Exam_2_20C_Solution

# Exam_2_20C_Solution - Math 20C Summer 2010 Exam 2 Solution...

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Unformatted text preview: Math 20C. Summer, 2010. Exam 2 Solution. 1. The radius of the circle is √ 4 = 2 meters. • Since the object is moving counterclockwise, κ = 1 2 meter − 1 • T = < − 1 , > • N = < , − 1 > • a = s ′′ T + κ ( s ′ ) 2 N • < 5 , − 8 > = s ′′ < − 1 , > + 1 2 ( s ′ ) 2 < , − 1 > = < s ′′ , − 1 2 ( s ′ ) 2 > • s ′′ = − 5 • 1 2 ( s ′ ) 2 = 8 • ( s ′ ) 2 = 16 • s ′ = 4. The object is traveling 4 meters per second and is slowing down 5 meters per second 2 2. f ( x,y ) = 2 x − y + 6 • f = c ⇐⇒ 2 x − y + 6 = c ⇐⇒ y = 2 x + 6 − c • f = 4 on y = 2 x + 2 • f = 6 on y = 2 x • f = 8 on y = 2 x − 2 • f x = ∂ ∂x (2 x − y +6) = 2 • f y = ∂ ∂y (2 x − y +6) = − 1 • ∇ f = ( 2 , − 1 ) • Figure A3 4 6 8 x 2 y 1 ∇ f Figure A3 3. V = πr 2 h • V r = 2 πrh • V h = πr 2 • V (3 , 2) = π (3 2 )(2) = 18 π • V r (3 , 2) = 2 π (3)(2) = 12 π • V h (3 , 2) = π (3 2 ) = 9 π • Tangent plane:...
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## This note was uploaded on 08/31/2011 for the course MATH 20C taught by Professor Helton during the Spring '08 term at UCSD.

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Exam_2_20C_Solution - Math 20C Summer 2010 Exam 2 Solution...

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