MATH 275 001 - Review Notes for Test #3 -- 11/12/04

This is the final form of this list. Last-minute additions in red.

The test will cover the material of Assignments #14 - #25.

Things to Know:

1. The Chain-Rules shtik.
2. The meaning of "explicit".
3. Simple limits of functions of several variables, and how their existence may fail owing to direction-of-approach variations.
4. Clairaut's Theorem, what it says.
5. Computation of arc length as an integral of speed.
6. Integration of f(x,y) or f(x,y,z) along a path with respect to arc length.
   • Appearance of a Riemann-Sum Term for such an integral.
   • Evaluating such integrals by means of a parametrization.
   • Importance of the explicit speed subcontract in evaluating such integrals via a parametrization. Other integrals have fudge factors analogous to speed.
7. Match up contour-map features with graph features.
8. Match up contour-map features with partial derivatives.
9. For functions of form f(x,y) = Ax + By + C:
   1. Contour map.
   2. Partial derivatives.
   3. Direction in xy-plane for maximum slope.
   4. Direction in xy-plane for maximum rate of change.
   5. Maximum slope of graph, computation thereof.
   6. Slope of the graph in a given direction.
10. Partial derivatives, computation thereof. It's MATH-170 review.
11. Partial derivatives, slope interpretation thereof.
12. Parametrization of the surface which is the graph of z = f(x,y). In class we used the vector-valued function R(x,y) = [x     y     f(x,y) ]^T, which right-thinking folks write as a column matrix.
13. Vectors tangent to the graph of z = f(x,y) arise from evaluating R_x and R_y at the point in question.
14. The Fundamental Cross Product (FCP).
    • This is the cross product of R_x and R_y.
    • Evaluating the FCP at a point (a,b) in the domain of f(x,y) produces a normal vector to the plane tangent to the graph of f(x,y) at the point of tangency (a, b, f(a,b)). This enables us to write a formula for this tangent plane in function form:

      TPF(x,y) = A(x - a) + B(y - b) + f(a,b),

      where A = f_x(a,b) and B = f_y(a,b).
    • Later we will mine the FCP for other goodies.
15. The Gradient Vector for f(x,y):
    • How grad(f)(a,b) is inherited from the tangent-plane function at (a,b).
    • The significance of the direction of grad(f)(a,b) in R².
    • The significance of the length of grad(f)(a,b) in R².
    • The relation of the direction of grad(f)(a,b) to the contour f(x,y) = f(a,b).
    • The direction of grad(f) when grad(f) is evaluated at a point on the contour. f_x(x,y) = 0.
    • The direction of grad(f) when grad(f) is evaluated at a point on the contour. f_y(x,y) = 0.
    • The direction of grad(f) when grad(f) is evaluated at a point on the contour. f(x,y) = 0 (or on any other contour of f).
    • The derivative of f in the direction of vector A (unitize).
    • Be able to find all the critical points of z = f(x,y).
    • Be able to put together a rough gradient-direction plot on which you can spot and classify critical points, that is, know how the gradient direction arrows behave around
      • maxima
      • minima
      • saddles
      • critical points which are "none of the above"
16. The second derivative of f in the direction of vector A.
    • The Hessian formula for the second directional derivative of f in the direction of vector A.
    • How MATH-108 completing the square shows the importance of the Hessian's determinant.
    • The Second Derivative Test for maxima and minima of functions of form z = f(x,y).
17. Double Integrals:
    • Riemann-sum terms
    • Upper and Lower Darboux Sums for simple cases.
    • How to use Fubini's Theorem to evaluate a double integral over
      • a rectangle
      • other regions
      by means of iterated integrals.
    • Reversing the order of integration.
18. Writing a double integral as an iterated integral in polar coordinates. This entails a "fudge factor" analogous to the speed in a ds line-interal evaluation.

Algebra, Trigonometry, and Calculus Facts
Updated: Wed Nov 10 11:38:17 MST 2004

Be sure you know arctan'.

The Grand Duchesses of Calculus want you to know the following sorts of things off the tops of your heads:

• Trig Facts.
• Derivatives and Antiderivatives.
• More Derivatives and Antiderivatives.

