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32A handout3 - MATH 32A/1 WINTER QUARTER 2009 HANDOUT 3...

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MATH 32A/1 WINTER QUARTER 2009 HANDOUT 3 CONTENTS HOMEWORKS #5, #6 THE TAYLOR EXPANSION APPENDIX TAYLOR EXPANSIONS IN TWO VARIABLES TAYLOR EXPANSIONS MAXIMA, MINIMA AND SADDLE POINTS APPENDIX MINIMA, MAXIMA AND SADDLE POINTS
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32A/1 - WINTER QUARTER 2009 HOMEWORK ASSIGNMENTS HOMEWORK #5 (due Friday February 20) PROBLEM 1 (p&p). Find ∂u ∂t , ∂u ∂x , 2 u ∂x 2 for the function u ( t, x ) = e x 2 / 4 t t . PROBLEM 2 (p&p). Find ∂u ∂t , ∂u ∂x , 2 u ∂x 2 , ∂u ∂y , 2 u ∂y 2 for the function u ( t, x, y ) = e ( x 2 + y 2 ) / 4 t t . PROBLEM 3 ( M ) . Do all derivatives in Problems 1 and 2 with MATHEMATICA. PROBLEM 4 (p&p). Find ∂u ∂x , 2 u ∂x 2 , ∂u ∂y , 2 u ∂y 2 for the functions u ( x, y ) = ( x 2 + y 2 ) a ( a an arbitrary constant) , u ( x, y ) = log( x 2 + y 2 ) . Hint: you can save work noticing that the last two derivatives are like the two first (interchanging x and y ) . PROBLEM 5 (p&p). Find ∂u ∂x 2 u ∂x 2 , ∂u ∂y 2 u ∂y 2 , ∂u ∂z 2 u ∂z 2 for the functions u ( x, y, z ) = ( x 2 + y 2 + z 2 ) a ( a an arbitrary constant) , u ( x, y, z ) = log( x 2 + y 2 + z 2 ) . Hint: You can save work here too; for instance the computations for the first two derivatives are just the same as the computations for the first two derivatives in Problem 5. Same for the other derivatives. PROBLEM 6 ( M ) . Do all derivatives in Problems 4 and 5 with MATHEMATICA. PROBLEM 7 (p&p). The volume V, the temperature T and the pressure P of a gas are related by Dieterici’s equation P ( V b ) e a/RV T = RT where a, b, R are physical constants. Show that if we use this equation to define V = V ( T, P ) as a function of T and P we have ∂V ∂P = RTV 2 ( V b ) P ( RTV 2 a ( V b )) . 1
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Simplifying differentials is dangerous. In one variable calculus, simplifying differentials (that is, considering df/dt as a quotient (which is not: it is the limit of a quotient) is usually harmless: for instance we can ”show” the chain rule for a function f ( x ( t )) as follows: df dx dx dt = df dt (simplify dx in numerator and denominator). This sort of thing produces spectacularly wrong results in mul- tivariable calculus. If we try to “show” the chain rule for a function f ( x ( u, v ) , y ( u, v )) simplifying differentials, ∂f ∂x ∂x ∂u + ∂f ∂y ∂y ∂u = ∂f ∂u + ∂f ∂u = 2 ∂f ∂u so we obtain the chain rule with an erroneous factor of 2. A more striking example is the next problem. PROBLEM 8 (p&p). The equation F ( x, y, z ) = 0 can be used to put any of the variables as function of the others: x = x ( y, z ) , y = y ( x, z ) , z = z ( x, y ) . Assuming that ( x, y, z ) is such that ∂F ( x, y, z ) ∂x = 0 , ∂F ( x, y, z ) ∂y = 0 , ∂F ( x, y, z ) ∂z = 0 , show ∂x ∂z ∂z ∂y ∂y ∂x = 1 (not 1 as if we simplify differentials). IMPLICIT FUNCTIONS Problems 1.5, 1.7 (both p&p), 1.6, 1.8 (both M ) . TAYLOR EXPANSIONS IN TWO VARIABLES, Problems 1, 2, 3, 4, 5 (all p&p). You can check your computations with MATHEMATICA as in TAYLOR EXPANSIONS.
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