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midterm 1 fall 2002 solutions
Course: MATH 21B, Fall 2002School: Harvard
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Harvard - MATH - 21B
Harvard - MATH - 21B
Name: _ Math 21b Midterm 1 Thursday, October 24th, 2002 Please circle your section: Tom Judson Katherine Visnjic (CA) MWF 9-10 Andy Engelward Jakub Topp (CA) MWF 10-11 Andy Engelward Erin Aylward (CA) MWF 11-12Question 1 2 3 4 5 6 7 TotalPoints 1
Harvard - MATH - 21B
Name: _ Math 21b Final Exam Tuesday, January 14th, 2002 Please circle your section: Tom Judson Katherine Visnjic (CA) MWF 9-10 Andy Engelward Jakub Topp (CA) MWF 10-11 Andy Engelward Erin Aylward (CA) MWF 11-12Question 1 2 3 4 5 6 7 8 9 10 TotalP
Harvard - MATH - 21B
St. Michael - MATH - 100
St. Michael - MATH - 100
Linear AlgebraJim Hefferon1 32 11 32 1x1 1 32 1x1 1 2 x1 3 16 82 16 82 1Notation R, R+ , Rn N C {. . . . . .} (a . b), [a . b] . V, W, U v, w 0, 0V B, D En = e1 , . . . , en , RepB (v) Pn Mnm [S] M N V W = h, g H, G t, s T,
Cornell - MATH - 2930
Math 293 Solutions to Problem Set 7. 2.3 #3. The equation x dx + t2 x = sin t is nonlinear because of the product x dx . Rewritdt dt ing it as dx = sin t - t2 we see that it is not separable because the right-hand side cannot dt x be factored as a pr
Cornell - MATH - 2930
Math 293 Solutions to Problem Set 8. Here are just the answers to the even-numbered problems. The answers to odd-numbered problems are in the book. More complete solutions will be posted after the problem sets are turned in. 4.7 #4. The general solut
Cornell - MATH - 2930
Math 293 Solutions to Problem Set 9.dy 1.3 #12. dx = y. The isoclines are obtained by setting dy dx equal to a constant c, the slope of the slope-indicator line segments all along the isocline. In the present case the isoclines are the horizontal li
Cornell - MATH - 2930
Math 293 Solutions to Problem Set 10. 10.2 #4. y + 9y = 0 has general solution y = c1 sin 3x + c2 cos 3x. Plugging in the condition y(0) = 0 gives c2 = 0, so y = c1 sin 3x. To apply the condition y () = -6 we compute y = 3c1 cos 3x, and then plugging
Cornell - MATH - 2930
Math 293 Solutions to Problem Set 11. 10.3 #1. Both terms of x3 + sin 2x are odd, so the whole function is odd. In other words, (-x)3 + sin(-2x) = -x3 - sin 2x = -(x3 + sin 2x). #2. sin2 (-x) = (- sin x)2 = (sin x)2 = sin2 x, so this function is even
Cornell - MATH - 2930
Math 293 Solutions to Problem Set 12. 10.5 #4. In general : The equation ut = kuxx with boundary conditions ux (0, t) = 0 = ux (L, t) and initial condition u(x, 0) = f (x) has solution2 2 2 a u(x, t) = 0 + an e-kn t/L cos(nx/L) 2 n=0where the a
Cornell - MATH - 293
Math 293 Practice Prelim #1 Formulas that may or may not be useful: x = sin cos y = sin sin z = cos dx dy dz = 2 sin d d d 1. Consider the integral2 0 0 4-x2Spring 2000xe2y dy dx . 4-y(a) Sketch the region of integration. (b) Evaluate
Cornell - MATH - 2930
Math 293 Practice Prelim #2 (The actual exam will probably be five problems rather than six.) 1. Find the solution of the initial value problem d2 x dx + 5x = 0 +2 2 dt dt with x(0) = 0 and dx (0) = 1. dtSpring 20002. Find the solution of the ini
Cornell - MATH - 2930
Math 293 Practice Prelim #3 (The actual exam will not be as long as this one.) Integration formulas which may be useful:Spring 2000cos ax sin ax sin ax cos ax x cos ax dx = x + + 2 a a a a2 cos ax sin ax cos ax x2 sin ax dx = -x2 + 2x 2 + 2 3 a a
Cornell - MATH - 2930
Math 293 1. In the integralSolutions to Prelim #11 1 1 0 y 1/3 x4 +11 3Spring 2000yy= x3dx dy the limits of integration say that(1,1)x goes from x = 0 to x = y , and then y goes from 0 to 1. Solving 1 x = y 3 for y gives y = x3 . So th
Cornell - MATH - 2930
Math 293 Prelim #3 Solutions 1. (a)Spring 2000- 3- 2-023(b)- 3- 2-023(c) bn = 1 1 cos(n+1)x + 0 sin x cos nx dx = 0 (sin(n + 1)x - sin(n - 1)x) dx = - n+1 cos(n-1)x 1 cos(n+1) 1 cos(n-1) 1 1 2 + n+1 + n-1 - n-1
Cornell - MATH - 2930
Math 293 Answers to Practice Prelim #1xe 1. 0 0 4-y dy dx . This is awkward to integrate in the given order, so switch the order to 4 4-y xe2y get 0 0 4-y dx dy. Then after the first inte2 4-x22ySpring 2000y 4 y = 4 - x2 2 xgration we have e8
Cornell - MATH - 2930
Math 293 Prelim #2 Solutions 1. The equation x xy -2 y2 x dy dx -1 -1 -2 -1Spring 2000dy dx+ y 2 = x 2 y 2 is separable since it can be rewritten as , from which we get yx2 2== y2x-xboth sides gives -y=- ln x + C ,3 2initial c
Cornell - MATH - 2930
Math 293, Spring 2000Instructor: Allen Hatcher Office: 553 Malott Hall Phone: 255-4091 Email: hatcher@math.cornell.edu Textbooks: Thomas & Finney, Calculus 9th edition, the same book as in Math 192. This will be the textbook for the first part of th
Cornell - MATH - 2930
PRELIM 1, MATH 293, SPRING 2006NUMBER OF STUDENTS: 284 MEAN: 17.5 MEDIAN: 18 STANDARD DEVIATION: 2.3706050number of students40302010091011121314 score1516171819201
SDSMT - MATH - 123