Test #3 will call on both trigonometry and calculus.

Grading Notes on Assignment #14
Last Update: Sun Oct 10 11:29:48 MDT 2004

• 125 points possible for this assignment:

  Problem	Points
  14.3: 6	10
  14.3: 8	20
  14.3: 8'	10
  14.3: 20	20
  14.3: 22	20
  14.3: 24	20
  14.3: 66	25

• The average score was about 88.
• 14.3: 6: You've got to use the right point on the surface.
• 14.3: 8: There's enough information available to dope out a rise and a run.
• 14.3: 20 Check QRFC and BUD.
• 14.3: 22 needs the MATH-170 chain rule.
• 14.3: 24 needs the MATH-170 Fundamental Theorem of Calculus, Part 1, page 396.

Grading Notes on Assignment #15
Last Update: Sun Nov 7 17:51:55 MST 2004

• 70 points possible for this assignment:

  Problem	Points
  14.2: 10	10
  14.2: 20	10
  14.2: 38	10
  14.2: 39	10
  14.3: 32	20
  14.3: 60	10

• The average score was about 58.
• 14.2: 38 needs l'Hôpital's Rule.

Grading Notes on Assignment #16
Last Update: Sun Nov 7 17:52:05 MST 2004

• 75 points possible for this assignment:

  Problem	Points
  16.2: 2	25
  16.2: 6	25
  16.2: 10	25

• The average score was about 52.
• It was hard to get folks to write out an explicit simplified-speed subcontract.
• 16.2: 6 needed to have its dx changed to a ds so that the answer comes out as

  ( (e² + 1)^3/2 - 2^3/2 )/3

• Know how to parametrize paths:
  • straight lines
  • circular arcs
• The substitution method came back from MATH-170. The "new limits" POV needs proliferation (as we do with writing double integrals in polar coordinates).

Grading Notes on Assignment #17
Last Update: Fri Sep 3 08:49:14 MDT 2004

• 105 points possible for this assignment:

  Problem	Points
  1	55
  2	30
  3	10
  4	10

• The average score was about 94.
• The dope on matrix multiplication.
• Problems 1(j) and (k): Matrix multiplication is Noncommutative. This means that the order of the factors is important.
• Problem 2: for questions on the angle between two vectors, the cosine is better.
• Click here for an alleged answer sheet.

Grading Notes on Assignment #19
Last Update: Thu Nov 4 16:48:22 MST 2004

• 95 points possible for this assignment:

  Problem	Points
  14.4: 24	10
  14.4: 28	10
  14.4: 30	10
  14.5: 2	10
  14.5: 9	10
  14.5: 10	15
  14.5: 14	15
  14.5: 16	15

• The average score was about 81.
• 14.4: 30: it says to go from A = (3, -1) to B = (2.96, -0.95). This means that delta z is given by

  f(B) - f(A) = f(new) - f(old) = f(to) - f(from) = f(2.96, -0.95) - f(3, -1)

  while dz is given by

  dz = f_x(A) dx + f_y(A) dy
     = f_x(old) (run in x from old to new) + f_y(old) (run in y from old to new)
     = f_x(3,-1) (-0.04) + f_y(3,-1) (+0.05)

Grading Notes on Assignment #20
Last Update: Sun Nov 7 18:15:54 MST 2004

• 90 points possible for this assignment:

  Problem	Points
  14.7: 3	30
  14.7: 7	30
  14.7: 9	30

• The average score was about 61.
• It's important to get the gradient-direction arrows on the partial-derivative zero contours.
• It's important to get the gradient-direction arrows in the "counties" fenced off by the partial-derivative zero contours.

Grading Notes on Assignment #21
Last Update: Sun Nov 7 18:18:12 MST 2004

• 75 points possible for this assignment:

  Problem	Points
  14.6: 11	25
  14.6: 12	25
  14.6: 16	25

• The average score was about 61.
• 14.7: 11: Directional derivative = 23/10
  Second directional derivative = -219/400
• 14.7: 12: Directional derivative = 0
  Second directional derivative = 2/5
• 14.7: 16: Directional derivative = -9*sqrt(14)/28
  Second directional derivative = 45/28
• The concavity and growth of the graph in the given direction.

