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final_review_key

# final_review_key - 9 = f(9 0 =-61 2 Problem 9(a x = 4 y = 4...

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MATH 2433 FINAL EXAM REVIEW KEY Problem 1. (a) ( i ) Center (1 , 1 , - 1), r = 17 ( ii ) ( x - 1) 2 + ( y - 1) 2 + ( z + 1) 2 = 17 (b) ( i ) - 32 ( ii ) 1 14 (3 , 2 , - 1) or 3 14 i + 1 7 j - 1 14 k ( iii ) 2 3 14 ( iv ) - 2 3 i - 2 3 j + 1 3 k Problem 2. (a) perpendicular (b) 2( x + 2) - ( y - 7) - 2( z - 4) = 0 (c) x = - 2 - 9 t, y = 7 - 18 t, z = 4 Problem 3. (a) v ( t ) = - e - t i + e t j - 2 k (b) a ( t ) = e - t i + e t j (c) L = Z 0 p e - 2 t + e 2 t + 2 dt = . Problem 4. (a) ( i ) 1 6 (17) 3 / 2 - 1 6 ( ii ) κ = 1 5 5 (b) ( i ) T = 2 3 cos (2 t ) i + 2 3 sin (2 t ) j + 5 3 k , N = - sin (2 t ) i + cos (2 t ) j ( ii ) κ = 4 9 ( iii ) a T = 0 , a N = 4 1

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Problem 5. (a) f xx = 1 x and f yx = - 1 x + 2 y . (b) - 9 5 (c) ∂f ∂t = ln t 2 + 1 + 4 s 2 e 2 t se t + ste t + - 1 2 t + 4 s 2 te 2 t 2 se t (d) 5( x - 2) + 7( y - 2) - ( z - 8) = 0 Problem 6. Let Problem 7. (a) ( i ) (0 , - 2) , (2 , 1) , ( - 2 , 1) ( ii ) (0 , - 2) local max , (2 , 1) saddle point , ( - 2 , 1) saddle point. (b) - 71 Problem 8. (a) absolute maximum f ( - 1 , 3) = f ( - 1 , - 3) = 11 and absolute minimum f (1 , 0) = 1 (b) absolute maximum f (1 , 1) = 4 and absolute minimum values
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Unformatted text preview: , 9) = f (9 , 0) =-61 2 Problem 9. (a) x = 4 , y = 4 , z = 2 (b) maximum ± 1 2 , 1 2 , ± √ 2 2 ² → 50 o and minimum ± 1 2 ,-1 2 , ± √ 2 2 ² , ±-1 2 , 1 2 , ± √ 2 2 ² → -50 o Problem 10. (a) 1 2 sin 16 (b) ( i ) Z 2 Z 2 x x 2 dy dx ( ii ) Z 4 Z √ y y/ 2 dxdy (c) Z 2 Z √ 4-x 2 (4-x 2-y 2 ) dy dx = Z π/ 4 Z 2 (4-r 2 ) r dr dθ = π Problem 11. (a) 4096 35 (b) Z π/ 2 Z 1 Z √ 2-r 2 r 2 r dz dr dθ (c) Z 2 π Z π/ 2 π/ 4 Z 1 ρ 2 sin φdρdφdθ Problem 12. (a) 1 4 (b)-1782 Problem 13. (a) 2 33 (b) 0 (c) 3 π 8 3...
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final_review_key - 9 = f(9 0 =-61 2 Problem 9(a x = 4 y = 4...

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