Math16A - Final Solutions Ex

# Math16A - Final Solutions Ex - Math 16A, Solutions to the...

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Unformatted text preview: Math 16A, Solutions to the Final Exam (1) The average rate of growth is ( W (5)- W (4)) / (5- 4) = 0 . 1 5 2- . 1 4 2 = 2 . 5- 1 . 6 = 0 . 9 grams per week. The instantenous rate of growth is the value of the derivative W ( t ) = 0 . 2 t at t = 4, so it is W (4) = 0 . 8 grams per week. (2) The functions whose first derivative equal 1 /x 2 are the antiderivatives- 1 /x + C 1 where C 1 is a constant. The antiderivatives of- 1 /x + C 1 are- ln( x )+ C 1 x + C 2 where C 2 is another constant. Hence the functions whose second derivative equals 1 /x 2 are precisely the functions f ( x ) =- ln( x ) + C 1 x + C 2 , where C 1 and C 2 are arbitrary constants. (3) We use the method of implicit differentiation, where y = y ( x ) is regarded as a function of x . Applying d/dx to both sides of the equation xy + y 3 = 14, we find d dx ( xy + y 3 ) = y + x dy dx + 3 y 2 dy dx = y + ( x + 3 y 2 ) dy dx = 0 , and hence dy/dx =- y/ ( x + 3 y 2 ). Substituting x = 3 , y = 2 into this expression, we find that the slope of the tangent line equals- 2 / (3 + 3...
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## This note was uploaded on 05/12/2010 for the course CHEM 3A taught by Professor Fretchet during the Fall '08 term at University of California, Berkeley.

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Math16A - Final Solutions Ex - Math 16A, Solutions to the...

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