tutorial4_sol

tutorial4_sol - ´ ´ = x 2 z i + (1-2 xyz ) j + k . div...

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MATH1211/10-11(2)/Tu4sol/TNK Department of Mathematics, The University of Hong Kong MATH1211 Multivariable Calculus (2010-11 2nd semester) Solution to Tutorial 4 Note : Some solutions given below are outline only. You may need to give more details in your solution. 1. x ( t ) = ( 3 t 2 , 2(2 t + 1) 3 / 2 ) , therefore x 0 ( t ) = (6 t, 6(2 t + 1) 1 / 2 ). Length of the path is L = Z 2 0 p (6 t ) 2 + 36(2 t + 1) dt = Z 2 0 p 36 t 2 + 72 t + 36 dt = 6 Z 2 0 | t + 1 | dt = 6 Z 2 0 ( t + 1) dt = 6 h t 2 2 + t i 2 0 = 24 . 2. x 0 ( t ) = ( t sin t, t cos t, 0); k x 0 ( t ) k = | t | = t (because t 0). Hence, when t > 0, T ( t ) = x 0 ( t ) k x 0 ( t ) k = (sin t, cos t, 0) , N = d T /dt k d T /dt k = (cos t, - sin t, 0) k (cos t, - sin t, 0) k = (cos t, - sin t, 0) , B = T × N = (sin t, cos t, 0) × (cos t, - sin t, 0) = (0 , 0 , - 1) . Note that ds dt = k x 0 ( t ) k = t . Hence curvature is κ ( t ) = ± ± ± ± d T ds ± ± ± ± = k d T /dt k ds/dt = k (cos t, - sin t, 0) k t = 1 t . Also d B ds = (0 , 0 , 0). Since d B ds = - τ N , the torsion τ is 0. 3. We need to verify that x 0 ( t ) = F ( x ( t )). In fact, since F ( x,y,z ) = ( z, - y, - x ), we have F ( x ( t )) = F (sin t,e - t , cos t ) = (cos t, - e - t , - sin t ) . On the other hand x 0 ( t ) = ² d dt sin t, d dt e - t , d dt cos t ³ = (cos t, - e - t , - sin t ) . Hence x 0 ( t ) = F ( x ( t )). 4. ∇ × F = ´ ´ ´ ´ ´ ´ ´ ´ i j k ∂x ∂y ∂z z ( x + y 2 ) x 2 yz ´ ´ ´ ´ ´ ´
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Unformatted text preview: ´ ´ = x 2 z i + (1-2 xyz ) j + k . div (curl F ) = div ( x 2 z i + (1-2 xyz ) j + k ) = ∂ ∂x x 2 z + ∂ ∂y (1-2 xyz ) + ∂ ∂z 1 = 2 xz-2 xz = 0 . Remark : Since F is in C 2 , by Theorem 4.4 on p.218 we know that div (curl F ) must be zero. The purpose of the verification here is to help students understand the computations involved in the expression div (curl F ). 5. Write F = ( F 1 ,F 2 ,...,F n ). Then ∇ · ( f F ) = ∇ · ( fF 1 ,fF 2 ,...,fF n ) = n X i =1 ∂ ( fF i ) ∂x i = n X i =1 ² ∂f ∂x i F i + f ∂F i ∂x i ³ = n X i =1 ² ∂f ∂x i F i ³ + n X i =1 ² f ∂F i ∂x i ³ = ( ∇ f ) · F + f ( ∇ · F ) = F · ∇ f + f ∇ · F . 1...
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