Calculus III

Dr. Mills' Math 251 (Engineering Calculus III) Web Page

Fall 2005

Syllabus
What Is Expected of You
Sample Graphs from MAPLE (MS Word Document)

NOTES [NEW]:

1. Section 16.8 and 16.9 Sample Problems [PDF]
2. MAPLE Graphs Corresponding to Sample Problems (16.8,16.9) [MSWord]

EXAM ANNOUNCEMENTS:

1. Exam 4 (12/8, Neckers 156, 7:30-8:30 pm): Chapter 16
2. Final Exam (12/15, Engr-A 408, 10:10 am - 12:10 pm): Cumulative

EXAM REVIEW GUIDES [NEW]:

1. Exam 4 -- The following text examples are pertinent.
   a. Sect. 2 -- 2, 6, 8
   b. Sect. 3 -- 3, 4, 5
   c. Sect. 4 -- 2, 3, 4
   d. Sect. 5 -- 3, 5
   e. Sect. 6 -- 7, 9, 10
   f. Sect. 7 -- 3, 4, 5
   g. Sect. 8 -- 1, 2
   h. Sect. 9 -- 1, 2

2. Final Exam -- The following text examples are pertinent.
   a. Chapter 12 [15 examples]
         i. Sect. 2 - 7
        ii. Sect. 3 - 3, 6
       iii. Sect. 4 - 3, 6
        iv. Sect. 5 - 3, 7, 10
         v. Sect. 6 - 4, 6, 8
        vi. Sect. 7 - 1, 2, 4, 7

   b. Chapter 13 [11]
         i. Sect. 1 - 2, 6
        ii. Sect. 2 - 1, 3, 6
       iii. Sect. 3 - 1, 4, 5
        iv. Sect. 4 - 3, 6, 7

   c. Chapter 14 [25]
         i. Sect. 1 - 4, 12
        ii. Sect. 2 - 3, 4, 9
       iii. Sect. 3 - 3, 4, 6, 8
        iv. Sect. 4 - 1, 2, 4
         v. Sect. 5 - 2, 5, 9
        vi. Sect. 6 - 4, 5, 6, 8
       vii. Sect. 7 - 3, 6, 7
      viii. Sect. 8 - 1, 2, 4

   d. Chapter 15 [18]
         i. Sect. 1 - 3
        ii. Sect. 2 - 3, 4
       iii. Sect. 3 - 3, 4
        iv. Sect. 4 - 1, 2
         v. Sect. 6 - 1, 2
        vi. Sect. 7 - 2, 3, 4
       vii. Sect. 8 - 2, 3, 4
      viii. Sect. 9 - 1, 2, 3

e. Chapter 16 [21]
See above.

HOMEWORK:

1. Section 13.3, Due 9/19: 5, 9, 13, 17, 23, 27, 39, 41
2. Section 13.4, Due 9/23: 7, 11, 15, 19, 23, 27, 31, 35
3. Section 14.1, Due 9/28: 3, 7, 13, 15, 17, 27, 29, 31, 35, 39, 43, 61
4. Section 14.2, Due 10/5: 3x (x=1,2,...,10)
5. Section 14.3, Due 10/7: 8, 19, 25, 31, 37, 41, 51, 63, 66
6. Section 14.4, Due 10/10: 5, 9, 15, 17, 22, 25, 31, 37
7. Section 14.5, Due 10/12: 3, 7, 11, 13, 19, 23, 27, 33, 40, 45, 47, 53a
8. Section 14.6, Due 10/14: 5, 9, 13, 17, 21, 25, 29, 33, 39, 43, 51
9. Section 14.7, Due 10/17: 5, 11, 17, 23, 27, 31, 37, 45
10. Section 14.8, Due 10/19: 5, 9, 13, 17, 19, 21, 23, 39
11. Section 15.1, Due 10/21: 3, 5, 9, 13, 17
12. Section 15.2, Due 10/24: 5, 11, 19, 21, 27, 33
13. Section 15.3, Due 11/2: 9, 17, 23, 27, 41, 47
14. Section 15.4, Due 11/4: 5, 11, 15, 19, 25, 31
15. Section 15.6, Due 11/7: 5, 8, 11, 17
16. Section 15.7, Also Due 11/7: 8, 13, 18, 25, 28, 48
17. Section 15.8, Due 11/9: 4, 9, 12, 21, 24, 29
18. Section 15.9, Due 11/14: 2, 6, 9, 10, 11, 14, 19
19. Section 16.1, Also Due 11/14: 4, 6, 11-14, 22, 26, 29-32
20. Section 16.2, Due 11/16: 5, 9, 13, 15, 19, 21, 39
21. Section 16.3, Due 11/18: 5, 10, 13, 17, 19, 31
22. Section 16.4, Due 11/28: 3, 7, 10, 12, 13, 19
23. Section 16.5, Also Due 11/28: 3, 7, 12, 17, 19, 25
24. Section 16.6, Due 12/2: 4, 8, 11-16, 20, 34, 38
25. Section 16.7, Also Due 12/2: 4, 8, 12, 16, 20, 24
26. Section 16.8, Due 12/5: 2, 4, 6, 8, 10 [NEW]
27. Section 16.9, Also Due 12/5: 4, 8, 12, 16 [NEW]

