consider the following.\n\nx = ln(t), y = 2t, z = t², (0, 2, 1)\n\nfind the equation of the normal plane of…

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}\rangle). The derivative (\vec{r}'(t)=\langle\frac{1}{t},2,2t\rangle). When (t = 1), (\vec{r}'(1)=\langle1,2,2\rangle).

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\rangle=0). Here, ((x_0,y_0,z_0)=(0,2,1)) and (\vec{n}=\vec{r}'(1)=\langle1,2,2\rangle). 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\rangle). When (t = 1), (\vec{r}''(1)=\langle-1,0,2\rangle).

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\rangle). 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\rangle=0). Using ((x_0,y_0,z_0)=(0,2,1)) and (\vec{n}=\langle4,-4,2\rangle), 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)