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# Homework lesson 8 Math 124

Fizza Aslam
UW
Math 124
David H. Collingwood

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Detailed description and solutions of homework 8.
COURSE
Math 124
PROF.
David H. Collingwood
TYPE
Bundle
PAGES
13
WORDS
KARMA
75 ?

## Popular in Math

This 13 page Bundle was uploaded by Fizza Aslam on Wednesday October 21, 2015. The Bundle belongs to Math 124 at University of Washington taught by David H. Collingwood in Fall 2015. Since its upload, it has received 226 views. For similar materials see Math 124 in Math at University of Washington.

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Date Created: 10/21/15
Homework Lesson 8 Lecture Question 1 In lecture 8 we discuss the process of graphical differentiation and look at a few simple examples Here is one of the first examples fx Determine whether the statements below are true or false a f 393 gt 0 False Because slope is negative b f 392 0 False because derivatives do not exist at sharp corners c f 391 gt 0 True because slope is positive d f 392 gt 0 False Because slope is negative for decreasing line e f 394 gt 0 False Because slope is still negative f f 395 lt 0 True Because slope is still negative Lecture Question 2 In lecture 8 we discuss a common exam problem using the function fxx2 5x7 The problem is to study tangent lines to the graph of fx that pass through the point P212 See the lecture for a picture Determine whether the statements below are true or false a There are exactly two different tangent lines to fx passing through P True As you can see in the picture two lines red and green are tangent to the function from point P Let s use definition of derivative to find the formula for y f a fahfa m f a 140 h ah2 5ah7 a2 5a7 f a 2 11m h gt0 h ai2ahh2 5a 5h1 a5a f a 2 11m h gt0 h I 1 2ahh2 5h f 00 I h2ah 5 f ltagt TBAT fa 1Igt112ah 5 f a 2a 5 Typical tangent line has equation y f39CaXx a fa Plug in the rules for fx and f x fa a2 5a7 fa 2a 5 y 2a 5x aa2 5a7 Impose fact line passes through 212 and solve for a 12 2a 5 2 01012 501 7 12 4a 2a2105aa2 5a7 0 a2 4aSORa24a 5 0 012 501 a 5 0 aa5 1a5 0 01 5a 1 O 9 So either a 5 or a 1 that means xcordinates are 5 and 1 Plugging in fa aquot2 5a 7 f 5 52 5 5 7 9 f 5 57 so Q 557 f1 12 51 7 9 f1 3 and R 13 Now for determining equations Let mq be slope of line PO and mr be slope of line PR Mq 5712 5 2 15 and Mr 3 121 2 93 3 Cq 57 15 5 18 and Cr 3 31 6 Hence the equations are Q 9 y 15x 18 and R 9 y 3x 6 Now this information can be used to find answers below b One ofthe tangent lines has equation y15x18 TRUE c One ofthe tangent lines has equation y18x15 FALSE d One ofthe tangent lines has equation y3x6 TRUE e One ofthe tangent lines has equation y15x6 FALSE Lecture Question 3 Cameron is located 2 miles due East of an intersection between two country roads He decides to walk in a straight line crosscountry toward a position 3 miles due North of the intersection it takes Cameron one hour to reach that location The formula for the distance between Cameron and the intersection at time t hours is dt 2 212 312 a b c At time t hours the formula for the instantaneous rate of change of the distance between 826t 2 41 t2 9t2 Note You can do this problem by writing out the definition of d39t and carefully doing the algebraic steps to calculate the limit Let s take the derivative of dt by definition to find out instantaneous rate of change at time t hours Cameron and the intersection is d39t dt h dt h J2 2t h2 3t h2 J2 2t2 3t2 2 lim h 0 h J41 t h2 9th2 J41 12 912 m d t 2 lg 2 1i h 0 h hm J41 t h2 9th2 41 t2 912 J41 t h2 9th2 41 t2 912 h0 h J41 t h2 9th2 41 t2 912 4 8t 8h8th4ti4h2 9 18th9h2 48 4ti 91 lim 139 hJ41 t h2 9t h2 J41 12 912 1 h 8 261 13h 1m h 0 hJ41 t h2 90 h2 J41 12 912 am 8 261 130 J41 t 02 90 02 J41 12 912 8 26t 2J41 12 912 d t f d39t2 mph then t0615 or 72117 Plug in 2 in the equation we found in part a Z 4 131 2J41 t2 912 2J41 12 9122 2 4 1312 46 32 162 3612 46 104 1692 721 1172 e 72 117 9 t0615 f d39t 1 mph Cameron is located at the point 2 lt 1J41 12 912 2 4 1312 4 81 412 912 16 104 1692 156t2 96t12 0 13t2 8t1O Now plugging this into quadratic equation t biVb2 4ac 2a t Z 8 i J 82 413x1 213 8 i x t Z 26 d39t 1 only when subtracting the discriminant so 8 xE t Z 26 hence t017445 Now considering parametric equations xt 22t 22017445 1652 yt 3t 3O17445 0523 hence the location of Cameron will be 16520523 Lecture Question 4 The graph shows the influence of the temperature T on the maximum sustainable swimming speed S of Coho salmon 3 ii lentils E v 139 I a What is the meaning of the derivative S 39T S39 T is the rate of change of the maximum sustainable speed of Coho salmon with respect to the temperature b Estimate the values of S 395 and S 3925 For T 5 0C it appears the tangent line to the curve goes through the points 5 15 and 10 20 So S 5 z 20 1510 5 1 cms C This tells us that at T 5 0C the maximum sustainable speed of Coho salmon is changing at a rate of 1 cms C In a similar fashion for T 25 0C we can use the points 20 35 and 25 25 to obtain S 25 z 25 3525 20 2 cms C As it gets warmer than 20 0C the maximum sustainable speed decreases rapidly Lecture Question 5 The 24 graphs below are labeled by letters from a to x For each of the following graphs of fx give the letter