experimental report calculus 1 discuss about newtons method

Solving for the maximal interest rate Initially, we have the formula of paying off a mortgage over a fixed period is: Substitute value onto the formula, we have equation We have first po

VIETNAM NATIONAL UNIVERSITY, HO CHI MINH CITY

HO CHI MINH UNIVERSITY OF TECHNOLOGY

(0000

Experimental Report Subject: Calculus 1

Group ID:

Lecturer teacher:

Group Information:

1 Group No: 02

2 Members:

Full name Student ID Hoang Nguyéén Đức 2252021 Anh

Trâân Ngọc Ánh Vân 2252900

Dương Nhật Thành 2252474 Nguyêên Mạnh Cường 2252099 Đoàn Nguyên Phúc 2152627

Class: CC16

Table of contents

TL Infr0CfÏOH co <5 << Hi HT lv 1

II Detailed expÏa'nafÏOTIS - so <5 << 0 in TH ng ng vn 2 Problem Í <5 <5 sọ cọ TT nà n0 2 Discuss about Newton’s method

Solving for the maximal interest rate

Problem 2 si nà n0 3 Theory to find max-min of a function on

Present code algorithm to find max-min of on

Apply code to find the shorftesf possible (ime co co 3153551551591 9595 4 Apply code to find the smallest length of the fold 5 Problem ỔỔ o- «cọ nà ni 6 Discuss the Trapezoidal metHo( - s5 55 5511 1 Y1 g1 ng 1 my, 6 Illustrate Trapezoidal method in MATLAB 7 Apply code in part a Evaluate the definite integraÌ - -.s« s««<««s« 7 Apply code in part b Compute the area of the p00IÏL s55 5< 5 sex 7 TIT MATLAB code and ExpÌanati0T -o- << <5 Y1 Y1 Y1 mg v1 men 9 Problem Í <5 <5 sọ cọ TT nà n0 9

Problem 2b Go 5S 0 TT TT TH 0 8 10

ProbilÌeim Í s5 55 0 TT TT TH 0 00 0m 12

Problem 2b Go 5S 0 TT TT TH 0 8 12

Problem SjÙ G55 cọ TT TH TT 0m 12

L Introduction

The vast subject of mathematics known as calculus includes concepts such as instantaneous rates of change, areas under curves, sequences, and series All of these subjects revolve around the concept of a limit, which involves analysing a function's behaviour at places that are progressively closer to a given point but never actually reach it The two primary applications of calculus are integral calculus and differential calculus

The calculus was independently developed by Sir Isaac Newton and the

German Gottfried Leibniz They were both working on motion difficulty studies at the end of the 17th century The argument between the men over who invented calculus first was fierce

The ability to predict the locations of the stars was found to be essential for assisting in nautical navigation Finding longitude while a ship was at sea was the most challenging task Any government would prosper if it could successfully send ships to the New World and successfully return them with cargo

II Detailed explanations

Problem 1

Discuss about Newton’s method

Newton method or Newton Raphson method is a technique for solving equations of the form by successive approximation The idea is to pick an initial guess such that is reasonably close to zero We then find the equation of the line tangent to at and follow it back to the axis at a dew guess The formula for this is

We then find the equation of the line tangent to at and follow it back to the axis to get anew (and improved!) guess from the formula

We keep on refining our guesses until we are close enough for whatever application we have in mind In general, we have the recursive formula

In typical situations, Newton's method homes in on the answer extremely quickly, roughly doubling the number of decimal points in each round Therefore, if your original guess is good to one decimal place, 5 rounds later you will have an answer good to digits

Solving for the maximal interest rate

Initially, we have the formula of paying off a mortgage over a fixed period is: Substitute value onto the formula, we have equation

We have first point

After using MATLAB to run Newton method, we get the answer is Therefore, the maximal interest rate the borrower can afford to pay is 8.10% per year

Problem 2

Theory to find max-min of a function on

In calculus, you might have learnt how to find the local maximum and local minimum values of a function using the first derivative test and second derivative test

As we know, the function has neither a local maximum value nor a local minimum value In this article, you will learn how to find the minimum and maximum values of

a function in a closed interval in detail

Step 1: Identify all critical points of the function f in the g1ven Interval, 1.e., these are the values of where either or fis not differentiable

Step 2: Choose the endpoints of the interval

Step 3: At all these points, i.e the values listed in the previous steps, i.e steps

1 and 2, evaluate the values of the given function

Step 4: Identify the minimum and maximum values of f out of the values estimated in the previous step

