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# Homework lesson 9 Math 124

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This 14 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 47 views. For similar materials see Math 124 in Math at University of Washington.

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Date Created: 10/21/15

Homework Lesson 9 Lecture Question 1 In lecture 9 we discuss common notation used for the derivative f yfx determine whether the statements below are true or false a The derivative of yfx is denoted f 39x TRUE modern notation b The derivative of yfx is denoted dydx TRUE classical notation Lecture Question 2 In lecture 9 we discuss some basic useful computational facts related to the derivative Determine whether the statements below are true or false The notation 39 means quottake the derivative of the expression in the parenthesisquot The quotdegreequot of a polynomial fx is the highest power of x appearing in fx a If a is a real number bigger than 1 then xa 39 a xa 1 TRUE 1st rule of derivatives 1 1 b 363 7 3x3 TRUE c 2X2 4X 7 2X 4 Let s take derivative 22x2 1 4 1x1 0 0 4X4 so answer is false 039 25 395 24 False Because the derivative of a constant is always zero e If fx is a polynomial of degree 7 then the degree off 39x is 6 True Because degree decreases by 1 for each derivative you take f If fx is a polynomial of degree 7 then the degree off quotx is 2 False Because degree decreases by 1 for each derivative you take So for the second derivative in this case degree should be 72 5 Lecture Question 3 In lecture 9 we discuss the derivative offx4x3 3x2 6x 12 The smallest possible slope of a tangent line to the graph of yfx is 274 The equation of the tangent line to the graph of yfx which has smallest possible slope is As we know that the typical tangent line has the formula y f ax a fa and we know that f a is 274 and a 14 Let s plug in this information 2 y ltx gt4ltgt ltgt eltgt12 27 27 1 y Tx1 6E 1 6 12 So our equation is 27 193 F7 Lecture Question 4 Differentiate the functionfx 260 F x 0 because derivative of a constant is always zero Lecture Question 5 Differentiate the functionfx X3 4x 1 F x 3x2 4 using 1st rule Lecture Question 6 Differentiate the function gx x21 2x gx x2 2x3 g x 2x 23x3391 g x 2x 6x2 Lecture Question 7 Differentiate the function gt 2tquot34 g t 2 E t g l 4 1 gt 2t 3 gt 7 211 Lecture Question 8 In this problem consider the curve y fx x27x 1 a For any real number a the formula forf 39 a fa a2 7a 1 f a 2a 7 b The equation of the tangent line to the curve yfx at the point Pafa is y f axa fa y 2a 7x aa2 7a 1 y2ax 2a2 7x7ampa2 7amp 1 y 7x2ax a21 y 72ax a21 c There are two tangent lines to the curve yfx with xintercept 8 The xcoordinates of these two points of tangency listed left to right are 5 and 11 Plugging point 80 in equation determined in part b O 72a8 a21 O 5616a a21 a2 16a 55 O 16 i 6 2 so a11 or a5 a Lecture Question 9 Differentiate the function A3 2 As 1ls396 A s 11 6s396391 A s 665397 66 A S 3 7 Lecture Question 10 Differentiate the function Ra 2a 12 Ra 4a2 4a 1 R a 8a4 Lecture Question 11 Differentiate the function SR 47139R2 SR 427m SR 2 BER Lecture Question 12 2 Differentiate the function 17 xE i 3v 1 1 vltx2x 3 l 2 vx2x6x 3 2 1 11 2 21 v 12gx6 x3 1 5 2 5 1 17 3X 3X 1 2 U 1 39 E 3x6 3x3 Lecture Question 13 Find an equation of the tangent line to the curve at the given point y x4 3x2 x 1 3 Let s find slope of tangent line