Find the exact length of the curve

[shamieh]

A little bit confused.

Find the exact length of the curve

$$y = \frac{1}{4}x^2 - \frac{1}{2}\ln x$$

$$1 \le x \le 2$$

Using the formula: $$y = \sqrt{1 + (\frac{dy}{dx})^2} \, dx$$

I obtained this:

$$\int ^2_1 \sqrt{ \frac{1}{2} + \frac{x^2}{4} + \frac{1}{4x^2}}$$

Now my problem is I'm stuck. If I bring the $$\frac{1}{2}$$ out I will have a $$\sqrt{\frac{1}{2}}$$ which won't really do me any good. Any suggestions?

 

[soroban]

Hello, shamieh!

Find the exact length of the curve.

$$y \:=\: \tfrac{1}{4}x^2 - \tfrac{1}{2}\ln x,\;1 \le x \le 2$$

Using the formula: .$$y \:=\: \sqrt{1 + (\tfrac{dy}{dx})^2} \, dx$$

I obtained this: $$\int ^2_1 \sqrt{\tfrac{1}{2} + \tfrac{x^2}{4} + \tfrac{1}{4x^2}}$$

You're doing great!

Note that: .\tfrac{x^2}{4} + \tfrac{1}{2} + \tfrac{1}{4x^2} \;=\;\tfrac{1}{4}\left(x^2 + 2 + \tfrac{1}{x^2}\right)

. . . . . . . =\;\tfrac{1}{4}\,\left(x + \tfrac{1}{x}\right)^2

Then: .\sqrt{\tfrac{1}{4}\left(x + \tfrac{1}{x}\right)^2} \;=\;\tfrac{1}{2}\left(x + \tfrac{1}{x}\right)And so you have: .\tfrac{1}{2}\int^2_1\left(x + \tfrac{1}{x}\right)\,dx

 

[shamieh]

Hello, shamieh!


Note that: .\tfrac{x^2}{4} + \tfrac{1}{2} + \tfrac{1}{4x^2} \;=\;\tfrac{1}{4}\left(x^2 + 2 + \tfrac{1}{x^2}\right)

. . . . . . . =\;\tfrac{1}{4}\,\left(x + \tfrac{1}{x}\right)^2

Then: .\sqrt{\tfrac{1}{4}\left(x + \tfrac{1}{x}\right)^2} \;=\;\tfrac{1}{2}\left(x + \tfrac{1}{x}\right)And so you have: .\tfrac{1}{2}\int^2_1\left(x + \tfrac{1}{x}\right)\,dx

Thanks, I see what you're saying but I'm confused on the algebra. Here is what I am getting, maybe you can tell me where I am going wrong.

Ok, here is what I am getting now:

$$\frac{1}{4} (2 + x^2 + \frac{1}{x^2})$$

Oh WOW I'm an idiot. Stared at my paper for 40 minutes trying to figure out how you factored out a 1/4 before I realized that 1/4 * 2/1. Thank you so much for your help Sororban!

 

[shamieh]

Oh I see now.