      b.  1/0.3  =  10/3  =  3.33333        c.  (pi x)/2
32)  a.  sec(theta)      b.  e^x       c.  2 ln(x)  		1.6 
Thu
9/10		 p. 59  33-37 ALL
39-43 ODD     [40, 42 are OPTIONAL]
51, 53, 55
Answers:
34)  Exponentiate both sides, get e^(ln y)  =  e^(-t + 5) , or:   y = e^(-t) e^5  , since added exponents come from multiplication of the original (bases raised to powers.)  And e^(-t)  =  1/(e^t), so final answer is
    y  =  e^5 / e^t
36)  Exponentiate both sides, get e^(ln [1 - 2y])  =  e^t  ,   or:  1 - 2y = e^t, solve for y,  get:  y  =  (1 - e^t) / 2
40)   a.  -ln(4) / 5      b.  -ln 80   c.  1 
42)  a.  -100 ln(1000) =  -690.77 
       b.   kt = ln(1/10)  = ln(1) - ln(10)  =  0 - ln(10),  so: t =  -ln(10) / k
       c.  (ln 2) t  =  ln (1/2) , so  t = ln(1/2) / ln(2)  =  [ ln(1) - ln(2) ] / ln(2)  =  [0 - ln(2)] / ln(2) = 
             = - ln(2) / ln(2)   =   -1		  
  p. 83  5-17 odd, 19, 21, 25, 29, 31, 33, 30, 32  
Answers:
30)  16      32)  -2		 2.1, 2.2 
Tue
9/15
   REVIEW For Test 1       Bring your Book.
      BEFORE COMING TO THIS CLASS, you should have ALREADY Reviewed all old HW.  If you have not looked at and reviewed all old HW before coming in, you will WASTE a lot of time trying to do so during the Review Session.   If you HAVE already reviewed, you will be able to use the in-class review session to answer your (few, we hope) remaining questions, and fine-tune your knowledge well for the test. 		  
Thu
9/17 		   TEST 1    Covers all material on which HW was assigned, up through Section 2.2.   See above for assignments that should be studied and understood.
                   Closed Book, No Notes.

     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. 		  
Tue
9/22		   All HW assigned before Test 1 is DUE.  I will have a stapler there so you can staple it into one big packet.  Make sure your name is on at least the first page of it.    
   p. 106 1,    
2 parts a) through g)  only            
3, 4,     51-60 ALL
Answers:  2)  a)  T   b) T   c) F    d)  T   e)  T   f)  T   g)  F                    
4)  a)  Each one-sided limit is 1 ,  but f(2) = 2            b)  Yes, limit exists and equals 1, because both one-sided limits agree with each other (each = 1)  Note:  fact that f(2) isn't 1 is irrelevant to the limits.
c)  4 for both limits     d)  Yes, overall limit is 4, since both one-sided limits equal 4.   Note:  f(4) = 4, also, but this is irrelevant to the limit question.
52)  a) 2    b) 2       54)  a)  0+ (from above)     b)  0- (from below) 
56)  a)  0+ (from above)     b)  0- (from below)    58)  9/2  from either direction (y = 9/2 is the horizontal asymptote)   60)   -1 (y = -1 is horizontal asymptote) 		 2.4
  p. 117  1-21 odd     2.5
  p. 28  5, 7,
11 parts a, b, d only    		  1.3
Thu
9/24		    No HW yet.   READ Section 2.7  2.7
Tue
9/29		 p. 136 5, 7, 15, 17, 19, 21, 12, 23, 24   For these,  use the Lim h--> 0 definition, NOT the shortcut rules.
Answers:  12)  f ' (x) = 1 - 4x     
 24)  g'(x)  =  ...  =  3x^2 - 3  , Set = 0 , and solve:  3x^2 - 3 = 0 leads to x = 1 and x = -1.  This is where there are horizontal tangent lines with slope (deriv.) = to 0.		 2.7 
  p. 152  11, 13, 15   For these,  use the Lim h--> 0 definition, NOT the shortcut rules. 3.1 
Thu
10/1 		 p. 167 1-12 ALL  
17-28 ALL
29, 31, 35,      41 a) and c) only   ,   43
Answers (including 2nd derivatives for 1-12 All,  even though the 2nd deriv's were assigned to be done in 
Monday's  HW assignment.)
2)  y '   = 2x + 1    and     y'' = 2 
4)  dw/dz  =  21 z^6 - 21 z^2 + 42 z      and     d2w/dz2  =  126 z^5 - 42 z + 42
6)  dy/dx  =  x^2 + x + 1/4    and     y'' = 2x + 1
8)  ds/dt  = 2t^(-2) - 8t^(-3)       and      d2s/dt2  =  -4t^(-3) + 24 t^(-4) 
10)  y'  =  -2 + 3x^(-4)      and   y''  =  -12x^(-5)
12)  I will use "w"  instead of "theta" for the variable:
       dr/dw  =  12w^(-2) + 12 w^(-4) - 4w^(-5)     and   d2r/dw2  =  -24w^(-3) - 48 w^(-5) + 20w^(-6) 

