# Math Help - Find the length of the curve

1. ## Find 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?

2. Originally Posted by nhatie
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?

3. Originally Posted by mr fantastic

if find f'(x) = 9( y^1/3)- (-9/32)(y^-1/3)
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?

4. Originally Posted by nhatie
[CENTER]
if find f'(x) = 9( y^1/3)- (-9/32)(y^-1/3)
This is wrong. the derivative of $y^n$ is $ny^{n-1}$ so the derivative of $y^{4/3}$ is $(4/3)y^{1/3}$ and the derivative of $3y^{4/3}$ is $3(4/3)y^{1/3}= 4y^{1/3}$

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

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?
Where in the world did the " $2\pi$" 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 $\int_a^b \sqrt{1+ f'^2} dy$

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

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 $(x- y)^2= x^2- 2xy+ y^2= x^2- 1/2+ y^2$ and then $1+ (x- y)^2= x^2+ 1/2+ y^2= x^2+ 2xy+ y^2= (x+y)^2$.

In other words, $1+ f'^2$ is also a perfect square and so we can get rid of the square root!

5. 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?