Trapezoidal rule to estimate arc length error

[Zack K]

Homework 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.

 

[Mark44]

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.

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.

 

[Zack K]

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.

 

[Mark44]

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.

 

[Zack K]

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 $\sqrt{1+\frac{4}{x^6}}=\sqrt{\frac{x^6+4}{x^6}}$

 

[Mark44]

It should be $\sqrt{1+\frac{4}{x^6}}=\sqrt{\frac{x^6+4}{x^6}}$

That's more like it.

$E_T\leq\frac{(b-a)^3}{12n^2}[max |f^{(2)}(x)|]$

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.

 

[Zack K]

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.