Arclength

[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.

[Maxos]

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

Then .....

[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.

[mathmike]

you can break it down into partial fractions

[Brad Barker]

i second the inverse trig antiderivative!

[Brad Barker]

you can break it down into partial fractions

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

[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?

[mathelord]

well done iggy
well done

[HallsofIvy]

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.