18)  dz/dx  =  [ (x^2 - 1)(2) - (2x + 1)(2x) ] / [x^2 - 1]^2  =  simplify... = [-2x^2 - 2x - 2] / [x^2 - 1]^2  
20)  f'(t) = [ (t^2 + t - 2)(2t) - (t^2 - 1)(2t + 1) ] /  (t^2 + t - 2)^2   = simplify... 
       = simplified = [t^2 - 2t + 1] /  (t^2 + t - 2)^2 . 
        Note, now you can factor and cancel, and get a final answer  of  1 / (t + 2)^2
        Also note that you could/should have factored and cancelled original problem before taking deriv.  Then, deriv would have been a lot easier, and would still give final answer above.
22)  -17 / (2x - 7)^2
24)  y '  =  [(2e^x - x)(2x + 3e^x) - (x^2 + 3e^x)(2e^x - 1) / [2e^x - x]^2  = 
           could simplify numerator by multiplying out and cancelling a bit, but it won't simplify much, so skip it.
26)  I will use w for a variable here instead of theta:   dr/dw  =  -w^(-3/2) + 1/sqrt(w)
28) [  e^r - r(e^r) ] /  (e^r)^2  Note:  (e^r)^2  =  e^(2r)    So it all simplifies to:  (1 - r) / e^r  when you factor out e^r and cancel.

 3.2 
Tue
10/6 		 p. 177  1-5 odd (SKIP :"average velocity", answer other parts.)
7, 13 		 3.3 
  p. 186  1-25 odd [Even ones optional]
29, 31, 33, 35, 37 		 3.4 
Thu
10/8		 p. 199  1-59 [ending in 1,5, 9]     3.5 
Tue
10/13		    REVIEW For Test 2       Bring your Book.
      BEFORE COMING TO THIS CLASS, you should have ALREADY Reviewed all old HW since Test 1.  If you have not looked at and reviewed all old HW before coming in, you will WASTE a lot of time trying to do so during the Review Session.		  
Thu
10/15		    TEST 2    Covers all material on which HW was assigned, up through Section    See above for assignments that should be studied and understood.   Closed Book, No Notes.
 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. 		  
Tue
10/20		 p. 199  101-107 odd  Parametric Chain rule:   dy/dx  =  dy/dt divided by  dx/dt  3.5 
Thu
10/22 		   NO CLASS    
Tue
10/27		 p. 209 7, 9, 13, 15, 17  Note:  these are normal uses of the chain rule (review from the earlier chain rule section.  They are NOT implicit derivatives.
19-35 odd   These ARE implicit derivatives, using the implicit chain rule, which says that d/dx of a "y" function = the normal y-derivative times dy/dx.  
37 - 43 odd    2nd derivatives by the implicit process.		 3.6
Thu
10/29		 p. 221 11-39 odd, 41-51 odd, 55-61 odd, 3.7
Tue
11/3		 p. 221 67-85, 89-95 3.7
p. 230 1-11 odd, 17-27 odd, 29-39 odd, 49-69 odd 3.8
Thu
11/5		 p. 251 19-37 odd
Board Problems:  A) Estimate  5th root of 34  =  (34)^1/5    [ Hint:  Use f(x) = x^1/5.   Nearby number with easy known 5th root is of course x = 32, since 5th root of 32 = 2. ]
                            B)  Estimate  square root of 63      [ Hint:  Use f(x) = x^1/2.   Nearby number with easy known square root is  x = ? ]
                            C)  Estimate  cube root of 998      [ Hint:  Use f(x) = ??   Nearby number with easy known value is  x = ??  ]

