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PROBLEM 1. To investigate the limit of f (x) = sinx + {\mathbf{10}}{\mathbf{5}}cosx as x → 0, set your graphing calculator or computer to display exactly four digits to the right of the decimal point. After calculating f (x) with x = 0.1; 0.001; 0.00001; 0.0000001,..., what do you conclude? (Your answer may depend on how your particular calculator works.) Now zoom in on the y−intercept of the curve y = f (x) sufficiently to show that the value of the limit is nonzero. What is it? PROBLEM 2. In a certain country, income tax is assessed as follows. There is no tax on income up to 10000. Any income over 10000 is taxed at a rate of 10%, up to an income of 20000. Any income over 20000 is taxed at 15%. 1. Sketch the graph of the tax rate R as a function of the income I. 2. How much tax is assessed on an income of 14000? 3. Sketch the graph of the total assessed tax T as a function of the income I. PROBLEM 3. Find the derivative of function: \mathbit{f}\left(\mathbit{x}\right)=e1x,x≠00,x=0 at \mathbit{x}_\mathbf{0}=\mathbf{0}
VIET NAM NATIONAL UNIVERSITY HO CHI MINH CITY HCMC UNIVERSITY OF TECHNOLOGY CALCULUS I PROJECT Complete day: Ho Chi Minh City – May 15, 2021 TABLE OF CONTENTS \ PROBLEM To investigate the limit of f (x) = sinx + 10−5 cosx as x → 0, set your graphing calculator or computer to display exactly four digits to the right of the decimal point After calculating f (x) with x = 0.1; 0.001; 0.00001; 0.0000001, , what you conclude? (Your answer may depend on how your particular calculator works.) Now zoom in on the y−intercept of the curve y = f (x) sufficiently to show that the value of the limit is nonzero What is it? Solution: Step 1: Find the limit of f (x) = sinx + 10−5 cosx lim f ( x )=lim ( sinx+ 10−5 cosx )=sin 0+ 10−5 cos 0=1 10−5 x →0 x→ Step 2: Calculating f(x) with x = 0,1; 0,001; 0,00001; 0,0000001 f ( 0,1 )=0,099 f ( 0,01 )=1,0099.10−3 f ( 0,00001 ) =2.10−5 f ( 0,0000001 ) =1,01.10−5 Step 3: Conclusion The value of the limit is nonzero Code: PROBLEM In a certain country, income tax is assessed as follows There is no tax on income up to $10000 Any income over $10000 is taxed at a rate of 10%, up to an income of $20000 Any income over $20000 is taxed at 15% Sketch the graph of the tax rate R as a function of the income I How much tax is assessed on an income of $14000? Sketch the graph of the total assessed tax T as a function of the income I Solution: 1.R(I) is tax rate as a function of the incom I($) Sketch the 0, if ≤ I ≤ 10 000 ($) graph: R= 10%, if 10 000 < I ≤ 20 000 ($) Code: 15%, if I > 20 000 ($) 2.Tax is assessed on an income of $14000: 10 00 ($) : no tax rate 14 00 ($): 10% tax rate => Tax is assessed = 10%.(14 000 – 10 000) = 400 ($)Code: Code: 2.T(I) is total assessed tax T as a function of incom I($) 0 , no tax, ≤ I ≤ 10 000 0,1.I – 10 000, 10 000 < I ≤ 20 000 0,15.I – 20 000, I > 20 000 Code: PROBLEM Find the derivative of function: { x f ( x )= e , x ≠ at x 0=0 0, x=0 Solution: { f ( x )= e x x ≠ 0 x=0 f '+¿ (0 )= x→0 +¿ x lim e −0 = x−0 ¿¿ lim x ¿¿ ¿ e x → 0+ ¿ =+ ∞ x f '−¿ (0 )= −¿ x→ x lim ¿¿ e −0 = x−0 lim x →0 −¿ ' Since f +¿ (0 )≠ f x e =−∞ x ' −¿ (0) ¿ ¿¿¿ ¿ does not exist Thus: f is not differentiable at x 0=0 Code: PROBLEM Find asymptotes of function y= Solution: arctan x x (1−x ) lim arctan ( x ) =± x→ ±∞ π ; lim x ( 1−x )=−∞ x→ ±∞ arctan x lim =0 => x→ ±∞ x (1−x ) => y=0 is a horizontal asymptote when x → ± ∞ arctan x =lim lim x →0 x (1−x ) x→ 2x =lim =2 (because x → 0: arctan2x ≈ 2x) 1−x x (1−x ) x → => x=0 isn’t a vertical asymptote lim +¿ x→ arctan2 x = x(1−x) ¿ lim x→1 +¿ arctan2 x x arctan = −¿ ¿ 1−x =−∞ ¿¿¿ => x=1 is a vertical asymptote Because x → ± ∞=0 => The function doesn’t have slant asymptote Code: PROBLEM If f ( x )=ln ( x +3 ) , find f (100) ( x ) Solution: f ' ( x )= = =¿ x +3 x+ f ' ' ( x )=(−1 ) ¿ f (100 ) ( x ) =(−1 ) (−2 ) … (−99 ) ¿ Code: +∞ PROBLEM Evaluate ∫ x 2+dx x +9 −∞ Solution: +∞ I=∫ −∞ dx x +4 x +9 F (x)=∫ dx dx =∫ x + x +9 ( x +2) + Let u=x+2 ⇒du=dx ¿∫ du u x +2 = arctan = arctan √5 √ √5 u +( √ 5) √ ( ) −2 I=∫ −∞ +∞ dx dx +∫ 2 x +4 x +9 −2 x +4 x +9 x +2 −2 x +2 + ∞ arctan + arctan √5 √ −∞ √ √ −2 ( )| −π π π ¿ 0−( + −0= √5 ) √ √5 ¿ ( ) ( )| Plan: We will