y 4x36x 141361 19 Now plug in slope and point in typical line equation to find yintercept ymxc9391c 9 c 6 Hence our equation is y 9x 6 Lecture Question 14 Find an equation of the tangent line to the curve at the given point y 6x2 x3 1 5 Let s find slope of tangent line y 12x 3x2 2 121 312 9 Now plug in slope and point in typical line equation to find yintercept ymxc9591c 9 c 4 Hence our equation is y 9x 4 Illustrate by graphing the curve and the tangent line on the same screen Note that x is 49 when y0 and y14 when x2 so the only possible graph is it 35 an 25 Lecture question 15 Find the first and second derivatives of the function fx 19x19 8x8 x f x 361x18 64x7 1 f x 6498x17 448x6 Lesson question 16 The equation of motion of a particle is s t3 3t where s is in meters and t is in seconds Assume t 2 O a Find the velocity and acceleration as functions of t We know that velocity is a derivative of distance because velocity is the rate of change of distance with respect of time Similarly acceleration is the rate of change of velocity with respect of time So here velocity is the first derivative and acceleration is the second derivative of the function 5 f s vt 312 3 at 6t b Find the acceleration after 8 5 Simply plugging in t8 in function at gives us acceleration ie 48msquot2 c Find the acceleration when the velocity is 0 Here vt 0 that means 312 320 gt ti1 Since the question assumes that t gt 0 so t can only be 1 Now plug in the function of acceleration we get a61 6 msquot2 Lecture Question 17 Where does the normal line to the parabola y x x2 at the point 1 0 intersect the parabola a second time Let s find slope of tangent line y 1 2x gt1 21 1 Slope of the normal line will be 1slope of tangent line so 11 1 Now plug in slope and point in typical line equation to find yintercept ymxc9011c 9 c 1 Hence our equation is y x 1 x x2x 1 gtx21 gtxi1 Whenx1 y1 12 gtyO Whenx1 y 1 12 gty 2 So the two points are 10 and 12 Our answer is the second point And our graph is er Lecture Question 18 Consider the following functionfx x2 16 a Find aformula forf39 Note thatxz 16lt Oforx2lt 16 ltgt x lt4 ltgt 4ltxlt4 So quot X2 16 if39X E l quot 2 if a 4 s X2 15 if 4 a z FIX 2x if 4 c x a 4 39 XE15 ifngql 2X ifx4 I 2 if lel 3 4 2x if lxll a 4 For what values ofx is the function not differentiable Enter your answers as a comma separated list f4 To show thatf 394 does not exist we investigate llirr w by computing the left and right hand derivatives f4 h f4 lim 4 h2 16 0 f 4 211351 h h 0 h 2 23511 8 h 2 8 f4h f4 4h216 0 f 4 111135 h hug h hllgllt8 h 8 Since the left and right limits are different f 4 doesn t exist Similarlyf 39 4 does not exist Thereforefis not differentiable at 4 or at 4 b Sketch the graph off and f fix Fix It 5 I I A I I I I ID 15 IU 5 5 1D 5 I 5 10 E 15 Lecture Question 19 Differentiate the following functionfx 1862 0 because derivative of a constant is zero Lecture Question 20 Findf 39Xfx equot 8X We know that equotx is equotx by rule so the answer is f x equotx 8 Compare the graphs off andf 39 and use them to explain why your answer is reasonable Do not get confused by the specific functions Just apply your prior knowledge about f and f f39x 0 when f has a horizontal tangent f39 is positive when f is increasing f39 is negative when f is decreasing Lecture Question 21 The equation of motion of a particle is s 2t3 8t2 2t 3 where s is in meters and t is in seconds Assume t 2 O a Find the velocity and acceleration