Questions and Answers #3 Analysis

Questions and Answers Sheet 3 Analysis Question #1 Graph the sets of points whose polar coordinates satisfy the given polar equation. The given polar equation is written as follows: π π − ≤ θ ≤ ,1 ≤ 𝑟 ≤ 2 2 2 Answer: Consider the polar coordinates as 𝑃(𝑟, 0). Here, ris the directed distance from origin Oto point Pand θ is the directed angle from initial ray to OP. π π From the given equation, the value of \theta varies from − to and r varies from 1 to 2. 2 2 Polar point traces a circle when ris fixed at constant value and it traces a straight line when \theta is fixed at constant value. From the analysis, draw the sets of points whose polar coordinates satisfy the given polar equation as shown in Figure 1. Question #2 Graph the sets of points whose polar coordinates satisfy the given polar equation. The given polar equation is written as follows: π π − ≤ θ ≤ , −1 ≤ 𝑟 ≤ 1 4 4 Answer: Consider the polar coordinates as 𝑃(𝑟, θ). Here, r is the directed distance from origin O to point Pand θ is the directed angle from initial ray to OP. π π From the given equation, the value of \theta varies from − to and r varies from 1 to 2. 2 2 Polar point traces a circle when r is fixed at constant value and it traces a straight line when θ is fixed at constant value. From the analysis, draw the sets of points whose polar coordinates satisfy the given polar equation as shown in Figure 1. Question #3 Let 𝑓(𝑥) = 𝑥 5 + 5𝑥 4 + 6𝑥 + 3. a. Conduct a sign analysis of f''''. That is, determine the sign off'''' on each interval. Make sure to show how you found the boundaries (endpoints) of the intervals and how you computed the sign in each interval. b. indicate the intervals, if any, on which f is concave up and the intervals, if any, on which f is concave down. c. Find the x-coordinates of any inflection points of f. If there are more than one, enter them as a commaseparated list. If there are none, enter ''''DNE'''' Answer: Given: 𝐹(𝑥) = 𝑥 5 + 5𝑥 4 + 6𝑥 + 3 differentioting the above ef wet (x) we heve 𝐹1 (𝑥) = 5𝑥 4 + 20𝑥 3 + 6 again differentiat wet (x),we have 𝐹11 (𝑥) = 20𝑥 3 + 60𝑥 2 (a.) for beundasies (endpoints) of he interval we apply. 𝐹 [11} (𝑥) = 0 i.e 20𝑥 3 + 60𝑥 2 = 0 ov, 20𝑥 2 (𝑥 + 3) = 0 The roots of the abore equation are 0,0 8 − 3 b) for concave up, 𝐹11 (𝑥) > 0 So,for 𝑥 ≻ 3, F(x) is concave up. (c.) Infection points are those points at which Function changes its concarity that is concave up to concave down ov vice-versa . In the secHon(a) ,we see that 𝐹11 (𝑥) cnange its sign once at -3 where concarity changes from concave down to concave up. So, the point of Inflection is -3. Question #4 The reason to find points of intersection of polar