• Support PF! Buy your school textbooks, materials and every day products via PF Here!

Trapezoidal rule to estimate arc length error

143
5
Problem Statement
State the integral (do NOT evaluate) to compute the arc length between x = 1 and x = 5 for the function ##y=\frac{1}{x^2}## (already done)

How many intervals are required to numerically compute the integral in to an accuracy of ##10^{-3}## using the trapezoidal rule?
Relevant Equations
##L=\int_a^b \sqrt{1+(\frac{dy}{dx})^2}dx##

##E_T\leq\frac{(b-a)^3}{12n^2}[max |f^{(2)}(x)|]##
I got the first part of it down, $$L=\int_1^5 \sqrt{1+(\frac{1}{x^2})}dx$$

I just want to know if it's right to make your ##f(x)=\sqrt{1+\frac{1}{x^2}}## then compute it's second derivative and find it's max value, for the trapezoidal error formula.
 
32,344
4,130
I got the first part of it down, $$L=\int_1^5 \sqrt{1+(\frac{1}{x^4})}dx$$
Your integrand isn't right. What is ##\frac d{dx}\frac 1 {x^2}##? In your integrand above, it looks like you just squared ##\frac 1 {x^2}##, and forgot to find its derivative.
Zack K said:
I just want to know if it's right to make your ##f(x)=\sqrt{1+\frac{1}{x^4}}## then compute it's second derivative and find it's max value, for the trapezoidal error formula.
 
143
5
Your integrand isn't right. What is ##\frac d{dx}\frac 1 {x^2}##? In your integrand above, it looks like you just squared ##\frac 1 {x^2}##, and forgot to find its derivative.
Yes sorry I realized that then fixed it.
 
32,344
4,130
Yes sorry I realized that then fixed it.
It's still wrong. You need to find the derivative of 1/x^2, square it, and add 1. That will go inside the radical for your arc length integrand.
 
143
5
It's still wrong. You need to find the derivative of 1/x^2, square it, and add 1. That will go inside the radical for your arc length integrand.
Sigh... now I was thinking of integration.

It should be ##\sqrt{1+\frac{4}{x^6}}=\sqrt{\frac{x^6+4}{x^6}}##
 
32,344
4,130
It should be ##\sqrt{1+\frac{4}{x^6}}=\sqrt{\frac{x^6+4}{x^6}}##
That's more like it.

Zack K said:
##E_T\leq\frac{(b-a)^3}{12n^2}[max |f^{(2)}(x)|]##
After you get the integral squared away, there are some tricks you can use to find the max of |f2(x)|. If the function (f2(x)) is increasing on an interval [a, b], its largest value will be at x = b. If the function is decreasing, it's largest value will be at the left endpoint of the interval, x = a.
 
143
5
That's more like it.


After you get the integral squared away, there are some tricks you can use to find the max of |f2(x)|. If the function (f2(x)) is increasing on an interval [a, b], its largest value will be at x = b. If the function is decreasing, it's largest value will be at the left endpoint of the interval, x = a.
Wow that actually makes so much sense. My biggest issue is once I got the second derivative, I got this large function which seemed annoying to evaluate with x=5 (given that this question could be on an exam). Thank you.
 

Want to reply to this thread?

"Trapezoidal rule to estimate arc length error" You must log in or register to reply here.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top