the arc length of the curve

Question

vinicius1 · Answer

$$y=\frac{2}{3}x^{\frac{3}{2}}+2$$ from the points (0,2) to $$(2,\frac{4}{3}(\sqrt{2}+2))$$

ganeshie8 · Answer

where are you stuck ?

ganeshie8 · Answer

$$\int\limits_a^b \sqrt{1+\left(\dfrac{dy}{dx}\right)^2}~~dx$$

vinicius1 · Answer

I'm kind of lost actually. Is that the formula? Thanks

vinicius1 · Answer

do I use the points given as boundaries to the integral?

anonymous · Answer

The limits you use depend on which variable you want to use for integration. $x$ would be the easiest.
$$y=\frac{2}{3}x^{3/2}+2~~\implies~~\frac{dy}{dx}=\cdots$$
and using ganeshie's integral, you would set $a=0$ and $b=2$.

Alternatively, if you'd like to try using $y$, you'd need to write the curve as a function of $y$:
$$y(x)=\frac{2}{3}x^{3/2}+2~~\iff~~x(y)=\cdots$$
take the derivative with respect to $y$:
$$x(y)=\cdots~~\implies~~\frac{dx}{dy}=\cdots$$
then the integral is
$$\int_c^d\sqrt{1+\left[\frac{dx}{dy}\right]^2}~dy$$
with $c=2$ and $d=\dfrac{4(\sqrt2+2)}{3}$.

vinicius1 · Answer

I got 2/3((3^1/2) -1) and its not the correct answer.

ganeshie8 · Answer

still looking for help @vinicius1

vinicius1 · Answer

Yes, its actually an online test and I got that answer above, but its not the right one. The one wolfram gives also doesn't seem correct at all. What am I missing?

vinicius1 · Answer

Oh, no. Forget it. I missed an exponent. lol