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
Could anyone help me please?
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
Could anyone help me please?
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 $\displaystyle y = \frac13(x^{3/2} - 3x^{1/2})$. Its derivative is $\displaystyle y' = \frac12(x^{1/2}-x^{-1/2})$. Then $\displaystyle y'^2 = \frac14(x+x^{-1}-2)$, and $\displaystyle 1+y'^{\,2} = \frac14(x+x^{-1}+2) = \bigl(\frac12(x^{1/2}+x^{-1/2})\bigr)^2$. So the integral for the length is $\displaystyle \displaystyle\int_0^3\tfrac12(x^{1/2}+x^{-1/2})dx = \Bigl[\tfrac13(x^{3/2} + 3x^{1/2})\Bigr]_0^3,$ from which you should get the answer $\displaystyle 2\sqrt3$.
Hello, snopy!
$\displaystyle \text{Find the length of the curve: }\:y \:=\:\frac{1}{3}(x-3)\sqrt{x}\:\text{ from }x = 0\text{ to }x = 3$
The function is: .$\displaystyle y \:=\:\frac{1}{3}(x-3)x^{\frac{1}{2}} \;=\;\frac{1}{3}\left(x^{\frac{3}{2}} - 3x^{\frac{1}{2}}\right)$
Differentiate: .$\displaystyle y' \;=\;\frac{1}{3}\left(\frac{3}{2}x^{\frac{1}{2}} - \frac{3}{2}x^{-\frac{1}{2}}\right) \;=\;\frac{1}{2}\left(x^{\frac{1}{2}} - x^{-\frac{1}{2}}\right)$
Square: .$\displaystyle (y')^2 \;=\;\frac{1}{4}(x - 2 + x^{-1}) \;=\;\frac{1}{4}x - \frac{1}{2} + \frac{1}{4}x^{-1} $
Add one: .$\displaystyle 1 + (y')^2 \;=\;1 + \frac{1}{4}x - \frac{1}{2} + \frac{1}{4}x^{-1} \;=\;\frac{1}{4}x + \frac{1}{2} + \frac{1}{4}x^{-1}$
. . . . . . . . . . . . . . $\displaystyle =\; \frac{1}{4}\left(x + 2 + x^{-1}\right) \;=\; \frac{1}{4}\left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right)^2$
Square root: . $\displaystyle \sqrt{1 + (y')^2} \;=\;\frac{1}{2}\left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right)$
Therefore: .$\displaystyle \displaystyle L \;=\;\tfrac{1}{2}\int^3_0\left(x^{\frac{1}{2}} + x^{-\frac{1}{2}}\right)dx $
Got it?