HW 8 Solutions

HW 8 Solutions - HW #8 Solutions Math E-21a Fall 2009...

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1 HW #8 Solutions – Math E-21a – Fall 2009 Problem 1: Problem 2: Problem 3: Problem 4:
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2 Problem 5: Problem 6 : [ Note : The figure shown is a rather poor one. Yours is probably a lot better.] Here’s an alternative : Use the parameterization from part (a) which gives z and a function of the parameter t , namely 5 4cos 3sin 8 z tt  . Then use ordinary Calculus I methods to find its maximum and minimum values, i.e. calculate the derivative, set it equal to 0, and solve for the critical point. This gives: 2 5( 4sin 3cos ) 0 (4cos 8) dz t t dt t t   .
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3 This can only be 0 when the numerator is 0 and this occurs where 4sin 3cos 0 tt  or, more simply, where sin 3 tan cos 4 t t t  . We can draw two triangles corresponding to the two possible angles where this can happen: or In the 1st case, we have
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This note was uploaded on 02/10/2011 for the course MATH E-21a taught by Professor - during the Fall '09 term at Harvard.

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HW 8 Solutions - HW #8 Solutions Math E-21a Fall 2009...

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