final 2005 sol

# final 2005 sol - HKUST MATH 102 Final Examination...

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Unformatted text preview: HKUST MATH 102 Final Examination Multivariable Calculus Answer ALL 8 questions Time allowed – 3 hours Directions – This is a closed book examination. No talking or whispering are allowed. Work must be shown to receive points. An answer alone is not enough. Please write neatly. Answers which are illegible for the grader cannot be given credit. Note that you can work on both sides of the paper and do not detach pages from this exam packet or unstaple the packet. Student Name: Student Number: Tutorial Session: Question No. Marks 1 /12 2 /12 3 /12 4 /14 5 /12 6 /12 7 /14 8 /12 Total /100 Problem 1 (12 points) Your Score: (a) If f ( x, y ) = ( x 3 + y 2 ) 1 3 , find f x (0 , 0) and f y (0 , 0). (b) Let z = f ( x, y ), where x = g ( t ) and y = h ( t ). (i) Show that d dt ∂z ∂x = ∂ 2 z ∂x 2 dx dt + ∂ 2 z ∂y∂x dy dt and d dt ∂z ∂y = ∂ 2 z ∂x ∂y dx dt + ∂ 2 z ∂y 2 dy dt . (ii) Use the formulas in part (i) to help find a formula for d 2 z dt 2 . Solution: (a) f x (0 , 0) = lim h → f (0 + h, 0)- f (0 , 0) h = lim h → ( h 3 + 0) 1 3- h = lim h → h h = 1 . f y (0 , 0) = lim h → f (0 , 0 + h )- f (0 , 0) h = lim h → (0 + h 2 ) 1 3- h = lim h → h- 1 / 3 → ∞ . Hence, f y (0 , 0) does not exist.-0.4-0.2 0.2 0.4 x-0.4-0.2 0.2 0.4 y-0.5 0.5 z-0.4-0.2 0.2 0.4 y (b) (i) Since z = f ( x, y ) ⇒ z x or z y are also functions of x and y . Therefore, by Chain rule: d dt ∂z ∂x = ∂ ∂x ( z x ) dx dt + ∂ ∂y ( z x ) dy dt = ∂ 2 z ∂x 2 dx dt + ∂ 2 z ∂y∂x dy dt , and d dt ∂z ∂y = ∂ ∂x ( z y ) dx dt + ∂ ∂y ( z y ) dy dt = ∂ 2 z ∂x∂y dx dt + ∂ 2 z ∂y 2 dy dt . x y t t x z, z , z y (ii) dz dt = ∂z ∂x dx dt + ∂z ∂y dy dt d 2 z dt 2 = ∂z ∂x d 2 x dt 2 + dx dt d dt ∂z ∂x + ∂z ∂y d 2 y dt 2 + dy dt d dt ∂z ∂y = ∂z ∂x d 2 x dt 2 + ∂ 2 z ∂x 2 dx dt 2 + 2 ∂ 2 z ∂y∂x dx dt dy dt + ∂z ∂y d 2 y dt 2 + ∂ 2 z ∂y 2 dy dt 2 . – 1 – Problem 2 (12 points) Your Score: (a) Find the equation of the tangent plane at the point (- 1 , 1 , 0) to the surface x 2- 2 y 2 + z 3 =- e- z . (b) Find the absolute extrema of the function z = f ( x, y ) = xy- 5 3 x- 3 y on the closed and bounded set R , where R is the triangular region with vertices (- 1 , 0), (- 1 , 4), and (5 , 0). Solution: (a) Let F ( x, y, z ) = x 2- 2 y 2 + z 3 + e- z = 0. This is a level surface and ∇ F = (2 x,- 4 y, 3 z 2- e- z ) . At (- 1 , 1 , 0), ∇ F = (- 2 ,- 4 ,- 1). This vector is normal to the level surface at the point (- 1 , 1 , 0)....
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## This note was uploaded on 03/13/2012 for the course MATH 102 taught by Professor Jimmyfung during the Spring '11 term at HKUST.

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final 2005 sol - HKUST MATH 102 Final Examination...

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