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Unformatted text preview: Bush 18.02A pset 1, part II solutions, fall 2009 Problem 1 The frog heads at an angle upstream. The resultant velocity w is straight across the river, i.e. perpendicular to the flow. speed =  w  = u 2 v 2 . Angle = sin 1 ( v/u ). This requires u &gt; v (otherwise the frog is pushed downstream.) Problem 2 Tetrahedron is ABCD . M 1 is the midpoint of AD and M 2 is the midpoint of BC . M 3 is the midpoint of AB and M 4 is the midpoint of CD . M 5 is the midpoint of AC and M 6 is the midpoint of BD . We need to show: midpoint of M 1 M 2 = midpoint of M 3 M 4 = midpoint of M 5 M 6 . Let P = midpoint of M 1 M 2 . As usual we will write vectors OP , OA as P , A , etc. Our goal is to write the vector P completely in terms of A , B , C , D . (We will see that P = 1 4 ( A + B + C + D ).) We have AP = AM 1 + M 1 P = 1 2 AD + 1 2 M 1 M 2 . Now, M 1 M 2 = AM 2 AM 1 = AB + 1 2 BC 1 2 AD . Substituting this in the expression for P gives AP = 1 2 AD + 1 2 ( AB + 1 2 BC 1 2 AD ) . P A = 1 2 ( D A ) + 1 2 ( B A ) + 1 4 ( C B ) 1 4 ( D A )....
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This note was uploaded on 05/14/2010 for the course 18.02A 18.02A taught by Professor Johnbush during the Winter '10 term at MIT.
 Winter '10
 JohnBush

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