ANSWERS:   A)  dy = (1/5)x^(-4/5) dx.     Deriv. evaluated at x = 32  equals  1/80.   dx = 2 ( = 34-32)      So  y - 2 = (1/80) (2) , or   y = approximately 2 1/40  as a mixed number.
                       B)  dy =    ?  ?             dx     Deriv. evaluated at x = ?  equals   1/16.     dx = -1                     So  y - 8 = (1/16) (-1)   or  y = approximately 7 15/16 as a mixed number.
                       C)  dy =    ?  ?             dx     Deriv. evaluated at x = ?  equals   1/300   dx = -2                     So  y - 10 = (1/300) (-2)   or  y = approximately 9 149/150 as a mixed number.		 3.10
p. 273 p. 273    15-31 [ending in 1, 5, 9]  but ones ending in 3 and 7 are strongly encouraged as optional ones if you have any trouble with the regular ones. 4.1
Tue
11/10		 REVIEW For Test 3       Bring your Book.    MAKE SURE YOU KNOW ALL DERIVATIVEs, including all you learned for Test 2.  All Derivative Rules are always on every test!
     
BEFORE COMING TO THIS CLASS, you should have ALREADY Reviewed all old HW since Test 2.  If you have not looked at and reviewed all old HW before coming in, you will WASTE a lot of time trying to do so during the Review Session.

NOTE:  THERE WERE SEVERAL CORRECTIONS made in class to the REVIEW ASSIGNMENT today.  Here they are for those who missed them:
 
On the QUESTION Sheet:  #3)   Problem should have said find values in ...   [-1/2, 2]    ( not +1/2, 2)
                             
                                      #4)  A)  Problem should have said find dy/dx   (not dy/dt)
                                      #4)  B)  Problem should have said use t = 1    (not x = 1)

   **************************

On ANSWER SHEET:  5D)  Answer should be (-2xy^3 - 2x^4 ) / y^5     [ not (-2xy^3 + 2x^4 ) / y^5 ]
                                    5E)  Because of the mistake in 5D), Answer to 5E) should have been -36           (not 28)

AND, Answers to 3) and 4) were missing from the Answer sheet.  Here they are:

ANSWER to 3):  Local maxes or mins at (0,0) and (1, -1).       [Note, the solution to dy/dx = 0 at x = -1 is NOT in the interval for the problem, so ignore it.]
                          Endpoints at (-1/2, -7/16) and (2, 8)
                         
                           SO:  Absolute max at (2,8).  
                                  Absolute min at (1, -1)   Since this is also a local min/max, it must be a local MIN.
                                   (0,0) is not an absolute anything, but since the other local extreme value is a local MIN, (0,0) is probably a local MAX.  But this info not required on the test.

ANSWER to 4)  A)  dy/dx  =  -1/ [ 2t(1 + t^2) ]                     B)  y -  pi/4  =  (-1/4) (x + 2
  
Thu
11/12
TEST 3    Covers all material on which HW was assigned, up through Section 4.1   See above for assignments that should be studied and understood.   Closed Book, No Notes.
 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. 
Tue
11/17		 HW due
p. 299 9, 11, 13 4.3, 4.4
Thu
11/19
       
       
       
       
       
       
       
       
       
       
       
       
       
       
       
       
       
       
   
   
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