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MIT18_024S11_soln-pset8

MIT18_024S11_soln-pset8 - Solutions for PSet 8 1(10.5:11...

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Solutions for PSet 8 1. (10.5:11) Parameterize the sides of the square C by maps s i : [0 , 1] R 2 by s 1 ( t ) = (1 t, t ); s 2 ( t ) = ( t, 1 t ); s 3 ( t ) = ( t 1 , t ); s 4 ( t ) = ( t, t 1) . With this parametrization: 1 1 1 1 d x + d y 1 + 1 1 1 1 1 1 + 1 = d t + d t + d t + d t | x | + | y | (1 t ) + t t + (1 t ) (1 t ) + t t + (1 t ) C 0 0 0 0 The first and the third summands are 0, and the second and fourth terms cancel each other, giving: d x + d y = 0 | x | + | y | C 2. (10.9:6) Writing the equation of the cylinder in complete square form: 2 a a ( x ) 2 + y 2 = 2 4 Thus looking from high above the xy -plane the picture looks like: The parametrization of the cylinders’ intersection with the xy -plane is: a a a s ( t ) = cos t + , sin t, 0 2 2 2 1
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We need to lift it up to sit on the sphere: ) = a a a s ( t cos t + , sin t, z ( t ) 2 2 2 where z ( t ) 0 and , a a 2 a 2 a a cos t + + sin t + z ( t ) 2 = a cos t + + z ( t ) 2 = a 2 2 2 2 2 2 This means, that a z ( t ) = 1 2 cos t Now ( y 2 , z 2 , x 2 ) · d( x, y, z ) C 2 π a 3 sin t = sin 2 t, 2(1
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