Calculus III HOMEWORK and SYLLABUS

Calculus III HOMEWORK and SYLLABUS - Calculus III HOMEWORK...

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Calculus III HOMEWORK and SYLLABUS , Fifth Edition, Professor Ehrlich's Section Unit I -- Lines, Planes and Space Curves Page 797 -- 3 - 6, 12, 13, 17, 20, 23 - 34, 40 [Solution to No. 40 -- (x - 25/3) 2 + (y - 1) 2 + (z + 11/3) 2 = 332/9 ] Page 805 -- 1,5, 11, 16, 21, 25, 26, 35, 37, 39, 41, 43 Page 812 -- 1 - 3, 5, 7, 9, 13, 17, 19, 23, 26, 27, 31, 34, 35, 37, 39, 41, 43, 51, 54 - 59 Page 820 -- 1, 5, 7, 11, 14, 15, 17, 18, 19, 25, 29, 33, 39, 43, 45 Page 829 -- 3 - 5, 7, 9, 12, 13, 15, 19, 21, 22, 24, 25, 29, 33, 35, 37, 39, 41, 43 - 45, 47, 51, 53, 57, 58, 63, 65, 67, 69, 71 Page 837 -- 1, 3, 7, 9, 11 - 13, 16, 21 - 28, 41, 45 Page 843 -- 7, 9, 11, 15, 25, 29, 31 - 43, 45, 47, 49, 51, 55, 57, 61, 65 Page 855 -- 1, 3, 11, 12, 14, 19 - 25, 35 Page 861 -- 3, 11, 17, 19, 23, 31, 33, 39, 47, 49, 50 Page 868 -- 3, 11, 13, 17, 19, 24, 31, 39, 52 (just calculate the curvature, but NOT the torsion) Download Takehome Problem for Exam I, due on Friday, September 17th [or on Monday, September 20th if Hurricane Ivan effects us] at Take Home Problem for Exam I Comments on Exam I -- Exam I will be on Friday, September 24th in the usual classroom at our usual time, 8th period. I have posted one question to hand in on the web site, which is indicative of the sort of problem I give on my tests. You may use calculators in my section, and also make up a one page formula sheet yourselves. The bulk of the test will focus on Section 12.5, in which we see the elementary concepts of 12.1 - 12.4 bear fruit. One question will reflect the review problems from the Thursday Problem Session listed below. Please bring paper in which to show your work on the test questions. A good list of problems to review for Exam I would be -- Page 831 -- 65 Page 846 -- 19, 20, 21, 22, 23, 25 Also, review Problems # 12, # 17(c), # 23 in the Thursday Problem Session curricular materials.
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Commment on Problem 65,page 831 -- if you like to work problems on the distance from a point to a plane using formula (9) on page 828, one should note how the plane is given with the d on the left hand side in this setup (cf. equation (8)). Thus if we look at the given plane equation x - 2y -2z = 1, it must be reimaged as x - 2y - 2z -1 = 0. Hence the formula for the distance in this case becomes dist = |2 - 2(8) -2(5) -1|/sqrt(1 + 4 + 4) = |-25|/sqrt(9) = 25/3. Solution for No. 20 -- We are given the point P = (1,2,-2) and a line x = 2t, y = 3 - t, z = -1 + 2t. If P were to lie on this line L, then 2 = y = 3 - t so that t = 1; but then x = 2(1) = 2, which does not equal the x- coordinate of P. Hence, the point P does not lie on the line. Now one way, then, to work this problem is to determine two points Q,R on L and then we have three noncollinear points P,Q,R in the plane and can solve the problem by using the method of finding the plane through three noncolinear points. Letting t take on any two distinct values, we produce two different points on L. But it is convenient to set t = 0 in the equation for L to find Q = (0,3,1) and then set t = 1 to find R = (2,2,4). Hence we now calculate the vectors PQ = <-1,1,3> and PR = <1,0,6> and then less pleasantly, calculate a plane normal using the determinant formula n = PQ x PR = <6,9,-1>. Hence, using this vector n and arbitrarily selecting P (of the three points P,Q,R) for use in formula 7 on page 826, we obtain an equation for the desired plane 6(x - 1) + 9 (y - 2) - (z + 2) = 0.
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