of the graph that looks most like it could be the graph of the derivative function f39x i V Lecture Question 6 Use the given graph of fx to sketch the graph of f 39 The graph of derivative is on the right side above This is because during the highlighted part in yellow the graph is decreasing so f should be negative During highlighted in green the graph is increasing so f should be positive and for the third part it should again be negative because the graph is decreasing again Lecture Question 7 The graphs of four derivatives are given below Match the graph of each function in ad with the graph of its derivative in IIV The functions for each graph of derivative is matched just below them For I the function in blue decreasing increasing and then decreasing so the graph is negative at first then positive and then again negative For II the function is increasing decreasing increasing and decreasing so the derivative moves back and forth from negative to positive For Ill function has straight lines which means constant derivatives and sharp corners where derivatives do not exist For IV the function is at first decreasing and then increasing so the derivative is in negative quadrant at first but in positive quadrant after passing through zero Lecture Question 8 This was tutorial exercise for question 7 Lecture Question 9 Trace or copy the graph of the given function f Assume that the axes have equal scales Then sketch the graph off 39 below it li39Ji I H39liquot i l xt Lecture Question 10 Trace or copy the graph of the given function f Assume that the axes have equal scales Then sketch the graph off 39 below it jFJt Lecture Question 11 Trace or copy the graph of the given function f Assume that the axes have equal scales Then sketch the graph off 39 below it 9 Jul II Jump on sharp corners and positive or negative constant when original graph is increasing or decreasing Lecture Question 12 Trace or copy the graph of the given function f Assume that the axes have equal scales Then sketch the graph off 39 below it l 3 Hill The graph of derivative is at right The graph of function is always increasing so the derivative should be in positive quadrant Also there is a tangent line to function at y0 so the derivative is undefined at y0 Lecture Question 13 Find the derivative of the function using the definition of derivative gx V9x x 1imcJxhgx 399 h 0 h V9 x h V9 x 9 x h9 x g x 11m h 0 h x9 x h9 X 9 9e h 99e g x 11m h 0h9 x h9 x 1 g x lim Momma 1 I x g 2 r c State the domain of the function oo9 State the domain of its derivative oo9 Lecture Question 14 A rechargeable battery is plugged into a charger The graph shows Ct the percentage of full capacity that the battery reaches as a function of time t elapsed in hours I percentage ef tll charge lEIEI BEI EDI 4D EEIf I 2 4 E E Ill 12 hours a What is the meaning of the derivative C 39t C 39t is the instantaneous rate of change of percentage of full capacity with respect to elapsed time in hours b Sketch the graph of C 39t I I I I I t 2 4 E E 1 El 12 The function seems increasing along with the positive slope of tangent line but becoming less positive with time and approaching zero So the graph of derivative should be in 1st quadrant decreasing and approaching zero c What does the graph of C 39t tell you The graph of C 39t tells us that the rate of change of percentage of full capacity is decreasing and approaching 0 Lecture Question 15 The figure shows the graphs of f f 39 and f quot Identify each curve Let s take a as function f Here b has an x intercept where a has a horizontal tangent line so this goes true Also when a is increasing b is always positive so that means a is our function fand b is f This causes c to be f Also the c has xintercept where b has horizontal tangent line so this makes are Statement true Lecture Question 16 The figure shows graphs of f f 39 f quot and f quot39 Identify each curve H bi t Ill HTF Here fd f c f b and f a If we see function d has 2 horizontal slopes and function c has x intercepts on those points Also when d is increasing decreasing and then increasing c is positive negative and again positive Now c has just one horizontal slope and b has xintercept on that point also b is negative and then positive because c is decreasing and then increasing In the end since b is only increasing function a is always positive Lecture Question 17 Use the given graph to estimate the value of each derivative a F O2 b F 1O c F 215 d F 315 e F 4O f F 51 Lecture Question 18 Consider the function y fx l2 x and the point P07 There are two lines tangent to the graph of the function which pass through P see picture below Let R and S be the points on the graph where the tangential lines pass through P and assume R has the smaller x coordinate a The formula for f 39a Let s use definition of derivative to find formula for f a fmhli fim ll h f 1 V2ah 2a V2ahV2a a 1m kw h V2ahV2a 2ah 2 a f a 11m h 0h2 ahV2 a 1 f a lim h 02ah2a 1 Ia f 2 Typical tangent line has equation y f ax a at 1 x a V2a y 22a Impose fact line passes through 07 and solve for a 1 7 O a V2a 2V2a 14x2 a a 4 2a 14x2 a2 a 42 392 1963 a2 8a 16 a2 188a 376 0 b i Vbz 4ac a 2a 188i 1882 41 376 a 2a 188 i V36848 a f b The x coordinate of R is 1979164 c The x coordinate of S is 189979164

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