Solving Arc Length Problem: y=x^5/6+1/10x^3, 1≤x≤2

[3.141592654]

Homework Statement

Find the arc length of the curve:

$$y=\frac{x^5}{6}+\frac{1}{10x^3}$$

1$$\leq$$x$$\leq$$2

Homework Equations

$$ds=\sqrt{dx^2+dy^2}$$

$$ds=\sqrt{1+\frac{dy}{dx}^2}dx$$

The Attempt at a Solution

$$\frac{dy}{dx}=\frac{5}{6}x^4-\frac{3}{10x^4}$$

$$ds=\sqrt{1+(\frac{5}{6}x^4-\frac{3}{10x^4})^2}dx$$

If I use Trig. Sub, I have $$tan \theta=(\frac{5}{6}x^4-\frac{3}{10x^4})$$.

However, I don't know how to solve this for X. If this is the right path, can anyone help with the next steps, and is there a better way to do this problem?

Thanks.

 

[Mark44]

Homework Statement

Find the arc length of the curve:

$$y=\frac{x^5}{6}+\frac{1}{10x^3}$$

1$$\leq$$x$$\leq$$2

Homework Equations

$$ds=\sqrt{dx^2+dy^2}$$

$$ds=\sqrt{1+\frac{dy}{dx}^2}dx$$

The Attempt at a Solution

$$\frac{dy}{dx}=\frac{5}{6}x^4-\frac{3}{10x^4}$$

$$ds=\sqrt{1+(\frac{5}{6}x^4-\frac{3}{10x^4})^2}dx$$

OK, hold it right there. Just carry out the multiplication, and add the 1. You'll see that you still have a perfect square so you can take its square root.

This is a perfect example of a problem that has been laboriously cooked-up so that it's doable. Most of the arc length problems come out with difficult integrands, so some effort has to go into one to make it not too difficult to integrate.

If I use Trig. Sub, I have $$tan \theta=(\frac{5}{6}x^4-\frac{3}{10x^4})$$.

However, I don't know how to solve this for X. If this is the right path, can anyone help with the next steps, and is there a better way to do this problem?

Thanks.