# Math_270_Test_5_Study_Guide_Answers - Math 270 Test 5 Study...

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-5 5 5 Math 270 Test 5 Study Guide 1. a) If I try to plug in 1, I get 0/0, so I can use L’Hopital’s Rule to find the limit. ( 29 ( 29 2 1 1 1 2 ln ln 1/ 1/1 1 lim lim lim 1 2 2(1) 2 1 x x x d x x x dx d x x x dx = = = = - - b) 0 0 ln lim ln lim 1/ x x x x x x + + = If I try to plug in 0, I get -∞ , so I can use L’Hopital’s Rule to find the limit. ( 29 ( 29 1 2 2 0 0 0 0 0 ln ln 1/ lim lim lim lim lim 0 1/ 1/ x x x x x d x x x x dx x d x x x x dx + + + + + - - - = = = = - = - - c) For this one, I can just plug in 1. 3 3 1 sin 1 sin1 lim 1 sin1 0.1585 1 x x x x - - = == - However, I made a mistake typing the problem. It should have been 3 0 sin lim x x x x - which gives me 0/0, so I can use L’Hopital’s Rule. In fact, I need to use L’Hopital’s Rule three different times because I keep getting 0/0. ( 29 ( 29 ( 29 ( 29 3 2 0 0 0 0 3 2 0 0 0 sin 1 cos sin 1 cos lim lim lim lim 3 3 sin sin cos 1 lim lim lim 6 6 6 6 x x x x x x x d d x x x x x x dx dx d d x x x x dx dx d x x x dx d x x dx - - - - = = = = = = = = 2. A particle has position parameterized by ( ) 3 6 , ( ) 1 8 x t t y t t = + = - . a) How fast is the particle moving? Velocity = ( 29 ( 29 2 2 2 2 6 8 100 10 dx dy dt dt + = + - = = b) Write the line as y = mx + b . / 8 4 / 6 3 dy dy dt m dx dx dt - - = = = = and the line goes through the point (3,1), so the equation is ( 29 4 3 1 3 y x - = - + which is the same as 4 5 3 y x - = + c) Give a different parameterization of the line with the particle starting at the same point at t = 0 but moving twice as fast. We need to start at the same point, so our parameterization is in the form ( ) 3 , ( ) 1 x t at y t bt

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## This note was uploaded on 05/18/2011 for the course MATH 270 taught by Professor Vakarietis during the Summer '08 term at University of Louisiana at Lafayette.

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Math_270_Test_5_Study_Guide_Answers - Math 270 Test 5 Study...

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