graph which may require further analysis beyond solving two equations simultaneously Take examples of two polar equations 𝑟 = 1 − 2 cos θ and 𝑟 = 1 . Answer: Finding the point of intersection of polar graphs may require further analysis beyond solving two equations simultaneously. Take examples of two polar equations 𝑟 = 1 − 2 cos θ and 𝑟 = 1 . Solve these equations to obtain the point of intersection 𝑟 = 1 − 2 cos θ 1 = 1 − 2 𝑐𝑜𝑠 θ cos θ = 0 π 3π θ= , 2 2 π 3π 2 2 So, point of intersection will be (1, ) . (1, ). After plotting these polar equation, the graph obtained is: From the above figure, there are three points of intersection: π 3π 2 2 (1, ) , (1, ) and (1,0) . Simultaneous points of intersection are obtained. There may be intersection points that do not occur with the same co-ordinates in the two graphs. Question #5 3 Consider the integral ∫0 3𝑥 2 𝑑𝑥 . Use Riemann sums with right endpoints to compute the exact value of the integral. Answer: Step 1 Right Riemann sums is right end point sub interval analysis for approximation ∫ 𝑓(𝑥)𝑑𝑥 = Δ𝑥[𝑓(𝑥1) + 𝑓(𝑥2) + ⋯ 𝑓(𝑥𝑛)] 𝑎 Δ𝑥 = 𝑏 − ℎ b - upper limit of integration a - lower limit of integration h - sub interval Step 2 Here, 𝑓(𝑥) = 3𝑥 2 So, by dividing the interval (0, 3) into 𝑛 = 1 sub intervals length 0 Δ𝑥 = 3 − 1 Δ𝑥 = 3 and Now we will evaluate the function at right end point 𝑓(𝑥3) = 27 ∫ 3𝑥 2 𝑑𝑥 = 3(27) = 81 Question #6 In a certain city, the temperature (in 𝐹 ∘ ) at t hours after 9 AM is modeled by the function 𝑇(𝑡) = 50 + π𝑡 14 sin ( ) . Find the average temperature during the period from 9 AM to 9 PM. 12 Answer: Step 1 The temperature function is given as π𝑡 𝑇(𝑡) = 50 + 14 sin ( ) 𝐹 ∘ 12 t is time in hours The analysis is started at 9 AM so lete a that instant 𝑡 = 0 At 9 PM, 𝑡 = 12 Step 2 The average value of a function is given by 𝑏 1 ∫ 𝑓(𝑥)𝑑𝑥 𝑏−𝑎 𝑎 12 1 𝑇𝑎𝑣 = ∫ [50 + 12−0 0 𝑓𝑎𝑣 = = = 1 12 1 12 [50𝑡 + 14 ( π𝑡 14 sin ( )] 𝑑𝑡 π𝑡 − cos 12 π 12 12 12 )] 0 12 π×12 π 12 [15(12 − 0) − 14 ( ) {(cos = 15 − 14 π [cos π − cos 0] ) − cos 0}] = 15 − 28 π 𝑇𝑎𝑣 = 6.0873∘ 𝐹 Question #7 An analysis of passenger traffic at Daytona Beach International Airport over the past 40 years shows that the number of passengers using the airport can be modeled by the function 𝑃(𝑥) = 134𝑥 3 − 8335𝑥 2 + 138,607𝑥 + 222,284 where x represents the number of years since 1980 and the domain is [0, 40] a) At what rate was passenger traffic changing in 2000? Don't forget to label and interpret the answer. Answer: Step 1 Subpart a 𝑃(𝑥) = 134𝑥 3 − 8335𝑥 2 + 138,607𝑥 + 222,284 differentiate with respect to x 𝑃′(𝑥) = 402𝑥^{2} − 16670𝑥 + 138,607 in year 2000 𝑥 = 20 years 𝑃′ (20) = 402(20)2 − 16670(20) + 138,607 = 160,800 − 333,400 + 138,607 −33,993 