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?
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 $\displaystyle \left(\frac{2}{3}\right)\frac{3}{32} y^{-1/3}= \frac{1}{16}y^{-1/3}$.
Where in the world did the "$\displaystyle 2\pi$" come from? You are calculating the arc-length of a path, not rotating anything around an axis! You seem to be grabbing formulas at random. The arclength of f(y), from y= a to b is $\displaystyle \int_a^b \sqrt{1+ f'^2} dy$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!