ps1_part_II_mysol

ps1_part_II_mysol - Bush 18.02A pset 1, part II solutions,...

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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 > 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.

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ps1_part_II_mysol - Bush 18.02A pset 1, part II solutions,...

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