In 2000 passengers traffic is decreasing at the rate of 33,993 passengers per year. Question #8 An analysis of passenger traffic at Daytona Beach International Airport over the past 40 years shows that the number of passengers using the airport can be modeled by the function 𝑃(𝑥) = 134𝑥 3 − 8335𝑥 2 + 138,607𝑥 + 222,284 where x represents the number of years since 1980 and the domain is [0, 40] b) Determine the extrema of the function on its domain. Answer: Subpart b Find critical point by solving 𝑃′ (𝑥) = 0 𝑃′ (𝑥) = 402𝑥 2 − 16670𝑥 + 138607 = 0 Using quadratic formula 𝑥= 16670±√(16670)2 −4(402)(138607) 2(402) by using calculator 𝑥 = 11.5 and 𝑥 = 29.96 approx For extrema find the value of P(x) at critical points and end points of given domain 𝑃(0) = 134(0)3 − 8335(0)2 + 138,607(0) + 222,284 = 222,284 𝑃(11,51) = 134(11.51)3 − 8335(11.51)2 + 138,607(11.51) + 222,284 = 917758 (rounded) 𝑃(29,96) = 496988 (rounded) (40) = 1006564 absolute maximum is at 𝑥 = 40, 𝑃(40) = 1006564 passengers absolute minimum is at 𝑥 = 0, 𝑃(0) = 222,284 passengers Question #9 An analysis of passenger traffic at Daytona Beach International Airport over the past 40 years shows that the number of passengers using the airport can be modeled by the function 𝑃(𝑥) = 134𝑥 3 − 8335𝑥 2 + 138,607𝑥 + 222,284 where x represents the number of years since 1980 and the domain is [0, 40] c) For what years did traffic increase? Answer: Step 1 Subpart c interval [0, 11.51] (11.51, 29.96) (29.96, 40] sing ofP( x) +ve −ve +ve So P(x) is increasing from 𝑡 = 0 to 𝑡 = 11.51 and 𝑡 = 29.96 to 𝑡 = 40 means Passengers traffic is increasing from 1980 to 1992 and from 2010 to 2020. Question #10 a) A low voltage transformer manufacturing line has a purchase order of 1700 piece, If the manufacturing line production rate is 22 piece per minute use dimensional analysis techniques to determine the time taken to produce the requested transformers to the nearest minut in hours and minutes.. Answer: Step 1 a) Given Purchase order = 1700 piece Production rate = 22 piece per minute So Time taken to produce the requested transformers = = 1700 Purchase order Production rate 22 = 77.27𝑚𝑖𝑛 =1 hour 17.27 min Question #11 b) Assume that the production time of a transformer can be given by the following formals: 1 𝑡 = 𝐺𝑚3 𝑣 −3 𝑙 2 Where m is the mass of the transformer in kg, v is the velocity of the manufacturing line in 𝑚/𝑠 and l is the length of the production line in meters. Find the diamensions of G Answer: b) Given 1 𝑡 = 𝐺𝑚3 𝑣 −3 𝑡 2 unit of 𝑚 = 𝑘𝑔 ; then dimension of 𝑚 = 𝑀 unit of 𝑣 = 𝑚/𝑠 ; then dimension of 𝑣 = 𝐿/𝑇 ubit of 𝑙 = 𝑚 ; then dimension of 𝑙 = 𝐿 dimension of 𝑡 = 𝑇 Now. 