벡터와 공간 전체 문제

$\vec T(t) = \left\langle -\sin t,\cos t\right\rangle$; $\vec T(\pi) = \left\langle 0,-1\right\rangle$

$\vec T(t) = \frac{1}{\sqrt{5}}\left\langle 2,\cos t,-\sin t\right\rangle$; $\vec N(t) = \left\langle 0,-\sin t,-\cos t\right\rangle$

$\vec T(t) = \left\langle -\sin t, 3/5\cos t, 4/5\cos t\right\rangle$; $\vec N(t) = \left\langle -\cos t,-3/5\sin t,-4/5\sin t\right\rangle$

$\vec T(t) = \frac{1}{\sqrt{a^2+b^2}}\left\langle -a\sin t,a\cos t,b \right\rangle$; $\vec N(t) = \left\langle -\cos t,-\sin t,0\right\rangle$

$\vec T(t) = \frac{1}{\sqrt{a^2+1}}\left\langle -a\sin (at),a\cos (at),1 \right\rangle$; $\vec N(t) = \left\langle -\cos t,-\sin t,0\right\rangle$

$a_{\text{T}} = \frac{4t}{\sqrt{1+4t^2}}$ and $a_{\text{N}} = \sqrt{4-\frac{16t^2}{1+4t^2}}$

At $t=0$, $a_{\text{T}} = 0$ and $a_{\text{N}} = 2$;

At $t=1$, $a_{\text{T}} = 4/\sqrt{5}$ and $a_{\text{N}} = 2/\sqrt{5}$.

At $t=0$, all acceleration comes in the form of changing the direction of velocity and not the speed; at $t=1$, more acceleration comes in changing the speed than in changing direction.

$a_{\text{T}} = \frac{-2/t^5}{\sqrt{1+1/t^4}}$ and $a_{\text{N}} = \sqrt{\frac{4}{t^6}-\frac{4/t^{10}}{1+1/t^4}}$

At $t=1$, $a_{\text{T}} = \sqrt{2}$ and $a_{\text{N}} = -\sqrt{2}$;

At $t=2$, $a_{\text{T}} = -\frac{1}{4\sqrt{17}}$ and $a_{\text{N}} = \frac{1}{\sqrt{17}}$.

At $t=1$, acceleration comes from changing speed and changing direction in “equal measure;” at $t=2$, acceleration is nearly $\vec 0$ as it is; the low value of $a_{\text{T}}$ shows that the speed is nearly constant and the low value of $a_{\text{N}}$ shows the direction is not changing quickly.

$a_{\text{T}} = 0$ and $a_{\text{N}} = 2$

At $t=0$, $a_{\text{T}} = 0$ and $a_{\text{N}} = 2$;

At $t=\pi/2$, $a_{\text{T}} = 0$ and $a_{\text{N}} = 2$.

The object moves at constant speed, so all acceleration comes from changing direction, hence $a_{\text{T}}=0$. $\vec a(t)$ is always parallel to $\vec N(t)$, but twice as long, hence $a_{\text{N}}=2$.

$a_{\text{T}} = 2$ and $a_{\text{N}} = 4t^2$

At $t=\sqrt{\pi/2}$, $a_{\text{T}} = 2$ and $a_{\text{N}} = 2\pi$;

At $t=\sqrt{\pi}$, $a_{\text{T}} = 2$ and $a_{\text{N}} = 4\pi$.

The object moves at increasing speed (increasing at a constant rate of acceleration), hence $a_{\text{T}}=2$. Since the object is increasing speed yet always traveling in a circle of radius 1, the direction must change more quickly; the amount of acceleration that changes direction increases over time.

$a_{\text{T}} = 0$ and $a_{\text{N}} = a$

At $t=0$, $a_{\text{T}} = 0$ and $a_{\text{N}} = a$;

At $t=\pi/2$, $a_{\text{T}} = 0$ and $a_{\text{N}} = a$.

The object moves at constant speed, meaning that $a_{\text{T}}$ is always 0. The object “rises” along the $z$-axis at a constant rate, so all acceleration comes in the form of changing direction circling the $z$-axis. The greater the radius of this circle the greater the acceleration, hence $a_{\text{N}}=a$.

$a_{\text{T}} = 0$ and $a_{\text{N}} = 5$

At $t=0$, $a_{\text{T}} = 0$ and $a_{\text{N}} = 5$;

At $t=\pi/2$, $a_{\text{T}} = 0$ and $a_{\text{N}} = 5$.

The object moves at constant speed, meaning that $a_{\text{T}}$ is always 0. Acceleration is thus always perpendicular to the direction of travel; in this particular case, it is always 5 times the unit vector pointing orthogonal to the direction of travel.

time and/or distance

curvature

Answers may include lines, circles, helixes

Answers will vary; they should mention the circle is tangent to the curve and has the same curvature as the curve at that point.

$\kappa$

$a_{\text{T}}$ is not affected by curvature; the greater the curvature, the larger $a_{\text{N}}$ becomes.

$s = 3t$, so $\vec r(s) = \left\langle 2s/3, s/3, -2s/3\right\rangle$

$s = 7t$, so $\vec r(s) = \left\langle 7\cos (s/7), 7\sin (s/7)\right\rangle$

$s = \sqrt{13}t$, so $\vec r(s) = \left\langle 3\cos (s/\sqrt{13}), 3\sin (s/\sqrt{13}), 2s/\sqrt{13}\right\rangle$