Normal Rectifying and Osculating Planes Examples 1

# 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 vector-valued 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)
\begin{align} \quad (0, -1, 1) \cdot (x - (-1), y - 0, z - \pi) = 0 \\ \quad -y + z - \pi = 0 \end{align} ## 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)
\begin{align} \quad \quad (0, -3, 1) \cdot (x - 1, y - 2, z - 3) = 0 \\ \quad \quad -3(y - 2) + (z - 3) = 0 \end{align} ## 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)
\begin{align} \quad \quad \vec{r'}(t) \times \vec{r''}(t) = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k}\\ -6\sin (2t) & 2e^{2t+1} & 4t - 1\\ -12 \cos (2t) & 4e^{2t+1} & 4 \end{vmatrix} = \begin{vmatrix} 2e^{2t+1} & 4t - 1\\ 4e^{2t+1} & 4 \end{vmatrix} \vec{i} - \begin{vmatrix} -6 \sin (2t) & 4t - 1\\ -12 \cos (2t) & 4 \end{vmatrix} \vec{j} + \begin{vmatrix} -6 \sin (2t) & 2e^{2t+1} \\ -12 \cos (2t) & 4e^{2t+1} \end{vmatrix} \vec{k} \\ \quad \quad = \left ( [4(2e^{2t+1})-(4t-1)(4e^{2t+1})], -[(-6 \sin (2t))(4) - (4t - 1)(-12 \cos (2t))], [(-6 \sin (2t))(4e^{2t+1}) - (2e^{2t+1})(-12 \cos (2t))] \right ) \end{align}

Plugging in $t = 0$ and we get:

(4)
\begin{align} \vec{r'}(0) \times \vec{r''}(0) = (12e, 12, 24e) \end{align}

Plugging this into the formula above and we get the equation of the osculating plane at $t = 0$ is:

(5)
\begin{align} \quad \quad (12e, 12, 24e) \cdot (x - 3, y - e, z - 0) = 0 \\ \quad \quad 12e(x - 3) + 12(y - e) + 24ez = 0 \end{align} 