𝑡 𝐺= 1 𝑚3 𝑣 −3 𝑡 2 Substituting the dimensions of each variables in the formula of G 𝑇 𝐺 = 1 𝐿 −3 𝑀3( ) 𝑇 (𝐿)2 1 𝐺 = 𝑀−3 𝐿𝑇 −3 1 So, the dimension of 𝐺 = 𝑀−3 𝐿𝑇 −3 Question #12 c) In b, if 𝑚 = 220 grams, 𝑙 = 10.2𝑚, 𝑣 = 2.3𝑚/𝑠 and time is 30.1 ms. Find G? Answer: c) Given, 𝑚 = 220 grams = 0.22𝑘𝑔 𝑙 = 10.2𝑚 𝑣 = 2.3𝑚/𝑠 𝑡 = 30.1𝑚𝑠 = 30.1 1000 𝑠 = 0.0301𝑠 we have 𝑡 𝐺= 1 𝐺= 𝐺= 𝑚3 𝑣 −3 𝑡 2 0.0301 1 (0.22)3 ×(2.3)−3 ×(10.2)2 0.0301×12.17 0.60×104.04 1 𝐺 = 0.00587𝑘𝑔−3 𝑚𝑠 −3 Question #13 Discuss the continuity of function 𝑓(𝑥) = 1 𝑥 3 +1 on 𝑅 Answer: Step 1 Consider the given function, 1 𝑓(𝑥) = 3 𝑥 +1 1 The function takes indeterminate form ( ) , if the denominator becomes zero. 0 So in order to determine the points of discontinuity, equate the denominator of the rational function to zero, and solve for the roots, ⇒ 𝑥3 + 1 = 0 ⇒ 𝑥 3 = −1 ⇒ 𝑥 = −1, 𝑖, −𝑖 Step 2 As per the given question, the analysis is limited to the set of real numbers. So the only value of interest is 𝑥 = −1 Note that the function becomes undefined at point 𝑥 = −1 , so the given function is continuous at all points except at 𝑥 = −1 Thus, the function is continuous in the set 𝑅 − {−1} Question #14 Solve the polynomial inequality 𝑥 2 − 2𝑥 + 1 > 0 and graph the solution set on a real number line. Express the solution set in interval notation. Answer: Step 1 The given polynomial inequality is 𝑥 2 − 2𝑥 + 1 > 0 To solve the inequality, find its critical points by equating it to zero. 𝑥 2 − 2𝑥 + 1 = 0 𝑥2 − 𝑥 − 𝑥 + 1 = 0 𝑥(𝑥 − 1) − 1(𝑥 − 1) = 0 (𝑥 − 1)(𝑥 − 1) = 0 𝑥 = 1, 1 Step 2 Now, make the sign analysis on Number line to determine about the inequality. If 𝑥 < 1 , then for the given polynomial the result is always positive and even if 𝑥 > 1 , then also for the given polynomial the result is positive. Therefore, for all values of x except 1 ,the polynomial is positive. Answer: Hence, the interval notation for the given polynomial inequality is (−∞, 1) ∪ (1, ∞) Question #15 Evaluate the given limits by using table of values. Use table of values to evaluate the given limit, show your Complete Solutions (This is all about Basic Calculus: Evaluating limits using Table of Values). a) lim ( sin 𝑡(𝑐𝑜𝑠𝑡−1) 𝑡⇒0 ) 𝑡2 Answer: Step 1 In mathematics, the limit of a function is a fundamental concept in calculus and analysis concerning the behavior of that function near a particular input. a) lim sin 𝑡(cos 𝑡−1) 𝑡⇒0 𝑡2 Where 𝑓(𝑡) = t −0.5 −0.75 −0.9 −0.99 −0.999 −0.9999 sin 𝑡(cos 𝑡−1) 𝑡2 f (t ) 0.2347 0.3261 0.3659 0.3850 0.3966 0.3868 Step 2 t 0.5 0.25 0.9 0.99 0.999 0.9999 f (t ) −0.2347 −0.3261 −0.3659 −0.3850 −0.3966 −0.3868 Therefore limiting value is 0. Question #16 Evaluate the given limits by using table of values. Use table of values to evaluate the given