# Find the arc length of the given function

1. Aug 3, 2005

### iggybaseball

This is my last homework problem and I feel that I almost have it solved. The problem is as followed:

$$f(x) = \sqrt{4-x^2}$$

Find the arc length of the given function from x=0 to x=2.

I know that I am supposed to use this formula to solve for arclength:

$$\int_{0}^{2} \sqrt{1 + (f\prime(x))^2} dx$$

Ok so I take the derivative of

$$f(x) = \sqrt{4-x^2}$$

and I get:

$$f\prime(x) =\frac{-x}{\sqrt{4-x^2}}$$

I then plug this into the equation get:

$$\int_{0}^{2} \sqrt{1 + (\frac{-x}{\sqrt{4-x^2}}) ^2}$$

$$\int_{0}^{2} \sqrt{1 + \frac{x^2}{4-x^2}}$$

I get stuck here because I can't integrate this. I can't use any integral tables and I even try some algebraic manipulation inside of the square root and get:

$$\int_{0}^{2} \sqrt{\frac{4}{4-x^2}}$$

Any help? Thank you.

2. Aug 3, 2005

### Maxos

Well, if $$f(x)=\arcsin (x) \Rightarrow f'(x)= \frac {1}{\sqrt{1-x^2}}$$

Then .....

3. Aug 3, 2005

### robphy

Visit integrals.com and enter Sqrt[4/(4-x^2)]
Don't just take the answer [In addition, remember it's just a computer.]
... learn from the answer to figure out how you could have gotten it yourself.

The answer has an (x/2) in it. Hmmm. maybe a substitution or two is in order.

4. Aug 3, 2005

### mathmike

you can break it down into partial fractions

5. Aug 3, 2005

i second the inverse trig antiderivative!

6. Aug 3, 2005

yeah...but then it's a problem of looking up integration tables again, and not necessarily finding anything, either.

7. Aug 3, 2005

### iggybaseball

Alright I think I got the solution. Is this right?

$$\int_{0}^{2} \sqrt{\frac{4}{4-x^2}}$$

$$\int_{0}^{2} \frac{\sqrt{4}}{\sqrt{4-x^2}}$$

$$2\int_{0}^{2} \frac{1}{\sqrt{4-x^2}}$$

then using an integral table.....

$$2\arcsin{\frac{x}{2}} |_{0}^{2}$$

$$2\arcsin{1}$$

Does that look right?

8. Aug 4, 2005

### mathelord

well done iggy
well done

9. Aug 4, 2005

### HallsofIvy

Staff Emeritus
Tninking in terms of formulas or looking things up in a table, you miss most of what's interesting about this problem.

Looking at something like $$2\int_{0}^{2} \frac{dx}{\sqrt{4-x^2}}$$ (DON'T forget the "dx"), the first thing that should come to mind is a trig substitution (I hate integral tables!). Let x= 2sin(θ) so that dx= 2 cos(θ) and 4- x2= 4- 4sin2(θ)= 2(1- sin2(θ))= 2 cos(θ). When x= 0, 0= 2sin(θ) so θ= 0. When x= 2, 2= 2sin(θ) so θ= $$\frac{\pi}{2}$$. The integral becomes $$2\int_0^{\frac{\pi}{2}}d\theta$$. That's pretty easy, isn't it!

Yet another way to do this is to put the formula in parametric equations. $$y= \sqrt{4- x^2}$$ is (almost) the same thing as y2= 4- x2 or x2+ y2= 4, a circle of radius 2.
Parametric equations for that are x= 2cos(t), y= 2 sin(t) where t ranges from 0 to $$\frac{pi]{2}$$. $$ds= \sqrt{(\frac{dx}{dt})^2+ (\frac{dy}{dt})^2} = 2dt$$. x goes from 0 to 2 as t goes from 0 to $$\frac{\pi}{2}$$ so the arc length is $$2 \int_0^{\frac{\pi}{2}}dt$$ which is also pretty darn easy!

Finally, it should be obvious that you are talking about 1/4 of a circle of radius 2. Find the circumference and divide by 4.

Last edited: Aug 4, 2005