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Unformatted text preview: B U Department of Mathematics Math 101 Calculus I Fall 2004 Second Midterm Calculus archive is a property of Bo˘gazi¸ci University Mathematics Department. The purpose of this archive is to organise and centralise the distribution of the exam questions and their solutions. This archive is a nonprofit service and it must remain so. Do not let anyone sell and do not buy this archive, or any portion of it. Reproduction or distribution of this archive, or any portion of it, without nonprofit purpose may result in severe civil and criminal penalties. 1.) Find the point(s) on the ellipse x 2 +4 y 2 = 4 nearest the point (1 , 0). (Hint: Minimize the square of the distance.) Solution: d = p ( x 1) 2 + ( y 0) 2 . Let L be the square of the distance d . D = d 2 = ( x 1) 2 + y 2 where 2 ≤ x ≤ 2 Now, x 2 + 4 y 2 = 4 ⇒ y 2 = 1 x 2 4 D = ( x 1) 2 + 1 x 2 4 and dD dx = 2( x 1) 2 x 4 = 0 x 1 x 4 = 0 ⇒ x = 4 3 We must show that x = 4 3 is a min. point....
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This note was uploaded on 11/16/2011 for the course MATH 102 taught by Professor Soysal during the Winter '09 term at Boğaziçi University.
 Winter '09
 SOYSAL
 Calculus

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