use symbolic expressions and other commands to solve this math problem in MATLAB pretty(X) prints symbolic output of X in a format that resembles typeset mathematics int(f, x = a b) computes the definite integral Code: PROBLEM Find the area of the region bounded by the curve y=x −x2 and y=x √ 1−x Solution: The points of intersection of y=x −x2 and y=x √ 1−x are defined by the equation 4 x−x 2=x √ 1−x ↔ x −2 x + x =x (1−x ) ↔ x=0 , x=1 1 The area of the region bounded by two curves: A=∫ (x √ 1−x−x+ x )=0.1 (unit of area) Code: PROBLEM Solve the differential equation ( x +2 x3 ) dx + ( y +2 y ) dy=0 Solution: Integrating both sides, we receive general solution ∫ ( x +2 x ) dx+∫ ( y+ y ) dy=C ↔ 4 x + x + y + y =C 2 2 ↔ x 2+ x + y 2+ y 4=2 C=C Code: PROBLEM Solve the differential equation y ' ' −5 y ' +6 y =e−x Solution: Step Solve the homogeneous equation y ' ' −5 y ' +6 y =0 The characteristic equation k 2−5 k +6=0 has real different roots k =2 and k 2=3 Step The homogeneous solution is y h=C e2 x +C e3 x Step Find a particular solution of nonhomogeneous equation y ' ' −5 y ' +6 y =e−x The particular solution has a form y h=x s e− x C Since α =−1 is not the root of characteristic equation then s=0 and y p=C e−x yp −¿ y ' p y 'p' → C= 12 y ' ' −5 y ' +6 y = C e− x = −C e−x = C e− x = 12 C e−x = e− x 2x 3x Step The general solution is y gen= y h + y p=C e +C e + −x e 12 Plan: We will need to first write the characteristic polynomial in order to find the roots, which are not necessarily real-valued by entering the code to demonstrate how to find the roots of the characteristic equation of the given homogeneous ODEs Thus, we will find the homogeneous solution from the characteristic equation Then we will need find a particular solution by entering the code to demonstrate how we obtain the coefficients of a particular solution using the method of undetermined coefficients for the ODE Finally, the general solution is equal to the sum of the homogeneous and particular solutions Code: PROBLEM 10 Solve the system of the 1-st order differential equations { dx =−x −2 y +2 e−t dt dy −t =3 x+ y + e dt Solution 1: dx =−x −2 y dt Let: dy =3 x +4 y dt { (−1 24 ) Matrix: |−1−λ Characteristic equation: | −2 4−λ → (−1− λ ) ( 4−λ )+ 6=0 ↔ λ 2−3 λ+2=0 ↔ λ=1, λ=2 Basic solution : −2 t −2 t =0 { With λ=1, we have : t +3 t =0 ↔ t 2=−t → Eigen vector v1 (1 ;−1), basic solution x= et −et [ ] −3 t −2t =0 { −3 t → Eigen vector v 2(1 :− ), basic solution ↔ t 2= With λ=2, we have : t 1+2 t =0 2 e2 t y= −3 t e [ ] Partial solution : C1 ( t ) e t +C (t)e2 t Let Y ∗¿ C1 ( t ) x+ C2 ( t ) y is a partial solution → Y ∗¿ −C ( t ) e t − C2 (t ) e2 t [ ] C'1 ( t ) e t +C '2 ( t ) e2 t =2 e−t C'1 ( t )=8 e−2 t C1 ( t ) =−4 e−2t ↔ ' → Put Y ∗¿ to the equations : −C'1 ( t ) e t − C'2 ( t ) e t=e−t C2 ( t )=−6 e−3 t C ( t )=2 e−3 t { { A et + B e t−2e−t General solution : Y = Ax+ By +Y∗¿ −A e t− B e 2t +e−t [ ] with A, B = const { Solution 2: dx =−x −2 y +2 e−t dt Given dy =3 x+ y + e−t dt { d , the given system can be written as dt D x =−x−2 y+ e−t → ( D+1 ) x+2 y=2 e−t and D y =3 x + y +e−t →−3 x + ( D−4 ) y=e−t Using the symbol D for To eliminate y, we operate equation by ( D−4 ) and multiply equation by −2 and then adding, we get: [ ( D−4 )( D+1 ) +6 ] x=( D−4 ) e−t −2 e−t=−2e−t −8 e−t −2 e−t → ( D 2−3 D+2 ) x=−12 e−t ; where D ≡ d2 dt2 Let x=x c + x p be the general solution of equation , where x c: complementary function and x p: particular integral To find x c: Let x=e mt be a trial solution of ( D2−3 D+2 ) x=0 ; where m is a constant Then Dx=me mt and D2 x =m2 e mt The auxiliary equation of is m2−3 m+2=0 ↔ m=1, m=2 Thus x c =C e t +C e t ; where C 1, C are arbitrary constants To find x p: x p= −12 e−t −12 e−t −12 e−t = = =−2 e−t 2 D −3 D+2 (−1 ) −3 (−1 ) +2 The general solution of is x=x c + x p =C1 e t +C e t−2e−t Then dx =C1 e t +2C e2 t + 2e−t dt Using , and , we have: C e t +2 C2 e t +2 e−t=−C1 et −C e2 t +2 e−t −2 y+2 e−t ↔ C e t +2 C2 e t +2 e−t +C e t +C e2 t −2 e−t −2 e−t =−2 y ↔ −2 y=2 C1 e t +3 C2 e 2t −2 e−t t 2t −t ↔ y=−C1 e − C e + e The required complete solution of the given system of equations are defined by { x=C et +C e2 t −2 e−t ; where C 1, C are arbitrary constants y=−C et − C e t +e−t Code: