How do i code an integral that needs to be differentiated. L=int1/0(2+t^2)dt

The answer I've gotten would be . As would then be . I'm just trying to figure out how to verify/code this into MATLAB. If anyone could help I'd be very appreciative.

Full problem solving: Find the length of the curve. r(t)=<2t, t^2, 1/3t^3>, 0<=t<=1

Code written thus far:

Answer 1

[Rahul]

Hi Scott,

I understand that you're working on calculating the arc length of a 3D parametric curve defined by 'r(t)' over a defined interval and are looking to verify and implement this in MATLAB. The provided snippet seems to be an evaluating the arc length by first computing the derivative and its magnitude.

To compute the arc length of a parametric curve in 3D space, you can evaluate the integral of the norm of its first derivative over the given interval. Since the components of the vector function are continuous and differentiable polynomials, MATLAB's Symbolic Math Toolbox offers an ideal way to compute this exactly and with minimal manual effort.

Using the Symbolic Math Toolbox, the process can be broken down into the following steps:

1. Define the vector function 'r(t)' symbolically
2. Differentiate 'r(t)' with respect to t
3. Compute the Euclidean norm (magnitude) of the derivative
4. Integrate this magnitude from 't = 0' to 't = 1' to obtain the arc length

% Step 1: Symbolic definition
syms t
r = [2*t, t^2, (1/3)*t^3];

% Step 2: Derivative
dr = diff(r, t);

% Step 3: Magnitude of derivative
dr_mag = sqrt(sum(dr.^2));

% Step 4: Definite integration over [0, 1]
arc_length_sym = int(dr_mag, [0, 1]);

% Convert to numeric if needed
arc_length_val = double(arc_length_sym);

% Display results
disp('Exact Arc Length (symbolic): ');

Exact Arc Length (symbolic): 

disp(arc_length_sym)

fprintf('Arc Length (numeric): %.6f\n', arc_length_val);

Arc Length (numeric): 2.333333

For this particular curve, the integrand simplifies to 't^2 + 2', and the definite integral over [0,1] evaluates to 7/3 (approximately 2.3333). This confirms that the symbolic engine is working correctly and simplifies algebraic expressions automatically.

For more information regarding the usage of various functions mentioned in the given code snippet, you can refer to the following documentation links:

• Definite and Indefinite Integrals using 'int' function: https://www.mathworks.com/help/symbolic/sym.int.html
• Algebraic Simplification using 'simplify' function: https://www.mathworks.com/help/symbolic/sym.simplify.html

[Walter Roberson]

@Rahul

I notice that the output of disp(arc_length_sym) is dulled out. This is an Answers bug that I am currently reporting to Mathworks. I would appreciate it if you did not edit the output while Mathworks investigates.

[Rahul]

Hi @Walter Roberson,

Sure, I will not be editing the answer any further.