This maximum value will be the absolute maximum or the greatest, whereas the minimum value will be the absolute minimum or the least value of the function Present code algorithm to find max-min of on

In our code, first we will find the derivative of by MATLAB

Then we will use solve() command to find critical points of

Next, we will find the value of at critical points, lower bounded and upper bounded by command subs(), assume lower limit and upper limit are

Finally, we will use basic sort algorithm to find the value to find max/min value In this example, | will sort for the min value

You can invert the sign from “<” to “>” to find the max value

Apply code to find the shortest possible time

Figure 1

We have is bounded by

We have the distance of the woman from A to B is and distance arc from B to Cis

Now time taken to row is and time taken to walk is , so we can compute the total time equation

Next, we will find the critical points and examine whether it is min value or not Now we will find second derivative of

So is the local maxima, we can declined

We will examine at the lower limit and upper limit

So the smallest time is 1.54s, that means she has to walk the entire arc ABC in order to minimize her travel

Apply code to find the smallest length of the fold

First, we need to find the interval of x

a(S \z-1a

As you can see, the lower limit of x 1s 4.5836 and the upper limit of x is 8 Next, we need to find equation with y and x

D

5”

v

We have triangle CDE and triangle BCA are similar Therefore,

Let

Now, we need to find the derivative of respect to

Then, we will find the value of at the critical point, at the upper and lower limit

Therefore, result to smallest In conclusion, to minimize the length of the fold, should be equal to six

Problem 3

Discuss the Trapezoidal method

In Calculus, ““Trapezoidal Rule” is one of the important integration rules The name trapezoidal is because when the area under the curve is evaluated, then the total area is divided into small trapezoids instead of rectangles This rule is used for approximating the definite integrals where it uses the linear approximations of the functions

Trapezoidal Rule is a rule that evaluates the area under the curves by dividing the total area into smaller trapezoids rather than using rectangles This integration works by approximating the region under the graph of a function as a trapezoid, and it calculates the area This rule takes the average of the left and the nght sum

Let be a continuous function on the interval Now divide the intervals into equal subintervals with each of width,

Such that Then the Trapezoidal Rule formula for area approximating the definite integral

Is given by:

Where,

If, R.H.S of the expression approaches the definite integral

Illustrate Trapezoidal method in MATLAB

So we have the trapezoidal formula is:

With

Therefore, first we will compute

Then we will create an array , then the sum in the middle part is calculated in

MATLAB by:

The before * means that we are multiply every elements in array with Then the sum using trapezoidal result is presented by:

That is our result

Apply code in part a Evaluate the definite integral

In this part, we have the lower limit is 1, the upper limit is 5, then we will input equation onto the Command Window

After MATLAB executes the trapezoidal formula, we have the result is 34.02, which is quite precisely from the calculator: 34.01882640944523

Apply code in part b Compute the area of the pool

Initially, we have matrix S include the width of the pool

And

Then we have the Trapezoidal formula:

So the area of the pool by using the Trapezoidal formula is 80.8

Problem 1

% Problem 1: Newton method

% File Problem1.m

% Clear the Workspace and Command Window

clear variables; clc;

% Define function f(x)

syms x;

f=1000* (1- (1+x)*(-360))-135000*x; % nter the Function here

g=diff(f); % The Derivative of the Function

% Limit the number of decimal places is 10

epsilon = 5#19^- (1Ø+1);

x@ = input('Enter the intial approximation: ');

for i=1:100

f0=vpa(subs(,x,x@)); % Calculating the value of function at x@

f@_der=vpa(subs(g,x,x@)); % Calculating the value of function derivative at

xo

y=x0-f0/fe_der; % The Formula

err=abs(y-xÐ) ;

if err<epsilon %checking the amount of error at each iteration

break

end

xO=y;

end

y =y - rem(y,19^-19); %Displaying upto required decimal places

fprintf('So the maximal interestrate is: %.2*%% per year\n' ,y*12*1@@) ;

Problem 2a

% Problem 2a: Find the shortest way for the woman

% File Problem2a.m

% Clear the Workspace and Command window

clear variables; clc;

syms theta

% Define borders

D = [9;pi/2];

assume (D(1)<=theta<=D(2))

% Define the S1 function and S2 function

s1 = 4#cos(theta);

s2 = 2*2*theta;

fprintf("From the figure, you have function of S1 = %s”", s1);

fprintf("\nAnd S2 = %s\n", s2);

