6) Find the length of the curve

I try many different way but I cant seem to find the rite answer! :( can you tell me the answer?

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- Apr 28th 2010, 08:57 PMnhatieFind the length of the curve
6) Find the length of the curve

I try many different way but I cant seem to find the rite answer! :( can you tell me the answer? - Apr 29th 2010, 04:11 AMmr fantastic
- Apr 29th 2010, 05:42 AMnhatie
then i have 2pi int from -343 to 512 [(3)(y^4/3) - (3/32)(y^2/3)][sqrt[ 9( y^1/3)- (-9/32)(y^-1/3)] ???

I am not sure what I am doing... can you show me how to solve and the right answer? - Apr 29th 2010, 08:16 AMHallsofIvy
This is wrong. the derivative of $\displaystyle y^n$ is $\displaystyle ny^{n-1}$ so the derivative of $\displaystyle y^{4/3}$ is $\displaystyle (4/3)y^{1/3}$ and the derivative of $\displaystyle 3y^{4/3}$ is $\displaystyle 3(4/3)y^{1/3}= 4y^{1/3}$

Similarly, the derivative of $\displaystyle \frac{3}{32}y^{2/3}$ is $\displaystyle \left(\frac{2}{3}\right)\frac{3}{32} y^{-1/3}= \frac{1}{16}y^{-1/3}$.

Quote:

then i have 2pi int from -343 to 512 [(3)(y^4/3) - (3/32)(y^2/3)][sqrt[ 9( y^1/3)- (-9/32)(y^-1/3)] ???

I am not sure what I am doing... can you show me how to solve and the right answer?

Now, with $\displaystyle f'= 4y^{1/3}- \frac{1}{16}y^{-1/3}$, $\displaystyle f'^2= 16y^{2/3}- (2)(4)(\frac{1}{16})y^{1/3}y^{-1/3}+ \frac{1}{256}y^{-2/3}$$\displaystyle = 16y^{2/3}- \frac{1}{2}+ \frac{1}{256}y^{2/3}$.

Notice that adding 1 to that just changes the "- 1/2" to "-1/2". And since that was a "perfect square" so is this: If x and y are such that 2xy= 1/2, then $\displaystyle (x- y)^2= x^2- 2xy+ y^2= x^2- 1/2+ y^2$ and then $\displaystyle 1+ (x- y)^2= x^2+ 1/2+ y^2= x^2+ 2xy+ y^2= (x+y)^2$.

In other words, $\displaystyle 1+ f'^2$ is also a perfect square and so we can get rid of the square root! - Apr 29th 2010, 10:21 AMnhatie
i have sqrt (4y^1/3 + 1/16 y^1/3)^2 => int -343 to 512 (4y^1/3 + 1/16 y^1/3)

then

65/16y^1/3 |from -343 to 512

(65/16 (-343)^1/3) - (65/16 (512)^1/3)

is it the anwer?