Finding Length

**chaddy** · February 12th 2008, 01:05 PM

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]?

**galactus** · February 12th 2008, 01:18 PM

Arc length for a parametric curve is given by:

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

**chaddy** · February 12th 2008, 02:10 PM

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) $

Is this the right answer?

**galactus** · February 12th 2008, 02:30 PM

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.

**chaddy** · February 12th 2008, 02:53 PM

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?

**galactus** · February 12th 2008, 03:01 PM

Yep.

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