limit, show your Complete Solutions (This is all about Basic Calculus: Evaluating limits using Table of Values). b) lim (𝑒 2𝑡 × 𝑒 𝑡 −1 𝑡⇒0 𝑡 ) Answer: Step 1 In mathematics, the limit of a function is a fundamental concept in calculus and analysis concerning the behavior of that function near a particular input. Step 2 b) lim 𝑡⇒0 t −0.5 −0.75 −0.9 −0.99 𝑒 2𝑡 (𝑒 𝑡 −1) 𝑡 f (t ) −0.5790 −0.2093 −0.1211 −0.0885 Again, t f (t ) 0.5 7.0536 0.75 8.8996 0.9 10.9013 0.99 12.4979 Limiting value is 1. Question #17 The electrical currents flowing in the antennas, during an MR imaging scan are given by: 3𝑖3 + 3𝑖1 − 5𝑖2 = 7.5 𝑖2 − 7𝑖3 + 2𝑖1 = −17.5 −10𝑖1 + 4𝑖2 + 5𝑖3 = 16 1. Write the above system of equations in the form 𝐴𝑥 = 𝑏 , where A is a 3 by 3 matrix and x and b are 3 by 1 vectors. 2. Find the determinant of A. 3. Determine 𝑖1 , 𝑖2 and 𝑖3 by using Gaussian elimination. Answer: Step 1  3 −5 3   i1   7.5   2 1 −7  i2  =  −17.5   −10 4 5  i3   16  above solution is form if 𝐴𝑥 = 𝑏 where  3 −5 3  A =  2 1 −7   −10 4 5   i1   7.5    x = i2  & b =  −17.5 i3   16  Step 2 Determinant of A  3 −5 3  D  2 1 −7   −10 4 5  𝐷(𝐴) = 3(5 − (−28)) − (−5)(10 − 70) +3(8 − (10)) = 3(33) + 5(−60) + 3(18) = 99 − 300 + 54 = 153 − 300 𝐷(𝐴) = −147 Step 3 Now we have to calculate and find 𝑖1 , 𝑖2 , 𝑖3 using gaussian elimination Solve it using gaussian elimination 7.5   3 −5 3  2 1 −7 −17.5   −10 4 5 16  𝑅1 3 ⇒ 𝑅1 5   2.5   1 −3 1   1 −7 −17.5  2  −10 4 5 16      𝑅2 − 2𝑅1 ⇒ 𝑅2 𝑅3 + 10𝑅1 ⇒ 𝑅3 5   2.5  1 − 3 1   0 13 −9 −22.5   3   0 − 38 15 41    3 𝑅1 13 3 ⇒ 𝑅2 𝑅1 + 𝑅3 + 5[3} 𝑅 2 38 𝑅 3 2 ⇒ 𝑅1 ⇒ 𝑅3 32  1 0 − 13  0 1 − 27  13  0 0 − 147  13 𝑅3 − 147 13 80  13   135  − 26   332  − 13  − ⇒ 𝑅3 𝑅1 + 𝑅2 + 32 13 27 13 𝑅3 ⇒ 𝑅1 𝑅3 ⇒ 𝑅2 16   1 0 0 − 21    0 1 0 − 9   14    0 0 1 46   21  ⇒ 𝑖1 = − 16 21 , 𝑖{2] = − 9 14 , 𝑖3 = 46 21 Question #18 Given 𝑓(𝑥) = 𝑥 𝑥 2 −9 −(𝑥 2 +9) 𝑓 ′ (𝑥) = 𝑓 ′′ (𝑥) = (𝑥 2 −9)2 2𝑥(𝑥 2 +27) (𝑥 2 −9)3 a) Provide a complete analisis of 𝑓(𝑥) and complete the passee. Show all work in the space provided below. 𝑓(𝑥) has x-intercept(s) of and y-intercept of It has two types of asymptotes: one type is asymptote(s) at and the other type is asymptote(s) at 𝑓(𝑥)decreases for and has (how many?) local extrema. The graph concaves up for and concaves down for The point(s) of inflection is at The domain of f(x) is and the range is Answer: Step 1 a) Given 𝑥 𝑓(𝑥) = 2 𝑓 ′ (𝑥) 𝑥 −9 −(𝑥 2 +9) = (𝑥 2 𝑓 ′′ (𝑥) = −9){2] 2𝑥(𝑥 2 +27) (𝑥 2 −9)2 x-intercept of 𝑓(𝑥) = (0, 0) y-intercept of 𝑓(𝑥) = (0, 0) Vertical asymptote 𝑥 = 3, 𝑥 = −3 Horizontal asymptote 𝑦 = 0 Step 2 𝑥 𝑓(𝑥) = 2 𝑥 −9 F(x) has zero local extrema points. The graph concaves up for −3 < 𝑥 < 0.3 < 𝑥 < ∞ and Concaves down for ∞ < 𝑥 < −3.0 < 𝑥 < 3 The