UNIVERSITY OF TORONTO DEPARTMENT OF MATHEMATICS MAT 235 Y – CALCULUS IIFALL-WINTER 2008-09ASSIGNMENT #1. DUE ON OCTOBER 3.PROBLEMS AND SOLUTIONS1.A brief review of single variable calculus. a) Evaluate ()20lim917ln9ln31lnlnx2xxx+→++−++x, if the limit exists. Solution:Let 2917ln9lnAxx=++and 231lnlnBxx=++. Now, 22ABABAB−−=+, where . 2222917ln9ln9(1lnln)8lnABxxxx−=++−++=xNotice that 2222917917917ln9lnln(9)ln9lnlnlnlnAxxxxxxxx=++=++=++and similarly, 2113 ln1lnlnBxxx=++. So, 2291711ln(931 )lnlnlnlnABxxxxx+=+++++. Notice also that and 0lim lnxx+→= −∞lnlnxx= −when 0 < x< 1 . Therefore, ()00228ln84limlim33391711ln(931 )lnlnlnlnxxxABxxxxx++→→−== −= −++++++. b) Let P( t, f( t) ) be any point on the curve 22( )(1)xf xx=+and let Qbe the point at which the y-axis intersects the line that is tangent to the given curve at the point P. Suppose that A( t) denotes the area of the triangle OPQ, where O= ( 0 , 0 ) . Is there an absolute maximum value for the area A( t) ? If so, find this absolute maximum value and determine the coordinates of all the points Pfor which that maximum is reached. Solution:An equation of the line that is tangent to the given curve at the given point is y – f( t ) = f′( t ) ( x– t ) . If x= 0 then the value of yin this equation is y= f( t ) – tf′( t ) . So, Qis the point with coordinates ( 0 , f( t ) – tf′( t ) ) and OPQis a triangle whose base has a length of ( )( )ftt ft′−and whose height has a length of t. Therefore, the area of the triangle OPQis 1( )(( )( ))2A ttftt ft′=−. Now, 22313( )(1)xfxx−′=+, then 242223231(13)( )()2(1)(1)(1)ttttA tttt−=−=+++2t. Finally, 32244(2)( )(1)ttAtt−′=+which shows that A( t) reaches its absolute maximum at 2t= ±. The absolute maximum value for the area of the triangle OPQis 8(2 )27A±=and this maximum area is reached only when the tangency point Phas coordinates 2(2 ,9±±).
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