Table of Contents

Normal Rectifying and Osculating Planes Examples 1
Recall from the Determining Equations of Normal, Rectifying, and Osculating Planes page that if $\vec{r}(t) = (x(t), y(t), z(t))$ is a vectorvalued function and is defined at $t = t_0$ then:
 The Normal Plane at $t_0$ is perpendicular to $\vec{r'}(t_0)$.
 The Rectifying Plane at $t_0$ is perpendicular to $[\vec{r'}(t_0) \times \vec{r''}(t_0)] \times \vec{r'}(t_0)$.
 The Osculating Plane at $t_0$ is perpendicular to $\vec{r'}(t_0) \times \vec{r''}(t_0)$.
Also recall that the equation of a plane in $\mathbb{R}^3$ can be expressed in the form $\vec{n} \cdot \vec{P_0P} = 0$ where $\vec{n}$ is a vector that is perpendicular to the plane (see above), and $P_0P = (x  x_0, y  y_0, z  z_0)$, noting that $P_0 (x_0, y_0, z_0)$ is the point on the curve generated by $\vec{r'}(t)$ at $t = t_0$.
We will now look at some more examples of finding the normal, rectifying, and osculating planes of these curves.
Example 1
Let $\vec{r'}(t) = (\cos t, \sin t, t)$. Find the equation of the normal plane at $t = \pi$.
We note that $t = \pi$ corresponds to the point $\vec{r}(t_0) = (1, 0, \pi)$. We now need to find a vector that is perpendicular to this normal plane. From above, we see that $\vec{r'}(\pi)$ is perpendicular to the normal plane at $t = \pi$.
We first differentiate $\vec{r}(t)$ to get $\vec{r'}(t) = (\sin t, \cos t, 1)$. Therefore the vector $\vec{r'}(\pi) = (0, 1, 1)$ is perpendicular to the normal plane at $t = \pi$. Plugging this into the formula above and we get that the equation of the normal plane is:
(1)Example 2
Let $\vec{r'}(t) = (1, 2t, 3t^2)$. Find the equation of the rectifying plane at $t = 1$.
We first note that $t = 1$ corresponds to the point $(1, 2, 3)$. We now need to find a vector that is perpendicular to this rectifying plane. From above, we see that $[\vec{r'}(1) \times \vec{r''}(1)] \times \vec{r'}(1)$ is perpendicular to the osculating plane at $t = 1$.
We first differentiate $\vec{r}(t)$ to get that $\vec{r'}(t) = (0, 2, 6t)$, and we differentiate against to get $\vec{r''}(t) = (0, 0, 6)$. Therefore $\vec{r'}(t) \times \vec{r''}(t) = (12, 0, 0)$, and $[\vec{r'}(t) \times \vec{r''}(t)] \times \vec{r'}(t) = (0, 72t, 24)$. Plugging in $t = 1$ and we get that $(0, 72, 24)$ or equivalently, $(0, 3, 1)$ is perpendicular to the rectifying plane at $t = 1$. Plugging this into the formula above and we get that the equation of the rectifying plane at $t = 1$ is:
(2)Example 3
Let $\vec{r'}(t) = (3 \cos (2t), e^{2t + 1}, 2t^2  t)$. Find the equation of the osculating plane at $t = 0$.
We note that $t = 0$ corresponds to the point $\vec{r}(0) = (3, e, 0)$. We now need to find a vector that is perpendicular to this osculating plane. From above, we see that $\vec{r'}(0) \times \vec{r''}(0)$ is perpendicular to the osculating plane at $t = 0$.
We first differentiate $\vec{r}(t)$ to get $\vec{r'}(t) = (6 \sin (2t), 2e^{2t + 1}, 4t  1)$, and then differentiate again to get $\vec{r''}(t) = (12 \cos (2t), 4e^{2t + 1}, 4)$. Taking the cross product between the first and second derivatives and we have:
(3)Plugging in $t = 0$ and we get:
(4)Plugging this into the formula above and we get the equation of the osculating plane at $t = 0$ is:
(5)