curvature

• Jun 10th 2008, 10:48 AM
chris25
curvature
find the curvature K of the curve....

r(t) = 4t i + 3 cos t j + 3 sin t k

i am confused after i find r'(t)...
• Jun 10th 2008, 11:56 AM
TheEmptySet
Quote:

Originally Posted by chris25
find the curvature K of the curve....

r(t) = 4t i + 3 cos t j + 3 sin t k

i am confused after i find r'(t)...

Here is the formula

$\displaystyle \kappa=\frac{|\frac{dr}{dt} \times (\frac{dr}{dt})^2|}{|\frac{dr}{dt}|^3}$

I have to go to work. I will finish later if no one else has.

Good luck.
• Jun 10th 2008, 12:07 PM
wingless
Quote:

Originally Posted by TheEmptySet
Here is the formula

$\displaystyle \kappa=\frac{|\frac{dr}{dt} \times \color{red}(\frac{dr}{dt})^2\color{black}|}{|\frac {dr}{dt}|^3}$

I have to go to work. I will finish later if no one else has.

Good luck.

As far as I know, it's $\displaystyle \kappa = \frac{\left | \frac{dr}{dt} \times \frac{d^2r}{dt^2}\right |}{\left | \frac{dr}{dt} \right |^3} = \frac{\left | \dot {r} \times \ddot {r} \right |}{\left | r \right |^3}$

See here if you wonder how to get the formula:
Curvature -- from Wolfram MathWorld
• Jun 10th 2008, 07:14 PM
TheEmptySet
Quote:

Originally Posted by wingless
As far as I know, it's $\displaystyle \kappa = \frac{\left | \frac{dr}{dt} \times \frac{d^2r}{dt^2}\right |}{\left | \frac{dr}{dt} \right |^3} = \frac{\left | \dot {r} \times \ddot {r} \right |}{\left | r \right |^3}$

See here if you wonder how to get the formula:
Curvature -- from Wolfram MathWorld

@ wingless You are correct Thanks that is what I get for trying to do it in a hurry.:(

$\displaystyle r(t)=<4t,3\cos(t),3\sin(t)$

$\displaystyle \frac{dr}{dt}=<4,-3\sin(t),3\cos(t)>$

$\displaystyle \frac{d^2r}{dt^2}=<0,-3\cos(t),-3\sin(t)>$

Now we take the cross product to get

$\displaystyle \begin{vmatrix} i & j & k \\ 4 & -3\sin(t) & 3\cos(t) \\ 0 & -3\cos(t) & -3\sin(t) \\ \end{vmatrix}=(9\sin^2(t)+9\cos(t))\vec i-(-12\sin(t)-0)\vec j+(-12\cos(t)-0)\vec k$

Simplifying and taking the magnitude we get

$\displaystyle =|<9,12\sin(t),-12\cos(t)>|=\sqrt{(9)^2+(12\sin(t))^2+(-12\cos(t))^2}$
$\displaystyle =\sqrt{81+144(\sin^{2}(t)+\cos^{2}(t))}=\sqrt{225} =15$

Now we need the magnitude of

$\displaystyle \bigg| \frac{dr}{dt}\bigg|=\sqrt{4^2+(-3\sin(t))^2+(3\cos(t))^2}=\sqrt{16+9(\sin^2(t)+\co s^2(t))}=\sqrt{25}=5$

Finally we get

$\displaystyle \kappa=\frac{15}{5^3}=\frac{3}{25}$

Yeah.

Sorry about the typo before. (Crying)