math53-fa09-mt2-Frenkel-soln

# math53-fa09-mt2-Frenkel-soln - Solutions to the Second...

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Unformatted text preview: Solutions to the Second Midterm Exam Multivariable Calculus Math 53, November 5, 2009. Instructor: E. Frenkel 1. The height of a mountain above the point ( x, y ) is given by the formula h ( x, y ) = x 2 + sin 2 ( xy ). A climber is standing on the mountain at the point (1 , , 1). (a) In the direction of what unit vector should the climber move (with respect to the x, y coordinate plane) so as to achieve the steepest ascent? We find: h x = 2 x + 2 sin( xy ) cos( xy ) y, h y = 2 sin( xy ) cos( xy ) x . Hence the gradient at the point (1 , 0) is vector h (1 , 0) = ( 2 , ) . The steepest ascent rate is achieved in the direction of the unit vector ( 1 , ) . (b) In the direction of what unit vectors should the climber move to achieve the ascent rate equal to 50% of the steepest ascent rate? The ascent rate in the direction of a unit vector u equals u vector h (1 , 0) = | vector h (1 , 0) | cos , where is the angle between u and vector h (1 , 0). In order to achieve the ascent rate equal to 50% of the steepest ascent rate the climber should move in a direction that has the angle such that cos = 1 / 2. Therefore = / 3, and we get u = ( 1 / 2 , 3 / 2 )...
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## math53-fa09-mt2-Frenkel-soln - Solutions to the Second...

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