Explanation Given: q = f ( p ) = 10000 e − 0.25 p Calculation: Differentiate the given function with respect to p. f ( t ) = d d t ( 10000 e − 0.25 p ) We know that d d x ( e − a x ) = − a ⋅ e − a x Here, a = − 0.25 f ' ( p ) = 10000 ( − 0.25 ) e − 0.25 p f ' ( p ) = − 2500 e − 0.25 p Substitute p = 2 in f ( p ) f ( 2 ) = 10000 e − 0.25 ( 2 ) f ( 2 ) = 10000 e − 0

Applied Calculus, 6e WileyPLUS + Loose-leaf

6th Edition

ISBN: 9781119408901

Author: Deborah Hughes-Hallett, Andrew M. Gleason, Patti Frazer Lock, Daniel E. Flath

Publisher: Wiley (WileyPLUS Products)

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Chapter 3.2, Problem 44P

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To find and explain f(2) and f'(2).

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Chapter 3.2, Problem 43P

Chapter 3.2, Problem 45P

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Question 2 Let F be a solenoidal vector field, suppose V × F = (-8xy + 12z², −9x² + 4y² + 9z², 6y²), and let (P,Q,R) = V²F(.725, —.283, 1.73). Then the value of sin(2P) + sin(3Q) + sin(4R) is -2.024 1.391 0.186 -0.994 -2.053 -0.647 -0.588 -1.851 1 pts

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Chapter 3 Solutions

Applied Calculus, 6e WileyPLUS + Loose-leaf

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To find and explain f ( 2 ) and f ' ( 2 ) .

Solution Summary: The author compares the values of f(2) and '(2)=-1500.

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To find and explain f(2) and f'(2).

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Chapter 3 Solutions