# Thread: Length of a curve

1. ## Length of a curve

I have a task to calculate the length of the curve. Does anyone know outcome?

I big idiot((

http://www.sdilej.eu/pics/ee9c6c6833...d2329b4eea.jpg

2. It looks as though you are using the correct method, and you almost have the correct answer (except that you have crossed out the calculation at a crucial stage).

The function is $y = \frac13(x^{3/2} - 3x^{1/2})$. Its derivative is $y' = \frac12(x^{1/2}-x^{-1/2})$. Then $y'^2 = \frac14(x+x^{-1}-2)$, and $1+y'^{\,2} = \frac14(x+x^{-1}+2) = \bigl(\frac12(x^{1/2}+x^{-1/2})\bigr)^2$. So the integral for the length is $\displaystyle\int_0^3\tfrac12(x^{1/2}+x^{-1/2})dx = \Bigl[\tfrac13(x^{3/2} + 3x^{1/2})\Bigr]_0^3,$ from which you should get the answer $2\sqrt3$.

3. Hello, snopy!

$\text{Find the length of the curve: }\:y \:=\:\frac{1}{3}(x-3)\sqrt{x}\:\text{ from }x = 0\text{ to }x = 3$

The function is: . $y \:=\:\frac{1}{3}(x-3)x^{\frac{1}{2}} \;=\;\frac{1}{3}\left(x^{\frac{3}{2}} - 3x^{\frac{1}{2}}\right)$

Differentiate: . $y' \;=\;\frac{1}{3}\left(\frac{3}{2}x^{\frac{1}{2}} - \frac{3}{2}x^{-\frac{1}{2}}\right) \;=\;\frac{1}{2}\left(x^{\frac{1}{2}} - x^{-\frac{1}{2}}\right)$

Square: . $(y')^2 \;=\;\frac{1}{4}(x - 2 + x^{-1}) \;=\;\frac{1}{4}x - \frac{1}{2} + \frac{1}{4}x^{-1}$

Add one: . $1 + (y')^2 \;=\;1 + \frac{1}{4}x - \frac{1}{2} + \frac{1}{4}x^{-1} \;=\;\frac{1}{4}x + \frac{1}{2} + \frac{1}{4}x^{-1}$

. . . . . . . . . . . . . . $=\; \frac{1}{4}\left(x + 2 + x^{-1}\right) \;=\; \frac{1}{4}\left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right)^2$

Square root: . $\sqrt{1 + (y')^2} \;=\;\frac{1}{2}\left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right)$

Therefore: . $\displaystyle L \;=\;\tfrac{1}{2}\int^3_0\left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right)dx$

Got it?

4. thanks )))

5. One thing that is generally true of "length" integrals is that they tend to be very hard if not impossible! That idea of choosing a function so that $1+ f'^2$ turns out to be a perfect square is a favorite of Calculus teachers.