ps1_part_II_mysol

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

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Bush 18.02A pset 1, part II solutions, fall 2009 Problem 1 The frog heads at an angle θ upstream. The resultant velocity w is straight across the river, i.e. perpendicular to the flow. speed = | w | = u 2 - v 2 . Angle θ = 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 ). P = 1 4 ( A + B + C + D ) (as promised).

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