Exam 2 Solution 2014 on Mathematical Methods of Economic Analysis

# Exam 2 Solution 2014 on Mathematical Methods of Economic Analysis

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Mathematical Economics Exam #2, November 4, 2013 1. Consider the function f ( t ) = t 2 - t . a ) Compute the tangent vector of f at any t . b ) Give an equation for the tangent line at the point 4 - 2 . Answer: a ) The tangent vector is given by the derivative d f dt = 2 t - 1 . b ) The point is f (2), so the tangent vector is 4 - 1 . There are several way to write the tangent line. One is that L = x y : x y = 4 - 2 + t 4 - 1 . It can also be written by eliminating t from the equations. For that, y = - 2 - t , so t = - (2 + y ). Substituting in x = 4 + 4 t yields - 4 = x + 4 y . 2. Let f : R 2 + R + be defined by f ( x, y ) = x 2 / 3 y 1 / 3 . a ) Find the level curves of f . b ) Use the Implicit Function Theorem to show that y ( x ) defined by f ( x, y ( x ) ) = q for q > 0 defines a C 1 function y . c ) Using y as in part (b), compute dy/dx . d ) What happens if q = 0 in part (b). In particular, can you still compute dy/dx ? Answer: a ) The level curves are { ( x, y ) R 2 + : x 2 / 3 y 1 / 3 = q } for q 0. b ) We examine whether f y = (1 / 3) x 2 / 3 y - 2 / 3 6 = 0 for x, y 6 = 0. This condition is satisfied since x 2 / 3 y 1 / 3 = q > 0. Because the partial derivative is non-zero, the

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