Mathlab bài tập lớn giải tích 1 bktphcm

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

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PROBLEM 1 To investigate the limit of f (x) = sinx + 10−5cosx 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?

Solution:

Step 1: Find the limit of f (x) = sinx + 10−5cosx

lim

x →0 f ( x )=lim

x→ 0

(sinx+10−5 cosx)=sin 0+ 10−5.cos 0=1 10−5 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:

R =

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.

Solution:

1.R(I) is tax rate as a function of the incom I($).

Sketch the

graph:

Code:

0, if 0 ≤ I ≤ 10 000 ($) 10%, if 10 000 < I ≤ 20 000 ($) 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($).

PROBLEM 3 Find the derivative of function:

f ( x )={e

1

x , x ≠ 0

0, x=0 at x0=0

0 , no tax, 0 ≤ I ≤ 10 000 0,1.I – 10 000, 10 000 < I ≤ 20 000 0,15.I – 20 000, I > 20 000

f ( x )={e

1

x x ≠ 0

0 x=0

f '+¿(0)= lim

x→0+e

1

−0

x−0= lim

x → 0+¿ =+∞

¿ ¿ ¿

¿ ¿

f '

− ¿ ( 0 ) = lim

x→ 0− ¿e

1

−0

x−0= lim

x →0−¿ =−∞

¿¿ ¿

¿¿

Since f '+¿(0)≠ f '

− ¿ (0) ¿ ¿ does not exist Thus: f is not differentiable at x0=0

Code:

PROBLEM 4 Find asymptotes of function y= arctan 2 x

x (1−x )

Solution:

 lim

x→ ±∞ arctan (2 x )=± π

2 ; lim

x→ ±∞ x (1−x )=−∞

=> lim

x→ ±∞

arctan 2 x

x (1−x ) =0

=> y=0 is a horizontal asymptote when x → ± ∞.

 lim

x →0

arctan 2 x

x (1−x ) =limx→ 0

2 x

x (1−x )=limx → 0

2

1−x=2 (because x → 0: arctan2x ≈ 2x)

=> x=0 isn’t a vertical asymptote.



lim

x→ 1+ ¿arctan2 x

x(1−x)= lim

x→1+

arctan2 x

x

1−x =

arctan 2

0 − ¿ =−∞¿

¿¿¿

¿

=> x=1 is a vertical asymptote.

 Because x → ± ∞=0 => The function doesn’t have slant asymptote

Code:

PROBLEM 5 If f ( x )=ln (2 x +3) , find f( 100 )

(x )

Solution:

f ' ( x )= 2

2 x +3=

1

x+3

2

=¿

f ''(x )=(−1) ¿

f( 100 )

(x )=(−1) (−2) … (−99) ¿

Code:

PROBLEM 6 Evaluate ∫

−∞

+∞

dx

x2+4 x +9

Solution:

I=∫

−∞

+∞

dx

x2+4 x +9

F (x)=∫ dx

x2+4 x +9=∫ dx

(x +2)2+5

Let u=x+2 ⇒du=dx

¿∫ du

u2+(√5)2=

1

√5arctan( √u5)= 1

√5arctan(x +2√5 )

I=∫

−∞

−2

dx

x2+4 x +9+∫

−2

+∞

dx

x2+4 x +9

¿ 1

√5arctan(x +2√5 )|−−2∞+

1

√5arctan(x +2√5 )|+∞

−2

¿0−(−π

2√5)+ π

2√5−0=

π

√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 7. 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 x−x2=x√1−x ↔ x4

−2 x3+x4=x2(1−x )↔ x=0, x=1

The area of the region bounded by two curves: A=∫

0

1 (x√1−x−x+x2)=0.1 (unit of area)

Code:

PROBLEM 8 Solve the differential equation (x +2 x3)dx +(y +2 y3)dy=0.

Solution:

Integrating both sides, we receive general solution

∫ (x +2 x3)dx+∫ (y+2 y3)dy=C

↔ 1

2x

2

+1

2x

4

+1

2y

2 +1

2 y

4

=C

↔ x2

+x4

+y2

+y4=2 C=C1

Code:

PROBLEM 9 Solve the differential equation y ' '−5 y'+6 y =e−x.