Note: Use the Maple package for those problems labeled with a graphing or CAS symbol.

EXTRA CREDIT:

1. Section 15.5, Due 11/18 (worth 15 points): 4, 8, 12, 16, 24, 26
2. Section 17.1, Due 12/15 (worth 25 points): 4, 8, 12, 16, 20, 24, 28, 32, 33, 34 [NEW]

Spring 2005

I am teaching third-semester calculus in the Spring of 2005. We will meet MWF from 8:00 to 8:50 in EGRA 208.

Answers to Quizzes 1-23:

Q1: The surface in question is a three-dimensional parabolically-shaped surface whose intersection with any plane of the form z=c, c some real number, is a parabola having equation y=4(x-3)^2. Two points on the plane that are of distance 5 from the origin are (3,0,4) and (3,0,-4).

Q2: Let OA represent the vector corresponding to the line segment running from point O to point A, etc. As OA + AC = OC and OB + BC = OC, and further as BC = -2AC, a little algebra shows us that 3AC = 3(OC - OA) = OB - OA, or OC = (2/3)OA + (1/3)OB, which is what we wish to show.

Q3: First, cos(Pi/6) = a/10, where v = <a,b,c> and |v| = 10, thus a = 5*sqrt(3). Further, cos(Pi/3) = b/10, thus b = 5. As a^2 + b^2 = 100 = |v|^2, c^2 = 0, and thus c = 0, so v = <5*sqrt(3),5,0>. For the bonus, with w = <s,t,u> and s = u = 5 (due to the conditions given), so that s^2 + u^2 = 50, it follows that t = 5*sqrt(2) or -5*sqrt(2), and thus the angle that w makes with the positive y-axis is Pi/4 or 3*Pi/4.

Q4: Vector PQ = <1,0,-3> while vector PR = <6,-2,1>, thus PR x PQ = <6,19,2> is a vector perpendicular to the plane defined by PQ and PR. The area of the triangle defined by these vectors is 0.5|PR x PQ| = 0.5*sqrt(401).

Q5: We pick a point on the plane x+2y-3z=1; the point (1,0,0) is convenient. Then, we use the formula given in the text to determine the distance between this point and the plane 3x+6y-9z-4=0. We have D=|3(1)+6(0)-9(0)-4|/sqrt(3^2 + 6^2 + (-9)^2)=1/(3*sqrt(14)).

Q6: For Problem 1, completing the square for both y and z yields the standard form x^2 + ((y-2)^2)/4 + (z-3)^2 = 1, an ellipsoid with center (0,2,3). Use the MAPLE package to obtain a graph of this surface. For Problem 2, we have r^2 + z^2 = x^2 + y^2 + z^2 = 49, thus the surface is a sphere centered at the origin, having radius 7.

Q7: The value t=3 is the value for which the x and y-components of the vector functions coincide. Thus, in order for the z-components to coincide, we must have cos(3*Pi*a)=1. This is true if a is an integer multiple of 2/3.

