Given a vector valued function defined by $r(t):=(2t)i+(t^2)j+(\ln t)k$,Find the unit vectors $\vec T,\vec N,\vec B$ at the point $P(2,1,0)$,then find the curvature and the equation of the osculating plane at the point.
Using the definitions we see that :
$$\vec T=\frac{dr\left(t\right)}{dt}=2i+2tj+\frac{1}{t}k$$At the point we have $2i+2j+k$
$$\vec N=\frac{\frac{d\vec T}{dt}}{\left|\frac{d\vec T}{dt}\right|}=\frac{1}{\sqrt{4+\frac{1}{t^{4}}}}(2j-\frac{1}{t^2}k)$$ At the point we have $\frac{1}{\sqrt{5}}(2j-k)$
$$\vec B=\vec T\times \vec N=\begin{vmatrix} i & j & k\\ 2 & 2t& \frac{1}{t}\\ 0 & \frac{2}{\sqrt{4+\frac{1}{t^{4}}}}&\frac{-1}{t^2\sqrt{4+\frac{1}{t^{4}}}} \end{vmatrix}$$ At the point we have $\frac{1}{\sqrt{5}}\left(-4i-2j+4k\right)$
For the osculating plane we need a normal to the plane and a point on the plane:
The normal vector is given by: $$\frac{dr\left(t\right)}{dt}\times\frac{d^{2}r\left(t\right)}{dt^{2}}=\begin{vmatrix} i & j & k\\ 2 & 2t& \frac{1}{t}\\ 0 & 2&\frac{-1}{t^2} \end{vmatrix}$$ Setting $t=1$ gives the equation of the plane: $$-4\left(x-2\right)-2\left(y-1\right)+4\left(z-0\right)=0$$
Or in a more compact form: $$2x+y-2z=5$$
And the curvature at the point is $\sqrt{5}$.
Can someone please check the process?
These three vectors are unit vectors, so they must have magnitude $1$. To begin, we will find $\vec{\mathbf{r}}\,'(t)$. $$\vec{\mathbf{r}}(t)=\left\langle2t,t^2,\ln(t)\right\rangle\implies\vec{\mathbf{r}}\,'(t)=\left\langle2,2t,\frac{1}{t}\right\rangle$$ We are assuming $t\gt0$. Next, we will find the magnitude of $\vec{\mathbf{r}}\,'(t)$. $$\left\lVert\vec{\mathbf{r}}\,'(t)\right\rVert=\sqrt{4+4t^2+\frac{1}{t^2}}=\sqrt{\frac{4t^4+4t^2+1}{t^2}}=\frac{2t^2+1}{t}$$ We can now find $\hat{\mathbf{T}}(t)$. This will be the first of the three unit vectors we will find. $$\hat{\mathbf{T}}(t)=\frac{\vec{\mathbf{r}}\,'(t)}{\left\lVert\vec{\mathbf{r}}\,'(t)\right\rVert}=\frac{t}{2t^2+1}\cdot\left\langle2,2t,\frac{1}{t}\right\rangle=\left\langle\frac{2t}{2t^2+1},\frac{2t^2}{2t^2+1},\frac{1}{2t^2+1}\right\rangle$$ We are interested in the point $P(2,1,0)$. We can clearly see that the vector function $\vec{\mathbf{r}}(t)=\langle2,1,0\rangle$ when $t=1$. $$\hat{\mathbf{T}}(1)=\left\langle\frac{2}{3},\frac{2}{3},\frac{1}{3}\right\rangle$$ Next, Let us find $\hat{\mathbf{T}}\,'(t)$. This is the first step to finding the unit normal vector $\hat{\mathbf{N}}(t)$. $$\hat{\mathbf{T}}\,'(t)=\left\langle-\frac{4t^2-2}{\left(2t^2+1\right)^2},\frac{4t}{\left(2t^2+1\right)^2},-\frac{4t}{\left(2t^2+1\right)^2}\right\rangle$$ Let us find its magnitude. $$\left\lVert\hat{\mathbf{T}}\,'(t)\right\rVert=\sqrt{\frac{\left(4t^2-2\right)^2+32t^2}{\left(2t^2+1\right)^4}}=\sqrt{\frac{4\left(2t^2+1\right)^2}{\left(2t^2+1\right)^4}}=\frac{2}{2t^2+1}$$ Now, let us find $\hat{\mathbf{N}}(t)$. $$\hat{\mathbf{N}}(t)=\frac{\hat{\mathbf{T}}\,'(t)}{\left\lVert\hat{\mathbf{T}}\,'(t)\right\rVert}=\frac{2t^2+1}{2}\cdot\left\langle-\frac{4t^2-2}{\left(2t^2+1\right)^2},\frac{4t}{\left(2t^2+1\right)^2},-\frac{4t}{\left(2t^2+1\right)^2}\right\rangle$$ This easily simplifies. $$\hat{\mathbf{N}}(t)=\left\langle-\frac{2t^2-1}{2t^2+1},\frac{2t}{2t^2+1},-\frac{2t}{2t^2+1}\right\rangle$$ Let us calculate $\hat{\mathbf{N}}(1)$. $$\hat{\mathbf{N}}(1)=\left\langle-\frac{1}{3},\frac{2}{3},-\frac{2}{3}\right\rangle$$ The final unit vector that we will find is $\hat{\mathbf{B}}(1)$. $$\hat{\mathbf{B}}(1)=\hat{\mathbf{T}}(1)\times\hat{\mathbf{N}}(1)=\left\langle-\frac{2}{3},\frac{1}{3},\frac{2}{3}\right\rangle$$ Next, let us find the curvature of the space curve when $t=1$. $$k(1)=\frac{\left\lVert\hat{\mathbf{T}}\,'(1)\right\rVert}{\left\lVert\vec{\mathbf{r}}\,'(1)\right\rVert}=\frac{2}{2(1)^2+1}\cdot\frac{1}{2(1)^2+1}=\frac{2}{3}\cdot\frac{1}{3}=\frac{2}{9}$$ The general equation of the osculating plane at the point $P(2,1,0)$ can be found with the unit binormal vector we found earlier as it is a normal vector to the plane [1]. $$ax+by+cz+d=0\implies-\frac{2}{3}x+\frac{1}{3}y+\frac{2}{3}z+d=0$$ We can easily solve for $d$ by using the point in question. $$-\frac{2}{3}(2)+\frac{1}{3}(1)+\frac{2}{3}(0)+d=0\implies d=1$$ We now have the general equation of the osculating plane and have finished solving the problem in its entirety. $$-\frac{2}{3}x+\frac{1}{3}y+\frac{2}{3}z+1=0$$