Solution:

Step 1 Solve the homogeneous equation y ' '−5 y'+6 y =0

The characteristic equation k2−5 k +6=0 has 2 real different roots k1=2 and k2=3

Step 2 The homogeneous solution is y h=C1e 2 x

+C2e 3 x

Step 3 Find a particular solution of nonhomogeneous equation y ' '−5 y'+6 y =e−x

The particular solution has a form y h=xs e−xC Since α=−1 is not the root of

characteristic equation then s=0 and y p=C e−x

−¿

5 y p

' = −C e−x

1 y ' ' p = C e−x

y ' '−5 y'+6 y = 12 C e−x = e−x

→ C= 1

12

Step 4 The general solution is y gen=y h+y p=C1e 2 x+C2e 3 x+ 1

12e

−x

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 dt=−x −2 y +2 e

−t

dy

dt=3 x+4 y +e

−t .

Solution 1:

Let: {dx dt=−x −2 y

dy

dt=3 x +4 y

Matrix: (−1 23 4)

Characteristic equation:|−1−λ −2

3 4−λ|

→ (−1− λ) (4−λ )+ 6=0 ↔ λ2−3 λ+2=0↔ λ=1, λ=2

Basic solution :

With λ=1, we have : {−2 t1−2 t2=0

3 t1+3 t2=0 ↔ t2=−t1→ Eigen vector v1(1 ;−1), basic solution

x=[ e t

−e t]

With λ=2, we have : {−3 t1−2t2=0

3 t1+2 t2=0 ↔ t2=−3

2 t1→ Eigen vector v2(1 :−3

2), basic solution

y=[−3e 2 t

2 e

2 t]

Partial solution :

Let Y ∗¿C1(t ) x+ C2(t ) y is a partial solution → Y ∗¿[ C1(t) e t+C2(t)e2 t

−C1(t )e t−3

2C2(t )e2 t]

Put Y ∗¿ to the equations : { C1' (t) et+C2'(t ) e2 t=2 e−t

−C1' (t )e t

−3

2C2' (t )e 2 t

=e−t ↔{ C1' (t )=8 e−2 t

C2' (t )=−6 e−3 t →{C1(t )=−4 e−2t

C2(t )=2 e−3 t

General solution : Y = Ax+By +Y∗¿[ A e t+B e 2 t

−2e−t

−A e t−3

2B e

2t

+e−t] with A, B = const.

Solution 2:

Given {dx dt=−x −2 y +2 e

−t

dy

dt =3 x+4 y + e

−t

Using the symbol D for dt d , the given system can be written as

D x=−x−2 y+ 2 e−t→ ( D+1) x+2 y=2 e−t  and Dy=3 x + 4 y +e−t→−3 x +( D−4) y=e−t 

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 )2 e−t−2 e−t

=−2e−t−8 e−t−2 e−t

→(D2−3 D+2)x=−12 e−t ; where D2≡ d

2

d t2

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 D2x =m2e mt

The auxiliary equation of  is m2−3 m+2=0 ↔ m=1, m=2

Thus x c=C1e t+C2e 2 t; where C1, C2 are arbitrary constants

To find x p: x p= −12 e−t

D2−3 D+2=

−12 e−t (−1)2−3 (−1)+2=

−12 e−t

6 =−2 e

−t

The general solution of  is x=xc+xp=C1e t+C2e2 t−2e−t 

Then dx dt =C1e t+2C2e 2 t

+2e−t

 Using ,  and , we have:

C1e t+2 C2e 2 t+2 e−t=−C1e t−C2e 2 t+2 e−t−2 y+2 e−t

↔ C1e t+2 C2e 2 t+2 e−t+C1e t+C2e 2 t−2 e−t−2 e−t=−2 y

↔−2 y=2 C1e t+3 C2e 2t−2 e−t

↔ y=−C1e t

−3

2C2e 2 t

+e−t

The required complete solution of the given system of equations are defined by

{ x=C1e t+C2e 2 t−2 e−t

y=−C1e t−3

2C2e

2 t

+e−t ; where C1, C2 are arbitrary constants

Code: