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Unformatted text preview: MAT1332 additional practice problems The practice problems here are in addition to the material given in the suggested exercises, the assignments and the tests. The problems here do NOT represent a sample exam. 1. Find the volume of the solid obtained by rotating the area bounded by y = 4 x x 2 , y = 3, x = 1 and x = 3 about the xaxis. Sol. See the following Figure 1, 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 y=3 y=4xx 2 Figure 1: The illustrative graph of Question 1. denote A 1 ( x ) = 4 x x 2 and A 2 ( x ) = 3, we know that The volume of the solid is given by V = V 1 V 2 = Z 3 1 π ( A 1 ( x )) 2 dx Z 3 1 π ( A 2 ( x )) 2 dx = Z 3 1 π (4 x x 2 ) 2 dx Z 3 1 π · 3 2 dx = Z 3 1 π (16 x 2 8 x 3 + x 4 ) dx Z 3 1 9 πdx = π ( 16 3 x 3 2 x 4 + 1 5 x 5 ) fl fl fl 3 1 9 πx fl fl fl 3 1 = 136 15 π. 1 2. Consider the system x = √ 3 x + 2 y y = 11 x √ 3 y (a) Show that (0 , 0) is the only equilibrium. (b) Write down the Jacobian matrix. (c) Show that the eigenvalues are λ = ± 5. (d) For each eigenvalue, find the corresponding eigenvectors. Sol. (a) Set x = 0 and y = 0, we have √ 3 x + 2 y = 0 11 x √ 3 y = 0 Solving the above linear system for x and y , we obtain the unique solution x = 0 and y = 0. (b) The Jacobian matrix is given by J = • √ 3 2 11 √ 3 ‚ . (c) Note that tr( J ) = 0, and det( J ) = 25, so the eigenvalues are given by λ = tr( J ) 2 ± r (tr( J )) 2 4 det( J ) = ± √ 25 = ± 5 . (d) For λ 1 = 5, we have • √ 3 5 2  11 √ 3 5  ‚ ( √ 3 + 5) R 1→ • 22 2( √ 3 + 5)  11 √ 3 5  ‚ 1 2 R 1→ • 11 √ 3 + 5  11 √ 3 5  ‚ R 1 + R 2→ • 11 √ 3 + 5   ‚ Set x 2 as the free variable, i.e., x 2 = t , from the first row, we have 11 x 1 + ( √ 3 + 5) x 2 = 0 to solve for x 1 , then x 1 = √ 3+5 11 t . Thus, the corresponding eigenvector is v 1 = • √ 3+5 11 1 ‚ t, ( t 6 = 0) . For λ 2 = 5, we have • √ 3 + 5 2  11 √ 3 + 5  ‚ ( √ 3 5) R 1→ • 22 2( √ 3 5)  11 √ 3 + 5  ‚ 1 2 R 1→ • 11 √ 3 5  11 √ 3 + 5  ‚ R 1 + R 2→ • 11 √ 3 5   ‚ Set x 2 as the free variable, i.e., x 2 = t , from the first row, we have 11 x 1 + ( √ 3 5) x 2 = 0 to solve for x 1 , then x 1 = √ 3 5 11 t . Thus, the corresponding eigenvector is v 2 = • √ 3 5 11 1 ‚ t, ( t 6 = 0) . 2 3. Evaluate R 4 1 e √ x dx , Sol. Let u = √ x , then du = 1 2 1 √ x dx , i.e. dx = 2 udu . The integration limits are changed from x = 1 and x = 4 to u = 1 and u = 2. So by integration by parts, we have Z 4 1 e √ x dx = Z 2 1 e u · 2 udu = 2 Z 2 1 ude u = 2 ue u fl fl fl 2 1 2 Z 2 1 e u du = 2 ue u fl fl fl 2 1 2 e u fl fl fl 2 1 = 2 e 2 ....
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This note was uploaded on 03/19/2011 for the course MAT 1332 taught by Professor Munteanu during the Spring '07 term at University of Ottawa.
 Spring '07
 MUNTEANU
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