Đang tải... (xem toàn văn)
VIETNAM NATIONAL UNIVERSITY – HO CHI MINH CITY HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY CALCULUS 1 – ASSIGNMENT SUBMISSION GROUP 8 VIETNAM NATIONAL UNIVERSITY – HO CHI MINH CITY HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY CALCULUS 1 – ASSIGNMENT SUBMISSION GROUP 8 VIETNAM NATIONAL UNIVERSITY – HO CHI MINH CITY HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY CALCULUS 1 – ASSIGNMENT SUBMISSION GROUP 8
VIETNAM NATIONAL UNIVERSITY – HO CHI MINH CITY HO CHI MINH CITY UNIVERSITY OF TECHNOLOGY CALCULUS – ASSIGNMENT SUBMISSION GROUP 8: …………………………………………………………………………………… …………………………………………………………………………………… Instructor: Lê Thái Thanh Group members Trần Đình Đăng Khoa Phan Anh Quân Lê Xuân Phúc Lê Minh Tuấn Student ID 2211649 2212809 2153698 2153943 Ho Chi Minh City, Vietnam – 2022 Question A cell phone plan has a basic charge of $35 a month The plan includes 400 free minutes and charges 10 cents for each additional minute of usage Write the monthly cost C as a function of the number x of minutes used and graph C as function of x for ≤ x ≤ 600 Solution: 10 cents = $0.1 35, 0.1𝑥 + 35, 𝐶 (𝑥 ) = { 𝑖𝑓 𝑥 ≤ 400 𝑖𝑓 𝑥 > 400 Where as: C(x) : cost ($) x : time used (minutes) GRAPH OF C(X) 100 95 90 80 C(X) ($) 70 60 50 40 35 35 30 20 10 0 100 200 300 X (MINUTES) 400 500 600 Question Evaluate lim(𝑒 𝑥 + 𝑒 −𝑥 − 2) cot 𝑥 𝑥→0 Solution: lim (𝑒 𝑥 + 𝑒 −𝑥 − 2) cot 𝑥 𝑥→0 = lim cot 𝑥 𝑥→0 = lim 𝑥→0 = lim 𝑥→0 = −𝑥 (𝑒𝑥 + 𝑒 −2) (𝑒 𝑥 + 𝑒 −𝑥 −2) tan 𝑥 (𝑒 𝑥 − 𝑒 −𝑥 ) 1+𝑥2 (𝑒 − 𝑒 −0 ) =0 1+02 (Applying L’Hôpital’s Rule) Question Sketch the graph of the function y = 𝑥 −3𝑥 x2 −1 Solution: • Domain of y: 𝑥 ≠ −1 𝑥 ≠ x − ≠ => { 𝐷 = 𝑅/{−1, 1} • Roots of y: 𝑥 −3𝑥 x2 −1 • = => 𝑥= 𝑥 − 3𝑥 = => [𝑥 = −√3 𝑥 = √3 Asymtotes: o Vertical: 𝑥3 − 3𝑥 = +∞ 𝑥→−1 x2 − 𝑥3 − 3𝑥 lim+ = −∞ 𝑥→−1 x2 − lim− => x = -1 is a vertical asymtote of y 𝑥3 − 3𝑥 lim = +∞ 𝑥→1− x2 − 𝑥3 − 3𝑥 lim = −∞ 𝑥→1+ x2 − => x = is a vertical asymtote of y o Horizontal 𝑥3 − 3𝑥 lim = −∞ 𝑥→−∞ x2 − 𝑥3 − 3𝑥 lim = +∞ 𝑥→+∞ x2 − => The function y has no horizontal asymtote • First order derivative of y: y’ = 𝑑 𝑑𝑥 𝑥 −3𝑥 ( x2−1 ) (𝑥 − 3𝑥 )′(x − 1) − (𝑥 − 3𝑥 )(x − 1)′ = (𝑥 − 1)2 (3𝑥 − 3)(x − 1) − (𝑥 − 3𝑥 )(2x) = (𝑥 − 1)2 (3𝑥 − 6𝑥 + 3) − (2𝑥 − 6𝑥 ) = (𝑥 − 1)2 𝑥4 + = (𝑥 − 1)2 => The function is monotonically increasing on its Domain • Graph: Question Given a function 