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)...
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.