assignment4_sol

# assignment4_sol - MATH1211/10-11(2)/Asol4/TNK Department of...

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Unformatted text preview: MATH1211/10-11(2)/Asol4/TNK Department of Mathematics, The University of Hong Kong MATH1211 (2010-11, Second semester) Suggested solution to Assignment 4 Note : Some solutions given below are outline only. You may need to give more details in your solution. The parts that are typed in red color are remarks for you to read; they are not a part of the solution. 1. (a) For any f : R 3 → R , we have ∇ · ∇ f = ∇ · ∂f ∂x 1 , ∂f ∂x 2 , ∂f ∂x 3 = ∂ 2 f ∂x 2 1 + ∂ 2 f ∂x 2 2 + ∂ 2 f ∂x 2 3 = ∇ 2 f. So it makes sense to think of ∇ 2 as ∇ · ∇ . (b) ∇ 2 ( fg ) = 3 X i =1 ∂ 2 ∂x 2 i ( fg ) = 3 X i =1 ∂ ∂x i f ∂g ∂x i + g ∂f ∂x i = 3 X i =1 ∂f ∂x i ∂g ∂x i + f ∂ 2 g ∂x 2 i + ∂g ∂x i ∂f ∂x i + g ∂ 2 f ∂x 2 i = ∇ f · ∇ g + f ∇ 2 g + ∇ g · ∇ f + g ∇ f = f ∇ 2 g + 2 ∇ f · ∇ g + g ∇ 2 f. Alternatively, we make use of part (a) and the following identity established in tutorial 5 Problem 5: ∇ · ( f F ) = f ∇ · F + F · ∇ f. ( † ) Since ∇ ( fg ) = f ∇ g + g ∇ f (do you know how to prove this?) , we have ∇ 2 ( fg ) = ∇ · ∇ ( fg ) (by (a)) = ∇ · ( f ∇ g + g ∇ f ) = f ∇ · ∇ g + ∇ g · ∇ f + g ∇ · ∇ f + ∇ f · ∇ g (by ( † )) = f ∇ 2 g + 2 ∇ f · ∇ g + g ∇ 2 f. (c) Again by using the identity ( † ), we have ∇ · ( f ∇ g- g ∇ f ) = f ∇ · ∇ g + ∇ g · ∇ f- g ∇ · ∇ f- ∇ f · ∇ g (by ( † )) = f ∇ 2 g- g ∇ 2 f. (by (a)) 2. Let r and h denote the radius and the height of the cylinder, with r = 2 , h = 3, measured in inches. (a) Volume of cylinder is given by V ( r,h ) = πr 2 h . Hence V r ( r,h ) = 2 πrh , V h ( r,h ) = πr 2 , and V r (2 , 3) = 12 π , V h (2 , 3) = 4 π . With dr = ± . 1 , dh = ± . 05, the approximate error in the calculated volume is dV = V r (2 , 3) dr + V h (2 , 3) dh = (12 π )( ± . 1) + (4 π )( ± . 05) , which has maximum value ± 1 . 4 π (in 3 ). (b) Surface area of the cylinder is give by A ( r,h ) = 2 πrh + 2 πr 2 . Hence A r ( r,h ) = 2 πh + 4 πr, A h ( r,h ) = 2 πr , and A r (2 , 3) = 14 π, A h (2 , 3) = 4 π . With dr = ± . 1 , dh = ± . 05, the approximate error in the calculated surface area is dA = A r (2 , 3) dr + A h (2 , 3) dh = (14 π )( ± . 1) + (4 π )( ± . 05) , which has maximum value...
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## This note was uploaded on 05/04/2011 for the course MATH 1211 taught by Professor Wang during the Spring '11 term at HKU.

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assignment4_sol - MATH1211/10-11(2)/Asol4/TNK Department of...

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