10.4 Unit tangents, unit normals, and curvature
- Let $\myv r(t)=3\cos t\,\uv i+3\sin t\,\uv j+2t\,\uv k$. (It's a spiral: Moving in what direction on average? What's the radius of the spiral?)
The particle is moving on the surface of a cylinder of radius 3, spiralling in the $\hat k$ direction.
- Calculate the unit tangent vector, $\uv T(t)$.
$\uv T(t)=\myv r'(t)/|\myv r'(t)|$. So... $$\myv r'(t)=\langle -3\sin t, 3\cos t, 2\rangle$$ $$|\myv r'(t)|=\sqrt{(-3\sin t)^2+(3\cos t)^2 + 2^2}=\sqrt{9+4}=\sqrt 13.$$ $$\uv T(t)=\frac{1}{\sqrt 13}\langle -3\sin t, 3\cos t, 2\rangle.$$
- Calculate the principle unit normal vector, $\uv N(t)$.
$\uv N(t)= \uv T'(t)/|\uv T'(t)|$. So... $$\uv T'(t)=\frac{1}{\sqrt 13}\langle -3\cos t, -3\sin t, 0\rangle.$$ $$|\uv T'(t)|=\frac{1}{\sqrt 13}\sqrt{ (-3\cos t)^2+ (3\sin t)^2\rangle}=\frac 3{\sqrt 13}.$$ $$\uv N(t) = \langle -\cos t, -\sin t \rangle$$
- Calculate the unit tangent vector, $\uv T(t)$.
- Given the following position function $\myv r(t)=(2+3t+3t^2)\uv i
+(4t+4t^2)\uv j+(-6\cos t)\uv k$, write
$\myv a=a_T \uv T+a_N\uv N$ at $t=0$. You will not need to do really complicated calculations.$$\myv r'(t)=\langle 3+6t,4+8t, +6\sin t\rangle$$ $$\myv a(t) = \myv r''(t)=\langle 6,8, +6\cos t\rangle$$ At $t=0$, $\myv r'=\langle 3,4,0\rangle$, and $|\myv r'|=5$. So $\uv T=\langle \frac 35, \frac 45,0\rangle$.
At $t=0$ $\myv a=\langle 6,8,1\rangle=\langle 6,8,0\rangle + \langle 0,0,1\rangle=10\uv T+\langle 0,0,1\rangle$. $\langle 0,0,1\rangle$ is a unit vector, and is perpendicular to $\uv T$, so this must be $\uv N$.
So our acceleration vector has tangential and normal components $a_T=10$ and $a_N=1$.
- A particle moves along the curve given by $\myv r(t)=
t\,\uv i+\frac{\sqrt 6}{2}t^2\,\uv j+t^3\,\uv k$.
- Find the speed of the particle.
- Find the tangential component of acceleration.
- Find the normal component of acceleration.
- Find the curvature of the curve.
- Find the speed of the particle.