as functions of t f vt 612 161 2 f 5 at 12 16 b Find the acceleration after 1 s a1 4 c Graph the position velocity and acceleration functions on the same screen Lecture Question 22 Differentiate gx 9 exxE gx 9 exx Using product rule fx gixi fx g x f x 900 906 96quot 36 96x Ix x l 1 l x g 9e 2x2x29e 9 00 ex9ex xE 9 2 give 2 93x L 2V Lecture Question 23 39 x5 Differentiate y x3x2 I I Q I 9XfX fxg x Usmg quotient rule gm goof r x3x2gtltx539 x5x3x 239 x3 x 22 x3 x 21 x 53x2 1 y x3 x 22 x3 x 21 x 53x2 1 y x3 x 22 x3x 2 3x3x15x25 y x3 x 22 2x3 15x2 7 y x3x 22 Lecture Question 24 Differentiate Assume k is a constant y m E 9xf x fxgrx 905 gx2 Using quotient rule I t ket1 11 ket t ket2 t ket0 11 ket t ket2 1 ket 339 t ket2 I Lecture Question 25 Differentiate ft 2 8t 8 39 39 39 gxf xfxglx Usmg quotlent rule gm goof fat 2 8 xE8t 88 y 8 V 8 VEXB sari ft 2 2V 8 V 64 8 4 t f 8 V32 64 4 t I f 8 r Lecture Question 26 Differentiate Assume k is a constant fv v v v39 va 39 92 fv39 v g 1 v1 9 f 39 Mar b vb bv m 2 v 5 goo lags v2 U 5 y 2 v v1 v 92 v2 v2 Lecture Question 27 Findf 39x andf quotx fx x ex 5 5 fx equot x5 x5 ex 5 3 5 fx 6quot Ex5 x5 6quot fx equot x 5 E E E x fx 6 2x2x2 2x2x2 e 15 1 5 E 5 E 5 fx equot 4 x2 x2gt Exz x2 equot H x 151 E E fx 6 4x25x2x2 Lecture Question 28 x2x1 Find an equation of the tangent line to the given curve at the specified point y 10 x2 x 1x2 1 x2 1x2 x 1 x2 x 12 I x2 x 12x x2 12x 1 I x2x12 2x32x22x 2x3 x22x1 31 x2x12 x24x1 31 x2x12 12411 y 12112 62 y 9 3 2 2 yy xc gtO 1c gtc So our equation for tangent line is y 2x Lecture Question 29 a The curve y 11 x2 is called a witch of Maria Agnesi Find an equation of the tangent line to this curve at the point 112 1 x21 11 x2 I 1x22 2x y 1x22 21 y 1122 r 2l y 4 2 1 1 yyIXC 71C gtC1 So eqn isy x 1 b Illustrate part a by graphing the curve and tangent line on the same screen Look at the point 12 and 20 on graph 1quot EH Lecture Question 30 If h2 5 and h392 9find a h 39 EHH i w xh39x hxx39 dx x x2 Now plug in the values x2 hx25 and h x29 d hx 2 9 51 23 ET 22 T Lecture Question 31 Iff and g are the functions whose graphs are shown let ux fxgx and vx fxgx l a Find u 1 u 1 f1g 1 g1f 1 From the graph we can see that f12 and g11 For f 1 let s consider two points 12 and 121 so f 1211122 For g 1 let s consider two points 11 and 02 so g 121O11 u 12 112 2 2 O b Find v 1 95f 5 f 5g395 V 5 2 95 From the graph we can see that f53 and g52 For f 5 let s consider two points 53 and 24 so f 5345213 For g 5 let s consider two points 52 and 20 so g 52O5223 2 3 1 3 2 5 22 4 Lecture Question 32 1 FInd equations of the tangent lines to the curve y that are parallel to the line x 2y 3 x 1x 1 x 1x 1 I x12 x11x11 31 x12 x1 x1 2 I y x 12 x 12 Let s reorganize the line eqn 2yx3 9 yx2 32 which means slope of the lines will be 12 so 2 m e4x22x1ex22x 3O gt x 1x3O Plugging in the curve we get Y0 for x1 and y2 for x3 And by plugging in typical line eqn we get our equations 0 1 1 1 1 1 20 C C 2 y 2x 2 2 1 3 7 1 7 2 C C 2 y 2x 2 Lecture Question 33 1 In this problem consider the curve y fx a The derivative functionf 39 x Z 1 139 11 J 1 if L 21 1 if b Find the point Pafa on the graph where the tangent line at P has yintercept 12 Plugging in the tangent line equation f 39x f39x yf ax afa 1 1 O a 1 2 2 152 1xE 1 45 1 2 2152 1xE l E1EZ1E2 2 21E21E 1em va21va 5 21E2 1 7 1 2xE E21xE2 12VEa2E aka av a1 2aa2 O1 3aa2 3i 2 gtfa a 1 1 3 inE

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