Grading Notes on Assignment #22
Last Update: Thu Nov 4 16:46:58 MST 2004

• 230 points possible for this assignment:

  Problem	Points
  14.7: 6	25
  14.7: 8	25
  14.7: 10	40
  14.7: 12	40
  14.7: 14	40
  14.7: 18	60

• The average score was about 151.
• To ensure getting your due credit, list the critical points explicitly, like maybe in a nice display box.
• Preserve factoring as much as you can.
• Locating critical points is aided in all these problems by making a rough graph showing the two curves f _x = 0 and f _y = 0 .
• ¿ Trailing Hessian?   Several papers did not use the Hessian formalism in solving the problems. Rather, they showed the Hessian as an afterthought. This is too bad, because several times, evaluating the Hessian matrix at a critical point obviates actually computing the determinant of the Hessian in order to find the sign of the determinant of the Hessian: if the main-diagonal entries have a zero product, then the determinant is "minus" the square of the off-diagonal entries common value...
• Preserve factoring as much as you can.
• 14.7: 18: In this problem it was easier get all the info on all the critical points from a gradient-direction plot than from the second-derivative test.

Grading Notes on Assignment #23
Last Update: Sun Nov 7 17:11:27 MST 2004

• 80 points possible for this assignment:

  Problem	Points
  15.1: 2	20
  15.1: 4	20
  15.1: 12	10
  15.2: 4	15
  15.2: 10	15

• The average score was about 65.
• For 15.1: 2 Upper and Lower Darboux sums for the partition of the region caused by slicing horizontally with y = 1 and vertically with x = 1. This gives four rectangles of height 1 and base 2, so "delta A" = 2. In the following table, M is the maximum value of the integrand in the rectangle, and m denotes the minimum:

  M = 4 m = -1	M = 2 m = -17
  M = 1 m = -2	M = -1 m = -18

  This makes for

  Upper Darboux Sum = 12
  Lower Darboux Sum = -76

• For 15.1: 4 Upper and Lower Darboux sums for the partition of the region caused by slicing horizontally with y = 2 and vertically with x = 1. This gives four rectangles of height 2 and base 1, so "delta A" = 2. In the following table, M is the maximum value of the integrand in the rectangle, and m denotes the minimum:

  M = 33 m = 8	M = 34 m = 9
  M = 9 m = 0	M = 10 m = 1

  This makes for

  Upper Darboux Sum = 172
  Lower Darboux Sum = 36

Grading Notes on Assignment #24
Posted: Tue Nov 9 06:55:17 MST 2004
Last Update: Tue Nov 9 08:56:06 MST 2004

• 105 points possible for this assignment:

  Problem	Points
  15.3: 8	15
  15.3: 12	15
  15.3: 18	15
  15.3: 24	15
  15.3: 26	15
  15.3: 38	10
  15.3: 40	10
  15.3: 42	10

• The average score was about 89.
• 15.3: 8: Use dy dx.
• 15.3: 12: Use dx dy.
• 15.3: 18: If you goof and miscopy the point as (3,0), then it's best to get a dx dy integral for which the value is 5.
• 15.3: 24: This seems to be ambiguous. The description seems to apply to two different regions. The class unanimously used the one bordered by the y-axis. Why not the one bordered by the x-axis?
• 15.3: 38-42: The limits on the answer's left-most integral sign must NOT contain any x or y, the dummy variables of integration!!

Grading Notes on Assignment #25
Updated: Wed Nov 10 11:40:46 MST 2004

• 90 points possible for this assignment:

  Problem	Points
  14.7: 16	30
  15.4: 2	10
  15.4: 4	10
  15.4: 6	10
  15.4: 10	15
  15.4: 12	15

• The average score was about 70.
• 14.7: 16: Five critical points. One way to sniff them all out is to break down and use a GRAPH of f_x = 0 (in red) and f_y = 0 (in blue) simultaneously to help locate critical points. In this problem, the graph is really easy.
• 15.4: 10: Symmetry eliminates y from the integrand.
• 15.4: 12: Polar-coordinate adepts recognize x² + y² as r² immediately.