𝑦 = 𝑦(𝑥) in the implicit form 𝑒 𝑦 + 𝑥𝑦 = 𝑒 Find 𝑦 ′ (0) if 𝑦(0) = Differentiate both sides: 𝑑 𝑥 𝑑 (𝑒 + 𝑥𝑦) = (𝑒) 𝑑𝑥 𝑑𝑥 𝑑 𝑦 𝑑 (𝑥𝑦) = => (𝑒 ) + 𝑑𝑥 𝑑𝑥 => 𝑑𝑦 𝑑 𝑑 => 𝑒 𝑦 ( (𝑦)) + 𝑥 [ (𝑦)] + 𝑦 [ (𝑥)] = 𝑑𝑥 𝑑𝑥 𝑑𝑥 => 𝑒 𝑦 ( 𝑑𝑦 𝑑𝑦 )+𝑥( )+𝑦 = 𝑑𝑥 𝑑𝑥 𝑑𝑦 ) (𝑒 𝑦 + 𝑥) = −𝑦 𝑑𝑥 𝑑𝑦 −𝑦 => ( ) = 𝑦 𝑑𝑥 𝑒 +𝑥 −𝑦 => 𝑦 ′ (𝑥) = 𝑦 𝑒 +𝑥 => ( Since, 𝑦(0) = => 𝑥 = and 𝑦 = Then, 𝑦’(0) = −1 −1 = 𝑒1 + 𝑒 +∞ Question Evaluate ∫2 +∞ √(𝑥 + 5)3 𝑥 𝑑𝑥 𝐼 = ∫ 𝑥 𝑑𝑥 √(𝑥2 + 5) Let 𝑡 = 𝑥 + => 𝑑𝑡 = 2𝑥𝑑𝑥 Lower limit: 22 +5=9 Upper limit: +∞ ∞ 𝐼=∫ 𝑑𝑡 𝑡 3/2 ∞ −3/2 = ∫ 𝑡 𝑑𝑡 𝑡 −2+1 = 2−3 + = − −𝑡 = −(0 − 9−2 ) = Because as 𝑡 → +∞, 𝑡 → => √𝑡 →0 Question Find the arc length of the curve 𝑦 = (3 − 𝑥)√𝑥 , where ≤ 𝑥 ≤ 1 𝑦 = (3 − 𝑥)√𝑥 = (3√𝑥 − 𝑥 3/2 ) 3 3 √𝑥 => 𝑦’ = ( − √𝑥) = − √𝑥 2 √𝑥 => + (𝑦′)2 = + ( =1+ =( 2√𝑥 4𝑥 𝑥 + − √𝑥 )2 + ( )2 + 2√𝑥 − √𝑥 ) √𝑥 2√𝑥 =( The arc length of the curve is: 𝑏 L = ∫𝑎 √1 + (𝑦′)2 𝑑𝑥 = ∫0 ( √𝑥 = √𝑥 + = √3 + √𝑥 ) 𝑑𝑥 𝑥 3/2 2√𝑥 + √𝑥 ) Question Solve the differential equation 𝑦 ′ + 5𝑦 = −3𝑒 2𝑥 Multiply each term by the integral factor 𝑒 5𝑥 𝑒 5𝑥 𝑑𝑦 + 5𝑦𝑒 5𝑥 = −3𝑒 7𝑥 𝑑𝑥 Integrate on each side ∫ 𝑑𝑦 5𝑥 [𝑒 𝑦] 𝑑𝑥 = ∫ −3𝑒 7𝑥 𝑑𝑥 𝑑𝑥 Integrate the left side 𝑒 5𝑥 𝑦 = ∫ −3𝑒 7𝑥 𝑑𝑥 Integrate the right side 𝑒 5𝑥 𝑦 = − 𝑒 7𝑥 + 𝐶 Divide each term in 𝑒 5𝑥 𝑦 = − 𝑒 7𝑥 + 𝐶 give 𝑒 5𝑥 and abbreviated 𝑦=− 3𝑒 2𝑥 + 𝐶 𝑒 5𝑥 Question Solve the differential equation 𝑦 ′′ + 6𝑦 ′ + 5𝑦 = 8𝑥𝑒 −𝑥 𝑑 𝑑2 5𝑦(𝑥) + 𝑦(𝑥) + 𝑦(𝑥) = 8𝑥𝑒 −𝑥 𝑑𝑥 𝑑𝑥 This differential equation has the form: y’’ + p*y’ + q*y = s, where p = 6, q = 5, s = −8𝑥𝑒 −𝑥 It is called linear inhomogeneous second-order differential equation with constant coefficients The equation has an easy solution We solve the corresponding homogeneous linear equation y’’ + p*y’ + q*y = s, First of all we should find the roots of the characteristic equation 𝑞 + (𝑘 + 𝑘𝑝) = In this case, the characteristic equation will be: 𝑘 + 6𝑘 + = - this is a simple quadratic equation The roots of this equation: 𝑘1 = −5 𝑘2 = −1 As there are two roots of the characteristic equation, and the roots are not complex, then solving the correspondent differential equation looks as follows: 𝑦(𝑥) = 𝐶1 𝑒 𝑥𝑘1 + 𝐶2 𝑒 𝑥𝑘2 𝑦(𝑥) = 𝐶1 𝑒 −5𝑥 + 𝐶2 𝑒 −𝑥 We get a solution for the correspondent homogeneous equation Now we should solve the inhomogeneous equation y’’ + p*y’ + q*y = s, Use variation of parameters method Suppose that C1 and C2 - it is functions of x The general