Finding Arc Length of y=x^{\frac{1}{2}}- \frac{1}{3}x^{\frac{3}{2}}+\lambda

[rock.freak667]

1.Homework Statement

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}

Homework Equations

s=\int_{x_1} ^{x_2} \sqrt{1+ (\frac{dy}{dx})^2} dx

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}

 

[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.

 

[Pere Callahan]

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

 

[rock.freak667]

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

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!