Q8: With r(t)=<cos(t),3t,2sin(2t)>, we have T(0)=(1/|r'(0)|)r'(0)=<0,3/5,4/5>, where r'(t)=<-sin(t),3,4cos(2t)>.

Q9: Problem 1 -- With r(t)=<t^2,5t,t^2-16t>, we have r'(t)=<2t,5,2t-16>, so that v(t)=|r'(t)|=sqrt(8t^2-64t+281). As dv/dt=(8t-32)/sqrt(8t^2-64t+281), t=4 is a critical point. Further, as dv/dt<0 for t<4 while dv/dt>0 for t>4, we conclude that the speed is a minimum at time t=4. Problem 2 -- Use of the given curvature formula yields k(x)=exp(x)/(1+exp(2x))^(3/2). By taking the derivative of this function and setting it equal to zero, we obtain the critical value x=ln(1/2)/2, and it is easy to see that the maximum curvature occurs for this value of x. Further, as x grows without bound, it is clear that k(x) goes to 0.

Q10: The domain of the function f(x,y)=sqrt(y-x^2)/(1-x^2) is the set of all points in the xy-plane lying on or above the parabola y=x^2, minus those points with x-coordinate either 1 or -1.

Q11: Problem 1 -- The limit of f(x,y)=2yx^2/(x^4+y^2) as (x,y) approaches (0,0) does not exist. To see why, note first that the limit of the function as one proceeds along the path for which y=0 (so that one walks along the path formed by the intersection of the curve with the xz-plane) is L1=0. However, as one traverses the path dictated by the intersection of the curves f(x,y) and y=x^2, one obtains a limit value L2=1. Thus, the limit does not exist. Problem 2 -- We observe that u_t=(ak)sin(kx)cos(akt) while u_x=k*cos(kx)sin(akt). Thus, u_tt=-[(ak)^2]sin(kx)sin(akt) while u_xx=[-k^2]sin(kx)sin(akt), and thus we conclude that u_tt=a^2*u_xx.

Q12: Problem 1 -- f_x=(1/2)f_y=1/(1+(x+2*y)^2), and as these functions are continuous in a region centered at (x,y)=(1,0), f(x,y) is differentiable at the point (1,0). As f(1,0)=p/4, the linearization is L(x,y)=(p-2)/4+(1/2)x+y. Problem 2 -- f_s=f_xx_s+f_yy_s so f_s=7(-1)+8(-5)=-47 when (s,t)=(1,2). Likewise, f_t=f_xx_t+f_yy_t so f_t=7(4)+8(10)=108 when (s,t)=(1,2).

Q13: First, note that grad(f)=(sin(x/y)/y)<-1,x/y>. Now, the vector u=<a,b> we want is a unit vector, so a^2+b^2=1. Further, as D_fu=af_x+bf_y, we ask whether there exist a, b such that -1=(1/6)*sin(p/6)(-a+(b*p/6)), or equivalently, whether there exist a, b such that a=12+(b*p/6), under the constraint a^2+b^2=1. The pertinent quadratic equation is Ab^2+Bb+C=0, where A=(36+p^2)/36, B=4p, and C=143. However, this quadratic equation has no real solutions, and thus the desired real-component unit vector does not exist.

Q14: f_x=(x+y)/x while f_y=1+ln(x). As f_x=f_y=0 when (x,y)=(1/e,-1/e), this point is a critical point. Further, f_xx=-y/x^2, f_yy=0, and f_xy=1/x so D=-1/x^2, which is always negative, and thus the critical point above corresponds to a saddle point.

Q15: The upper left corners of the rectangles are (-1,2), (0,2), (1,2), (2,2), (-1,1), (0,1), (1,1), and (2,1). As sin(p*y)=0 for all integers y, the only contribution comes from the -2*x^2 terms. The Riemann sum is S=-2(1+1+1+1+4+4)=-24.

Q16-Q23: Answer Key

Links:

1. Syllabus
2. Lecture Notes (Lecture notes for the entire course are now available. The latest update was made on 3/21/05.)
3. 3-D Graph Paper (please use this for homework problems that require a three-dimensional graph)

Most recently updated on 07 Dec 05 at 1:50 p.m.
Contact Dr. Mills at dmills@math.siu.edu.

Comments: Donald D. Mills