% Time to taken to row t1 and time to walk t2 and sum of it

t1 = $s1/2; t2 = s2/4;

fprintf ("Now time taken to row is ti = %s", t1);

fprintf(" and t2 = %s",t2);

fprintf("\nTotal time = ti + t2 = %s\n", T);

% Use function smallest to find the smallest time

[ans,angle] = smallest(D, T);

fprintf("So the smallest time is %.2f s\n", subs(T,theta,ans)); fprintf("That means she has to walk entire arc ABC in order” +

to minimize her travle theta = %.2f rad", angle);

function [d,angle] = smallest(D, T)

syms theta; assume(D(1) <= theta <= D(2));

dT = diff(T);

S = solve(dT == 9, theta);

Ta = subs(T,D(1)); Tb = subs(T,D(2));

Ttheta = subs(T,S);

if (Ta < Tb) && (Ta < Ttheta)

d = Ta;

angle = D(1);

else if (Tb < Ttheta)

d = Tb;

angle = D(2);

else

d = Ttheta;

angle = S;

end

end

end

Problem 2b

% Problem 2b: Find x to minimize y

% File Problem2b.m

% Clear Workspace and Command Window

clear variables; clc;

syms x;

D = [4.5836 8];

% Define function y

y = sqrt(x^2 + 4*x^2/(x-4));

fprintf("We have the bounded of x is [%.2f %.2f].\n”,D(1),D(2)); fprintf("And the equation of y is %s.\n", y)3

% Using function to find the smallest value

[y, x] = smallest(D, y);

% Print output

fprintf( "After calculating in the function smallest, "+

“the minimize length y is %.2f when x = %.2f.", y, x); function [y,x] = smallest(D, y)

syms x; assume(x>@);

dY = diff(y);

S = vpasolve(dY == @,x, [D(1) D(2)]);

Ya = subs(y,D(1));

Yb = subs(y,D(2));

Ylocal = subs(y,S);

if (Ya < Yb) && (Ya < Ylocal)

y = Ya;

x = D(1);

else if (Yb < Ylocal)

y = Yb;

x = D(2);

else

y = Ylocal;

x = S3

end

end

Problem 3a

% Calculate the trapezoidal rules

% File Problem3a.m

% Clear the Workspace and Command Window

clear variables; clc;

% Define f(x)

syms f(x);

% Enter value of lower, upper limit and equation f(x)

a = input("Enter lower limit a: “);

b = input("Enter upper limit b: "};

f(x) = input("Enter equations: ");

S = trapezoidal(a,b,f(x)); % Call function trapezoidal

fprintf("The value of integration is %.2f\n", S$);

function out = trapezoidal(a,b,f)

syms x;

n = 100; % Number of itterations

h = (b-a)/n;

i = 1:1:n-1;

S = subs(f,x,ati.*h);

out = (h./2).*(subs(f,x,a)+2.*sum(S)+subs(f,x,b));

end

Problem 3b

% Compute the area of the pool

% File Problem3b.m

% Clear the Workspace and Command Window

clear variables; clc;

% Define the widths of the swimming pool in matrix S

S = [@ 6.2 7.2 6.8 5.6 5.0 4.8 4.8 0];

% Find delta x

deltax= (16-0)/8;

% Define the trapezoidal formula T = deltax/2*sum(S)/2

T = deltax/2*sum(S)*2;

% Print some statements

fprintf("Initially, we have matrix S concludes: \n");

disp(S);

fprintf("And deltax = (b-a)/N = (16-9)/8 = %.2f", deltax);

fprintf("\nThe trapezoidal formula: S = deltax/2*sum(f(k+1)+f(k)) = "+

"deltax/2*sum(S)*2 = %.2f", T);

10

11%

JA Dac

Vv INXS

Problem 1

II

V_References Y-.—^- =

[1] Christopher G Gibson Elementary geometry of differentiable curves: an

undergraduate introduction Cambridge, UK ; New York : Cambridge University Press, 2001

[2] Shoichiro Nakamura Numerical analysis and graphic visualization with

MATLAB Upper Saddle River, N.J : Prentice Hall PTR, ©2002

[3] Peter L, K., 2021 MATLAB for Begineers: A Gentle Approach Revised Edition Peter I Kattan (September 24, 2009)

[4] J Edwards (1892) Differential Calculus London: MacMillan and Co pp 143 ff Cajori, Florian (2007-01-01) A History of Mathematical Notations Torino: Cosimo, Inc ISBN 9781602067141

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