Finding Length

• Feb 12th 2008, 01:05 PM
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]?
• Feb 12th 2008, 01:18 PM
galactus
Arc length for a parametric curve is given by:

$\displaystyle \int_{a}^{b}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{d t})^{2}}dt$
• Feb 12th 2008, 02:10 PM
So would it be $\displaystyle \int_{a}^{b}\sqrt{(2t)^{2}+(3t^2)^{2}}dt$
$\displaystyle \int_{a}^{b}\sqrt{4t^{2}+9t^4}dt$
$\displaystyle \int_{a}^{b}t\sqrt{4+9t^2}dt$
$\displaystyle u = {4+9t^2}du$
and $\displaystyle du= 18t$
So $\displaystyle (1/18)\int_{a}^{b}\sqrt{u}du$
$\displaystyle (1/27)(u)^(3/2)$
$\displaystyle (1/27)(4+9b^2)^(3/2)-(1/27)(4+9a^2)^(3/2)$

Is this the right answer?
• Feb 12th 2008, 02:30 PM
galactus
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.
• Feb 12th 2008, 02:53 PM
$\displaystyle (1/27)(4+9(2)^2)^{3/2}-(1/27)(4)^{3/2}$
$\displaystyle (1/27)(40)^{3/2}-(1/27)(4)^{3/2}$
$\displaystyle (1/27)(4+9(2)^2)^{3/2}-(1/27)4^{3/2}$
$\displaystyle \frac{40^{3/2}-8}{27}$