Simplifying an integration problem involving a radical.

**bkbowser** · February 12th 2013, 11:57 AM

$r(t)=\vec{i}+t^2\vec{j}+t^3\vec{k}$

The problem wants me to find the arc length of r(t) as t moves from 0 to 1.

definition of the length of a curve (were $| r(t)^\prime|$ refers to the magnitude of the vector $r(t)^\prime$ );
$L=\int_{a}^{b} | r(t)^\prime|dt$

so $r(t)^\prime=\langle 0 + 2t + 3t^2\rangle$ and $L=\int_{0}^{1} [0^2+(2t)^2+(3t^2)^2]^\frac{1}{2}dt$

I'm pretty lost on this one. I've tried a few different things and haven't gotten any farther. Have I just made an early mistake that I'm not seeing?

**Prove It** · February 12th 2013, 03:50 PM

$\displaystyle \begin{align*} r'(t) \end{align*}$ will be a VECTOR, not a scalar value. $\displaystyle \begin{align*} r'(t) = \left < \frac{d}{dt} \left( 1 \right) , \frac{d}{dt} \left( t^2 \right) , \frac{d}{dt} \left( t^3 \right) \right > = < 0, 2t, 3t^2 > \end{align*}$ . So

$\displaystyle \begin{align*} \left| r'(t) \right| &= \sqrt{ 0^2 + \left( 2t \right)^2 + \left( 3t^2 \right)^2 } \\ &= \sqrt{ 0 + 4t^2 + 9t^4 } \\ &= \sqrt{ 9t^4 + 4t^2 } \\ &= \sqrt{ t^2 \left( 9t^2 + 4 \right) } \\ &= t \, \sqrt{9t^2 + 4} \end{align*}$

So your arclength will be found by $\displaystyle \begin{align*} \int_0^1{ t \, \sqrt{9t^2 + 4} \, dt } = \frac{1}{18}\int_0^1{18t \, \sqrt{9t^2 + 4}\,dt} \end{align*}$ . Let $\displaystyle \begin{align*} u = 9t^2 + 4 \implies du = 18t\, dt \end{align*}$ and note that $\displaystyle \begin{align*} u(0) = 4 \end{align*}$ and $\displaystyle \begin{align*} u(1) = 13 \end{align*}$ , then the integral becomes

$\displaystyle \begin{align*} \frac{1}{18}\int_0^1{18t\, \sqrt{ 9t^2 + 4}\,dt} &= \frac{1}{18}\int_4^{13}{ \sqrt{u}\,du } \\ &= \frac{1}{18} \int_4^{13}{u^{\frac{1}{2}}\,du} \\ &= \frac{1}{18} \left[ \frac{2}{3} u^{\frac{3}{2}} \right] _4 ^{13} \\ &= \frac{1}{27} \left( 13^{\frac{3}{2}} - 4^{\frac{3}{2}} \right) \\ &= \frac{1}{27} \left( 13\, \sqrt{13} - 8 \right) \end{align*}$

**bkbowser** · February 12th 2013, 04:48 PM

Originally Posted by Prove It

$\displaystyle \begin{align*} r'(t) \end{align*}$ will be a VECTOR, not a scalar value. $\displaystyle \begin{align*} r'(t) = \left < \frac{d}{dt} \left( 1 \right) , \frac{d}{dt} \left( t^2 \right) , \frac{d}{dt} \left( t^3 \right) \right > = < 0, 2t, 3t^2 > \end{align*}$ . So

$\displaystyle \begin{align*} \left| r'(t) \right| &= \sqrt{ 0^2 + \left( 2t \right)^2 + \left( 3t^2 \right)^2 } \\ &= \sqrt{ 0 + 4t^2 + 9t^4 } \\ &= \sqrt{ 9t^4 + 4t^2 } \\ &= \sqrt{ t^2 \left( 9t^2 + 4 \right) } \\ &= t \, \sqrt{9t^2 + 4} \end{align*}$

So your arclength will be found by $\displaystyle \begin{align*} \int_0^1{ t \, \sqrt{9t^2 + 4} \, dt } = \frac{1}{18}\int_0^1{18t \, \sqrt{9t^2 + 4}\,dt} \end{align*}$ . Let $\displaystyle \begin{align*} u = 9t^2 + 4 \implies du = 18t\, dt \end{align*}$ and note that $\displaystyle \begin{align*} u(0) = 4 \end{align*}$ and $\displaystyle \begin{align*} u(1) = 13 \end{align*}$ , then the integral becomes

$\displaystyle \begin{align*} \frac{1}{18}\int_0^1{18t\, \sqrt{ 9t^2 + 4}\,dt} &= \frac{1}{18}\int_4^{13}{ \sqrt{u}\,du } \\ &= \frac{1}{18} \int_4^{13}{u^{\frac{1}{2}}\,du} \\ &= \frac{1}{18} \left[ \frac{2}{3} u^{\frac{3}{2}} \right] _4 ^{13} \\ &= \frac{1}{27} \left( 13^{\frac{3}{2}} - 4^{\frac{3}{2}} \right) \\ &= \frac{1}{27} \left( 13\, \sqrt{13} - 8 \right) \end{align*}$

Thank you. That makes perfect sense.

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