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]?

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- Feb 12th 2008, 01:05 PMchaddyFinding 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 PMgalactus
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 PMchaddy
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 PMgalactus
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 PMchaddy
Yes, so it would be

$\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}

$

Correct? - Feb 12th 2008, 03:01 PMgalactus
Yep. (Clapping)