단위접선·법선벡터

First, some preliminaries:

$$\begin{aligned} \mathbf{r}(t)&=(t^3,t,e^{ct}) & \mathbf{r}(5)&=(5^3,5,e^{5c})\\ \mathbf{v}(t)&=(3t^2,1,ce^{ct}) & \mathbf{v}(5)&=(3\cdot 5^2,1,ce^{5c})\\ \mathbf{a}(t)&=(6t,0,c^2e^{ct}) & \mathbf{a}(5)&=(6\cdot 5,0,c^2e^{5c}) \\ \frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}(t)&= (6,0,c^3e^{ct})& \frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}(5)&= (6,0,c^3e^{5c})\\ && \mathbf{v}(5) \times \mathbf{a}(5)&=(c^2e^{5c},15ce^{5c}(2-5c),-30) \end{aligned}$$

Second, we figure out what value of $c$ makes $\tau(5)=0$.

$$\begin{aligned} 0&=\tau(5)=\dfrac{(\mathbf{v}(5) \times \mathbf{a}(5))\cdot\frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}(5)}{|\mathbf{v}(5) \times \mathbf{a}(5)|^2}\\ 0&=(\mathbf{v}(5)\times\mathbf{a}(5))\cdot \frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}(5)\\ &=(c^2e^{5c},15ce^{5c}(2-5c),-30)\cdot (6,0,c^3e^{5c})\\ &=6c^2e^{5c}(1-5c)\\ c&=0 \text{ or } c=\frac15 \end{aligned}$$

If $c=0$, then $\mathbf{r}(t)=(t^3,t,1)$, and so the entire curve is contained inside the plane $z=1$. (Its torsion is zero everywhere — not just at $t=5$.)

Consider the case $c=\frac15$. When $t=5$, our curve (and its osculating circle) passes through the point $\mathbf{r}(5)=(5^3,5,e)$. The normal vector to the plane of the osculating curve is the binormal vector $\hat{\mathbf{B}}(5)=\frac{\mathbf{v}(5)\times \mathbf{a}(5)}{|\mathbf{v}(5)\times \mathbf{a}(5)|}$. Since we don't need the normal vector to the plane to be a unit vector, we can take as the normal vector to the plane simply $\mathbf{v}(5) \times \mathbf{a}(5)$, or $(e/25,3e,-30)$. Then, an equation of the plane containing the osculating circle is $(e/25)x+(3e)y-30z=-10e$. An equivalent equation for this plane is $(1/25)x+3y-(30/e)z=-10$.