Question

consider the following.

x = ln(t), y = 2t, z = t², (0, 2, 1)

find the equation of the normal plane of the curve at the given point.

find the equation of the osculating plane of the curve at the given point.

Explanation:

Step1: Find the value of \(t\)

Given \(x = \ln(t)\), \(y = 2t\), \(z=t^{2}\) and the point \((0,2,1)\). Since \(x=\ln(t)=0\), then \(t = 1\) (because \(\ln(1)=0\), and when \(t = 1\), \(y=2\times1 = 2\), \(z=1^{2}=1\)).

Step2: Find the derivative of the position - vector function

The position - vector function \(\vec{r}(t)=\langle\ln(t),2t,t^{2}
angle\). The derivative \(\vec{r}'(t)=\langle\frac{1}{t},2,2t
angle\). When \(t = 1\), \(\vec{r}'(1)=\langle1,2,2
angle\).

Step3: Find the equation of the normal plane

The equation of the normal plane to the curve \(\vec{r}(t)\) at the point \((x_0,y_0,z_0)\) with normal vector \(\vec{n}\) is given by \(\vec{n}\cdot\langle x - x_0,y - y_0,z - z_0
angle=0\). Here, \((x_0,y_0,z_0)=(0,2,1)\) and \(\vec{n}=\vec{r}'(1)=\langle1,2,2
angle\). So, \(1\cdot(x - 0)+2\cdot(y - 2)+2\cdot(z - 1)=0\), which simplifies to \(x + 2y+2z-4 - 2=0\), or \(x + 2y+2z=6\).

Step4: Find the second - derivative of the position - vector function

\(\vec{r}''(t)=\langle-\frac{1}{t^{2}},0,2
angle\). When \(t = 1\), \(\vec{r}''(1)=\langle-1,0,2
angle\).

Step5: Find the binormal vector \(\vec{B}(1)\)

The binormal vector \(\vec{B}(t)=\frac{\vec{r}'(t)\times\vec{r}''(t)}{\vert\vec{r}'(t)\times\vec{r}''(t)\vert}\). First, find \(\vec{r}'(1)\times\vec{r}''(1)=

$$\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\1&2&2\\ - 1&0&2\end{vmatrix}$$

=\vec{i}(4 - 0)-\vec{j}(2 + 2)+\vec{k}(0 + 2)=\langle4,-4,2
angle\).
The equation of the osculating plane to the curve \(\vec{r}(t)\) at the point \((x_0,y_0,z_0)\) with normal vector \(\vec{n}=\vec{r}'(t)\times\vec{r}''(t)\) (evaluated at \(t\)) is \(\vec{n}\cdot\langle x - x_0,y - y_0,z - z_0
angle=0\). Using \((x_0,y_0,z_0)=(0,2,1)\) and \(\vec{n}=\langle4,-4,2
angle\), we have \(4\cdot(x - 0)-4\cdot(y - 2)+2\cdot(z - 1)=0\), which simplifies to \(4x-4y + 2z+8 - 2=0\), or \(2x-2y+z=-3\).

Answer:

Equation of the normal plane: \(x + 2y+2z=6\)
Equation of the osculating plane: \(2x-2y+z=-3\)