# Length of a Plane Curve

1. Feb 22, 2004

### oooride

Okay I'm understanding this for the most part but am having trouble with what I believe is an algebra step.

Find the length of an arc curve y=x^(3/2) from (1,1) to (2,2sqrt2)
For second part I need to do it in terms of x.

Okay..

Rewritten,

x = g(y) = y^(2/3)
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So,

L= int{ctod} sqrt(1 + [g'(y)]^2) dy

= int{1 to 2sqrt2} sqrt(1 + [2/3*y^(-1/3)]^2) dy

= int{1 to 2sqrt2} sqrt(1 + [4/9*y^(-2/3)] dy

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Because,

g'(y)= 2/3*y^(-1/3)

[g'(y)]^2 = 4/9*y^(-2/3)
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Now,
For inside the square root

1 + [g'(y)^2] = 1 + 4/9*y^(-2/3) = 1 + 4/9y^(2/3)

= 9y^(2/3) + 4 / 9y^(2/3)

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Back to the Formula,

L = int{1 to 2sqrt2) sqrt(9y^(2/3)+ 4 / 9y^(2/3))

= int {1 to 2sqrt2} (1/3)(1/y^(1/3)) sqrt(9y^(2/3) + 4) dy
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Where and how do you get (1/3)(1/y^(1/3)) from the statement before?

I know I have to use substitution but I'm confused on that step...

Last edited: Feb 22, 2004
2. Feb 22, 2004

There is no substitution.

$$\int^1_{2\sqrt{2}}\sqrt{\frac{9y^{2/3}+4}{9y^{2/3}}}\,dy$$
$$\int^1_{2\sqrt{2}}\frac{\sqrt{9y^{2/3}+4}}{\sqrt{9y^{2/3}}}\,dy$$
$$\int^1_{2\sqrt{2}}\frac{\sqrt{9y^{2/3}+4}}{3y^{1/3}}}\,dy$$
$$\int^1_{2\sqrt{2}}\frac{1}{3}y^{-1/3}\sqrt{9y^{2/3}+4}\,dy$$

Let's hope I got all the LaTeX right...

3. Feb 22, 2004

### oooride

Oh okay... I see it now thanks! :)