1. ## arc length

The problem

x= t^3 y=3/2 t^2 0 <= t <= sqrt(3)

using the formula

integral sqrt(f'(t)^2 + g'(t)^2) dt

so

int 0 to sqrt(3) sqrt((3t^2)^2 + (3t)^2) dt

so sqrt(9t^4 + 9t^2) dt

where would I go from here??

2. Originally Posted by ur5pointos2slo
The problem

x= t^3 y=3/2 t^2 0 <= t <= sqrt(3)

using the formula

integral sqrt(f'(t)^2 + g'(t)^2) dt

so

int 0 to sqrt(3) sqrt((3t^2)^2 + (3t)^2) dt

so sqrt(9t^4 + 9t^2) dt

where would I go from here??

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

try a substitution.

3. Originally Posted by danny arrigo

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

try a substitution.
I am sorry but I do not understand where you came up with that integral from..

4. Originally Posted by danny arrigo

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

try a substitution.
Your integral is $\int_0^{\sqrt{3}}\sqrt{9t^4+9t^2}\,dt$ right? So $\int_0^{\sqrt{3}}\sqrt{9t^2\left(t^2+1\right)}\,dt$. Now bring the $9t^2$ out from the square root.

5. Originally Posted by danny arrigo
Your integral is $\int_0^{\sqrt{3}}\sqrt{9t^4+9t^2}\,dt$ right? So $\int_0^{\sqrt{3}}\sqrt{9t^2\left(t^2+1\right)}\,dt$. Now bring the $9t^2$ out from the square root.

oooo ok yes thank you.