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# FUNDAMENTAL THEOREM OF CALCULUS MATH 111

Rice University

GPA 3.86

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This 4 page Class Notes was uploaded by Jayde Lang on Monday October 19, 2015. The Class Notes belongs to MATH 111 at Rice University taught by Staff in Fall. Since its upload, it has received 17 views. For similar materials see /class/224933/math-111-rice-university in Mathematics (M) at Rice University.

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Date Created: 10/19/15

Rice University Math 111 Fall 2008 Department of Mathematics Extra Credit with answers You can spend as much time as you want on solving these problems and you may use the text and your class notes as much as you want 1 encourage you to try the problems on your own and look at similar examples in the text if they don7t come out on the first try What you write should be your own work and not collaborative work with other students in the class You may compare answers with others but not solutions Answers will be posted soon after the deadline Friday Dec 12th 12noon The nal exam will be similar in length and level of difficulty The allotted time will be 2h30min and you7ll only be allowed to use your index card during the exam Some of the problems 1 2 3 4 5 ad 6 7 are actual problems from the Fall 2007 Final Exam kindly provided by Kelly McKinnie 1 a Let be a function Write the definition of the derivative of Hint This is a limit Answer f1hff 7 f I V133 h b Let be a function whose domain is the whole real line Assume you know the equation of the tangent line to the graph of at 5 is given by y 731 8 What is f 5 Answer fS 73 the slope of the tangent line 2 Calculate the following limits 3 lim VI 71 1 7 l39 7 113 12 7 9 lim 9 csc 59 9H0 d lim 6 ln 1 mam Answers a This limit doesn7t exist the function xz 7 1 is not de ned for z lt 1 b 16 c 15 d 0 3 1712 3 Write an equation of the line which is tangent to the graph of 2 71 at the point Answer fZ 43 and the equation of the line is y 1 7 4 Using implicit differentiation to calculate 2 given that my 6 Answer We think of I as an independent variable and of y as depending on I and take the derivative with respect to I on both sides This yields dy 7 gm dy zgwie mam We take the terms with on one side and nd dy E 71 ze y7ieixy dz y y We slove for Ly and nd Ly 73 d1 d1 1 5 Calculate the rst derivative of each of the following functions a cosl sinz b 91 31 4W3 0 hz cot I d lnsin2 z e 11 use logarithmic differentiation for e Answers a fz 7 sinl sin I cosz g z 61 4133 43z 4W3 2 2 cosz 78111 zicos z 1 hI y 2 ii 2 7csc21 sinz s1n z s1n I d l 2cosz mz 2 2s1nzcosz 2cotz s1n 1 si e lnnz lnz zlnz Taking derivatives on both sides we get 1 7 f nz7lnzz lnzl Solving for nz we get nz nzlnz 1 zx nz 1 6 Find the global maximum and the global minimum for the function 12 1075 on the interval 17 4 Answer 2755 1n the interval 17 4 we have the critical point z 2 We check the values of f at the critical point z 2 and the endpoints z 14 f1 17 f2 12 global minimum f4 20 global maximum 7 Each part of this problem concerns the function hz 16 3 a List the critical points of b Determine the open intervals on the zaxis on which hz is increasing and those where it is decreasing c Classify the critical points of hz as local minimum7 local maximum or neither Answer a h z 6 3 7 316 31 17 31 6 3 Critical point z 1 3 b 6 3 gt 0 for all I so the sign of the derivative is the same as the sign of1 7 31 17 31 gt 0 for z lt 137 thus h is increasing on 70013 17 31 lt 0 for z gt 137 thus h is decreasing on 13700 1 13 is a local maximum 8 Simplify the following expressions 3 log216 Answer log2 16 log224 4log2 2 4 13 e2ln 3 Answer 621quot 3 elquot 32 32 9 we used 61quot a 9 Two straight roads intersect at right angles At 10 AM a car passes through the inter section headed due east at 30mih At 11 AM a truck heading due north at 40 mih passes through the intersection Assume that the two vehicles maintain given speed and directions At what rate are they separating at 1 PM Answer Let Tt be the distance traveled by the truck north of the intersection and Ct be the distance traveled by the car east of the intersection and Dt the distance between the two At t1PM Tt 80 T t 40 Ct 90 and C t 30 Using the Pythagorean theorem we have D2t T2 t C2 t and solving for D we get that at 1PMDt10m We take the derivative with respect to t in the above equation and obtain 2D tD t 2TtTt 2CtC t Put in all the data and solve for Tt We get Tt mi 10 You need a tin can in shape of a right circular cylinder of volume 167T 07213 What ra dius and height would minimize its total surface area including top and bottom Answer Let T be the base radius and h the can heighti Then the total area of the can is TAarea top area bottom side area 27rT2 27rThi We use the volume information V 7T7 2h 167T to solve for h and get h if TAT 27T7 2 27mg 27T7 2 327 T T To nd the T that minimizes the total surface area we nd the critical points of TA 7 327T 47mg 7 327T 72 TA T 47rT 2 0 7 implies T 2 We recover h 1475 4 This are the dimensions that minimize the total area of the can

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