260S12Ex2solns

# This already implies that α 0 however if α 0 then w

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This already implies that α 0. However, if α = 0 then w ( x ) 2 = 0 so w ( x ) = constant. But w (0) = 0. Thus w ( x ) 0. This gives the trivial solution u ( x, t ) 0 which we discard. Consequently α < 0 so we write α = - λ 2 . Thus w ( x )+ λ 2 w ( x ) = 0 whose general solution is w ( x ) = A cos λx + B sin λx . The boundary condition w (0) = 0 implies A = 0 while the boundary condition w ( π ) = 0 implies B sin λπ = 0. We exclude the possibility that B = 0 since this gives us the trivial solution u ( x, t ) 0. Consequently, sin λπ = 0, so λ = k = 1 , 2 , . . . and α = - k 2 so the solution of dT/dt = αT is T ( t ) = Ce - k 2 t . Collecting our results we have the special solutions u k ( x, t ) = C k sin( kx ) e - k 2 t , k = 1 , 2 , . . . . B–4. Say the equation f ( X ) := f ( x, y, z ) = 0 implicitly defines a smooth surface in R 3 (an example is the sphere x 2 + y 2 + z 2 - 4 = 0). Let P R 3 be a point not on this surface. Assume Q is a point on the surface that is closest to P . Show that the vector from P to Q is orthogonal to the tangent plane to the surface at Q . [ Suggestion: Let X ( t ) be a smooth curve in the surface with X (0) = Q . Then Q is the point on the curve that is closest to P .] Solution: Using the curve X ( t ), let h ( t ) := X ( t ) - P 2 . Since h ( t ) is minimized at t = 0, then h (0) = 0. Now h ( t ) = X ( t ) - P, X ( t ) - P so h ( t ) = X ( t ) , X ( t ) - P + X ( t ) - P, X ( t ) = 2 X ( t ) , X ( t ) - P . At t = 0 this gives 0 = X (0) , Q - P , 4

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that is, the vector Q - P is perpendicular to the vector X (0) that is tangent to the surface at Q . Since ths is true for any tangent vector at Q , the vector Q - P is perpendiculat to the whole tangent plane at Q .
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