inflection points of 𝑓(𝑥) = (0, 0) The domain of 𝑓(𝑥) = 𝑥 < −3 or −3 < 𝑥 < 3 or 𝑥 > 3 Range of 𝑓(𝑥) = (−∞, ∞) Question #19 Given 𝑓(𝑥) = 𝑥 𝑥 2 −9 −(𝑥 2 +9) 𝑓 ′ (𝑥) = 𝑓 ′′ (𝑥) = (𝑥 2 −9)2 2𝑥(𝑥 2 +27) (𝑥 2 −9)3 b) State the asymptotic and end behaviours. Answer: Step 1 b) 𝑓(𝑥) = 𝑥 𝑥 2 −9 Vertical asymptote 𝑥 = 3, 𝑥 = −3 Horizontal asymptote 𝑦 = 0 The point where the denominator equal to zero Then, 𝑥 = −3 lim − ( 𝑥⇒3 𝑥 𝑥 2 −9 )=∞ Since limit is infinity then 𝑥 = −3 is a vertical asymptote. lim + ( 𝑥⇒3 𝑥 𝑥 2 −9 )=∞ Since limit is infinite then 𝑥 = 3 is a vertical asymptote. Question #20 Use Newton's Method to determine 𝑥1 and 𝑥2 for the function 𝑓(𝑥) = 𝑥 3 − 7𝑥 2 + 8𝑥 − 3 and the value of 𝑥0 = 5 Answer: Step 1 In numerical analysis, Newton's method, also known as the Newton–Raphson method, is a root-finding algorithm which produces successively better approximations to the roots (or zeroes) of a real-valued function. The most basic version starts with a single-variable function f defined for a real variable x, the function's derivative f ', and an initial guess 𝑥0 for a root of f. If the function satisfies sufficient assumptions and the initial guess is close, then 𝑥1 = 𝑥0 − 𝑓(𝑥0 ) 𝑓 ′(𝑥0 ) is a better approximation of the root than 𝑥0 . Geometrically, (𝑥1 , 0) is the intersection of the x-axis and the tangent of the graph of f at (𝑥0 , 𝑓(𝑥0 )) : that is, the improved guess is the unique root of the linear approximations at the initial point. The process is repeated as 𝑥𝑛+1 = 𝑥𝑛 − 𝑓(𝑥𝑛 ) 𝑓 ′(𝑥𝑛 ) Step 2 𝑥1 , 𝑥2 for the function 𝑓(𝑥) = 𝑥 3 − 7𝑥 2 + 8𝑥 − 3 , and the value of 𝑥0 = 5 Step 3 𝑓(𝑥) = 𝑥 3 − 7𝑥 2 + 8𝑥 − 3 𝑓 ′ (𝑥) = 3𝑥 2 − 14𝑥 + 8 𝑓(𝑥0 ) = 𝑓(5) = 53 − 7 × (5)2 + 8 × 5 − 3 = −13 𝑓 ′ (5) = 3 × 52 − 14 × 5 + 8 = 13 𝑥1 = 𝑥0 − = 5− −13 𝑓(𝑥0 ) 𝑓 ′(𝑥0 ) 13 =6 Step 4 𝑓(𝑥) = 𝑥 3 − 7𝑥 2 + 8𝑥 − 3 𝑓 ′ (𝑥) = 3𝑥 2 − 14𝑥 + 8 𝑓(𝑥1 ) = 𝑓(6) = 63 − 7 × 62 + 8 × 6 − 3 =9 𝑓 ′ (6) = 3 × 62 − 14 × 6 + 8 = 32 𝑥1 = 𝑥0 − = 6− 9 𝑓(𝑥0 ) 𝑓 ′(𝑥0 ) 32 = 5.7188 Step 5 𝑓(𝑥) = 𝑥 3 − 7𝑥 2 + 8𝑥 − 3 𝑓 ′ (𝑥) = 3𝑥 2 − 14𝑥 + 8 𝑓(𝑥0 ) = 𝑓(5.7188) = 5.71883 − 7 × (5.7188)2 + 8 × 5.7188 − 3 = 0.8479 𝑓 ′ (5) = 3 × 5.71882 − 14 × 5.7188 + 8 = 26.0498 𝑥1 = 𝑥0 − 𝑓(𝑥0 ) 𝑓 ′(𝑥0 ) 0.8479 = 5.7188 − 26.0498 = 5.6862 Step 6 𝑓(𝑥) = 𝑥 3 − 7𝑥 2 + 8𝑥 − 3 𝑓 ′ (𝑥) = 3𝑥 2 − 14𝑥 + 8 𝑓(𝑥0 ) = 𝑓(5.6862) = 5.68623 − 7 × (5.6862)2 + 8 × 5.6862 − 3 = 0.0107 𝑓 ′ (5) = 3 × 5.68622 − 14 × 5.6862 + 8 = 25.3919 𝑥1 = 𝑥0 − 𝑓(𝑥0 ) 𝑓 ′(𝑥0 ) 0.0107 = 5.6862 − 25.3919 = 5.6858 Step 7 𝑓(𝑥) = 𝑥 3 − 7𝑥 2 + 8𝑥 − 3 𝑓 ′ (𝑥) = 3𝑥 2 − 14𝑥 + 8 𝑓(𝑥0 ) = 𝑓(5.6858) = 5.68583 − 7 × (5.6858)2 + 8 × 5.6858 − 3 =0 𝑓 ′ (5) = 3 × 5.68582 − 14 × 5.6858 + 8 = 25.3834 𝑥1 = 𝑥0 − 𝑓(𝑥0 ) 𝑓 ′(𝑥0 ) 0 = 5.6858 − 25.3834 = 5.6858 The roots are 5.6858, 5.6858