Calculus Pre-lim 2 w. Solutions

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

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

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