Calculus Pre-lim 2 w. Solutions

# Calculus Pre-lim 2 w. Solutions - Math 111 Prelim 2 Name...

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Math 111 Prelim 2 Oct 25, 2007 Name: Instructor: Section: INSTRUCTIONS — READ THIS NOW This test has 7 problems on 9 pages worth a total of 100 points. Look over your test package right now . If you ﬁnd any missing pages or problems please ask a proctor for another test booklet. Write your name, your instructor’s name, and your section number right now . Show your work. To receive full credit, your answers must be neatly written and logically organized. If you need more space, write on the back side of the preceding sheet, but be sure to label your work clearly. This is a closed book exam. No calculators are allowed. You may bring a single 3x5 index card. This exam is 90 minutes long. Academic integrity is expected of all students of Cornell University at all times, whether in the presence or absence of members of the faculty. Understanding this, I declare I shall not give, use, or receive unauthorized aid in this examination. Signature of Student

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Math 111 (Fall 2007) Prelim 2 (Oct 25, 2007) 2 Question 1. (8 points) Solve the Initial Value Problem ± y ( x ) = 1 1+ x 2 y (1) = 0 . Since d dx tan - 1 ( x ) = 1 1+ x 2 , y ( x ) = tan - 1 ( x ) + C . To figure out the value of C , we use the condition y (1) = 0 : 0 = y (1) = tan - 1 (1) + C = π 4 + C. Therefore, y ( x ) = tan - 1 ( x ) - π 4 . Question 2. (14 points) Find the equation of the tangent line to the curve ye y = x 2 at the point ( e, 1). The slope of the tangent line is dy dx | ( e, 1) . Implicit differentiation gives: d dx ( y ) = d dx ( x 2 ) = 2 x. But, d dx ( y ) = dy dx · e y + y · e y dy dx . So, dy dx · e y + y · e y dy dx = 2 x Hence, dy dx = 2 x e y (1 + y ) Evaluating at ( e, 1) , dy dx | ( e, 1) = 2 e e 1 (1+1) = 1 e . Using the point-slope form for the equation of a line, we conclude that the tangent line to the curve at the point ( e, 1) is y = 1 e ( x - e ) + 1 .
Math 111 (Fall 2007) Prelim 2 (Oct 25, 2007) 3 Question 3. (16 points) (a) (8 points) Show that for any three positive real numbers a, b, c , the following equality holds: (log a b )(log b c )(log c a ) = 1 . We use the identity log x y = ln y ln x for any x, y > 0 . (log a b )(log b c )(log c a ) = ln b ln a · ln c ln b · ln a ln c = ln a ln a · ln b ln b · ln c ln c = 1 .

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## This note was uploaded on 04/07/2008 for the course MATH 1110 taught by Professor Martin,c. during the Spring '06 term at Cornell.

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Calculus Pre-lim 2 w. Solutions - Math 111 Prelim 2 Name...

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