# Arc length help

1. Mar 12, 2008

### rock.freak667

1.The problem statement, all variables and given/known data

$$y=x^{\frac}{1}{2}}- \frac{1}{3}x^{\frac{3}{2}}+\lambda$$

For $0 \leq x \leq 3$
Show that the arc length,s=$2\sqrt{3}$
2. Relevant equations
$$s=\int_{x_1} ^{x_2} \sqrt{1+ (\frac{dy}{dx})^2} dx$$

3. The attempt at a solution

$$\frac{dy}{dx}=\frac{1}{2\sqrt{x}} - \frac{1}{2}x^{\frac{1}{2}}$$

$$(\frac{dy}{dx})^2=\frac{1}{4x}-\frac{1}{2}+\frac{x}{2}$$

$$s=\int_{0} ^{3} \sqrt{\frac{1}{4x}+\frac{x}{2}+\frac{1}{2}} dx$$

and now that integral seems a bit too odd to get the answer $2\sqrt{3}$

2. Mar 13, 2008

### Gib Z

You made a tiny mistake when you squared the derivative, the coefficient of the linear term should be 1/4.

You integral, when corrected, then becomes

$$s= \int^3_0 \sqrt{ \frac{1}{4x} + \frac{x}{4} + \frac{1}{2} } dx$$

$$= \frac{1}{2} \int^3_0 2\sqrt{ \frac{1}{4x} + \frac{x}{4} + \frac{1}{2} } dx$$

$$= \frac{1}{2} \int^3_0 \sqrt{x + 2 + \frac{1}{x} } dx$$

I think I've already given out too much =]. Now just ignore that the bottom limit of the integral is x=0, or be rigorous and take limits instead.

3. Mar 13, 2008

### Pere Callahan

It might also be a good idea to think about completing the square ...

4. Mar 13, 2008

### rock.freak667

I was trying to do that but I kept on squaring it wrongly...even though I squared it out like 9 times...Seems I have to keep my eye out for those tiny mistakes...Thanks Gib Z & Pere Callahan!