# Thread: length of the curve: where is my mistake?

1. ## length of the curve: where is my mistake?

Ok, so let me start with the information that is given. I need to find the length of the curve.
x=t^3
y=3t^2/2
0≤ t ≥ square root (3)
So that's the information that is given. Now I believe the correct equation to use is this Mathwords: Arc Length of a Curve (parametric form)
(sorry for the link, but i have no idea how to type something like that)
Ok. so first I take the derivative of each of the functions, which gives:
y'=3t^2
x'=3t
then i square each of them, which gives:
9t^4
9t^2
So putting that into the fuction, it gives
(9t^4 +9t^2)^ 1/2

then taking the integral of that, i get
2(9t^4 +9t^2)^ 3/2/ 3(36t^3 +18t)

and evalutating at square root 3, i get 3.42857....
but the answer in the book says 7.
can anyone help me out here and point out what i did wrong? i may be way off, as i'm just learning this, but i need to know how to do this.

oh for the love of god none of my notation worked. i will edit this and make it better.
ok, that should be be better

2. your integral is incorrect.

note ...

$\sqrt{9t^4 + 9t^2} = \sqrt{9t^2(t^2 + 1)} = 3t\sqrt{t^2+1}$

$S = \frac{3}{2} \int_0^{\sqrt{3}} 2t\sqrt{t^2+1} \, dt$

using $u = t^2 + 1$ ...

$\frac{3}{2} \int_1^4 \sqrt{u} \, du$

now integrate.

3. Thanks for the help. Could you possibly explain this a bit though?
So you factor out 3/2, and get 2t. i'm confused as to how you changed the integral to be from 1 to 4, and what happened to the 2t?

4. Originally Posted by isuckatcalc
Thanks for the help. Could you possibly explain this a bit though?
So you factor out 3/2, and get 2t. i'm confused as to how you changed the integral to be from 1 to 4, and what happened to the 2t?
skeeter has made the substitution $u = t^2 + 1$. Are you familiar with solving integrals by using the technique often refered to as a "u-substitution" ....?

5. Originally Posted by mr fantastic
skeeter has made the substitution $u = t^2 + 1$. Are you familiar with solving integrals by using the technique often refered to as a "u-substitution" ....?
well... somewhat. if you could explain it though, that would be great, i apparently don't know it well enough, haha.

6. Originally Posted by isuckatcalc
well... somewhat. if you could explain it though, that would be great, i apparently don't know it well enough, haha.
Then perhaps you should go back and revise integration.

u-substitution - Google Search

7. so i integrate and get 2/3U^3/2
substitute t^2+1
2/3 (t^2+1) 3/2

but i'm still not getting 7 when solving for 4 and 1.

8. Originally Posted by isuckatcalc
so i integrate and get 2/3U^3/2
substitute t^2+1
2/3 (t^2+1) 3/2

but i'm still not getting 7 when solving for 4 and 1.
When you have a definite integral there is absolutely no need to go back to the old variable!

Just solve $\frac{3}{2} \int_1^4 \sqrt{u} \, du = \frac{3}{2} \cdot \frac{2}{3} [u^{3/2}]_1^4 =$ etc.

9. 4 and 1 are u-values ... you do not back substitute $t^2+1$ for $u$

$\frac{3}{2} \cdot \frac{2}{3}\left[u^{\frac{3}{2}}\right]_1^4 = 4^{\frac{3}{2}} - 1^{\frac{3}{2}} = 8 - 1 = 7$

10. nevermind, thanks guys