solution is: 𝑦(𝑥) = 𝐶1 (𝑥)𝑒 −5𝑥 + 𝐶2 (𝑥)𝑒 −𝑥 where C1(x) and C2(x) by the method of variation of parameters, we find the solution from the system: 𝑦1 (𝑥) 𝑑 𝑑 𝐶1 (𝑥) + 𝑦2 (𝑥) 𝐶2 (𝑥) = 𝑑𝑥 𝑑𝑥 𝑑 𝑑 𝑑 𝑑 𝐶1 (𝑥) 𝑦1 (𝑥) + 𝐶2 (𝑥) 𝑦2 (𝑥) = 𝑓(𝑥) 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 The free term f = -s, or 𝑓(𝑥) = 8𝑥𝑒 −𝑥 So, the system has the form: 𝑒 −𝑥 𝑑 𝑑 𝐶2 (𝑥) + 𝑒 −5𝑥 𝐶 (𝑥) = 𝑑𝑥 𝑑𝑥 𝑑 𝑑 𝑑 𝑑 𝐶1 (𝑥) 𝑒 −5𝑥 + 𝐶2 (𝑥) 𝑒 −𝑥 = 8𝑥𝑒 −𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 𝑑𝑥 Solve the system: 𝑑 𝐶 (𝑥) = −2𝑥𝑒 4𝑥 𝑑𝑥 𝑑 𝐶 (𝑥) = 2𝑥 𝑑𝑥 -it is the simple differential equations, solve these equations 𝐶1 (𝑥) = 𝐶3 + ∫(−2𝑥𝑒 4𝑥 )𝑑𝑥 𝐶2 (𝑥) = 𝐶4 + ∫ 2𝑥 𝑑𝑥 Substitute found C1(x) and C2(x) to 𝑦(𝑥) = 𝐶1 (𝑥)𝑒 −5𝑥 + 𝐶2 (𝑥)𝑒 −𝑥 The final answer: 𝑦(𝑥) = 𝐶3 𝑒 −5𝑥 + 𝐶4 𝑒 −𝑥 +𝑥 𝑒 Where c3 and c4 is a constants 𝑥 𝑦(𝑥) = (𝐶1 + 𝐶2 𝑒 −4𝑥 + 𝑥 − ) 𝑒 −𝑥 2 −𝑥 𝑥𝑒 −𝑥 𝑒 −𝑥 − + Question Solve the system of 1-st order differential equations 𝑥 ′(𝑡) = 𝑥(𝑡) + 2𝑦(𝑡) + { ′ 𝑦 (𝑡) = 2𝑥(𝑡) − 𝑦(𝑡) + Let 𝑃 ∶ 𝑥 ′ (𝑡) = 𝑥(𝑡) + 2𝑦(𝑡) + } 𝑄 ∶ 𝑦 ′ (𝑡) = 2𝑥(𝑡) − 𝑦(𝑡) + • 𝑄 – two times 𝑃 => 𝑦’ – 2𝑥’ = −5𝑦 (1) • Differentiate 𝑄 => 𝑦’’ = 2𝑥’ – 𝑦’ (2) From (1) and (2) => 𝑦’’ = 5𝑦 • Characteristic equation: 𝑘 − 5𝑘 = => 𝑘 = (double root) 5𝑡 => 𝑦 = 𝐶₁𝑒 + 𝐶₂𝑡𝑒 5𝑡 5𝑡 => 𝑦’ = 5𝐶₁𝑒 + 𝐶₂𝑒 5𝑡 + 5𝐶₂𝑡𝑒 5𝑡 We also have 𝑥 = −𝑦−𝑦′ +2 (𝑄) Substitute 𝑦 and 𝑦’ in 𝑄 We obtain 𝑥 = −3𝐶₁𝑒 5𝑡 − 3𝐶₂𝑡𝑒 5𝑡 − 𝐶₂𝑒 5𝑡 + Question 10 An aquarium 𝑚 long, 𝑚 wide, and 𝑚 deep is full of water Find the work needed to pump half of the water out of the aquarium (Use the fact that the density of water is 1000 𝑘𝑔/𝑚3 ) We got: 𝑊 = 𝐹 𝑠 In this case, we divide upper half of the aquarium into horizontal slices and use Intergral to calculate Work we need to pump We use set the range from which is the top of the Aquarium to its half position which is ½ So we got our General Formula for Work: W = ∫01/2 (𝑑𝑒𝑛𝑠𝑖𝑡𝑦) (𝑎𝑟𝑒𝑎 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑠𝑙𝑖𝑐𝑒𝑠) (𝑔𝑟𝑎𝑣𝑖𝑡𝑦) (𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒) In which: - Density of water = 1000 𝑘𝑔/𝑚3 - Area of the aquarium is the area of each slices o 𝐴 = 𝐿𝑒𝑛𝑔𝑡ℎ 𝑊𝑖𝑑𝑡ℎ = 2.1 = 𝑚2 - Gravity: The force from the Earth ( 𝑔 = 9.8 𝑚/𝑠 ) - Distance: We give Distance a variable 𝑥 to calculate the work with intergral ∫ 1000 × × 9.8 × 𝑥𝑑𝑥 1/2 1000 × × 9.8 × ∫0 𝑥𝑑𝑥 19600 × ( × From this calculation we got 𝑊 = 2450 𝐽 𝑥 | 02 )