1. ## Finding Length

What is the length of the plane curve defined by r(t) = <t^2 + 1, t^3 - 2> for t is an element of [0,2]?

2. Arc length for a parametric curve is given by:

$\int_{a}^{b}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{d t})^{2}}dt$

3. So would it be $
\int_{a}^{b}\sqrt{(2t)^{2}+(3t^2)^{2}}dt
$

$
\int_{a}^{b}\sqrt{4t^{2}+9t^4}dt
$

$
\int_{a}^{b}t\sqrt{4+9t^2}dt
$

$
u = {4+9t^2}du
$

and $
du= 18t
$

So $
(1/18)\int_{a}^{b}\sqrt{u}du
$

$
(1/27)(u)^(3/2)
$

$
(1/27)(4+9b^2)^(3/2)-(1/27)(4+9a^2)^(3/2)
$

4. Not quite. Aren't your limits of integration 0 to 2?. Use those, not a and b. That was just for an illustration of the formula.

5. Yes, so it would be
$
(1/27)(4+9(2)^2)^{3/2}-(1/27)(4)^{3/2}
$

$
(1/27)(40)^{3/2}-(1/27)(4)^{3/2}
$

$
(1/27)(4+9(2)^2)^{3/2}-(1/27)4^{3/2}
$

$
\frac{40^{3/2}-8}{27}
$

Correct?

6. Yep.