단위접선·법선벡터

$a_{\text{T}} = \frac{4t}{\sqrt{1+4t^2}}$ 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.

$\hat{\mathbf{T}}$ is tangent to the curve, while $\hat{\mathbf{N}}$ is perpendicular to it.

Using the right-hand rule and $\hat{\mathbf{B}}=\hat{\mathbf{T}} \times \hat{\mathbf{N}}$, $\hat{\mathbf{B}}$ points out of the page (towards the reader).

To see this, point the fingers of your right hand in the direction of $\hat{\mathbf{T}}$, and curl them inwards until they are in the direction of $\hat{\mathbf{N}}$. To do this, your thumb must be pointing towards you, not away from you. Your thumb shows the direction of $\hat{\mathbf{T}} \times \hat{\mathbf{N}}$.