Answer to Question #149794 in Calculus for Haider

Question

Find T,N,B if possible for the following curves
(a)r(t)=(t³/3)i+(t²/2)j,t>0
(b) r(t)=(t³/3)I+(t²/2)j,t>=0

Answer 1

The unit tangent vector $\overrightarrow{T}$ , the unit normal vector $\overrightarrow{N}$  and the unit binormal vector $\overrightarrow{B}$  are three mutually perpendicular vectors used to describe a curve in two or three dimensions. This moving coordinate system is attached to the curve and describes the shape of the curve independent of any parameterization.

If the curve is given parametrically by

$$x=x(t)\quad\quad y=y(t)$$

the position, unit tangent, unit normal, and unit binormal vectors are

$$\overrightarrow{r}(t)=x(t)\cdot \overrightarrow{i}+y(t)\cdot \overrightarrow{j}\\[0.3cm] \overrightarrow{T}=\frac{\overrightarrow{r}'(t)}{\left|\overrightarrow{r}'(t)\right|} \\[0.3cm] \overrightarrow{N}=\frac{\overrightarrow{T}'(t)}{\left|\overrightarrow{T}'(t)\right|} \\[0.3cm] \overrightarrow{B}=\overrightarrow{T}(t)\times\overrightarrow{N}\\[0.3cm]$$

1 case :

$$\overrightarrow{r}(t)=\frac{t^3}{3}\cdot\overrightarrow{i}+\frac{t^2}{2}\cdot\overrightarrow{j},\,\,\,t>0\\[0.3cm] \overrightarrow{r}'(t)=\frac{d}{dt}\left(\frac{t^3}{3}\cdot\overrightarrow{i}+\frac{t^2}{2}\cdot\overrightarrow{j}\right)=\frac{3t^2}{3}\cdot\overrightarrow{i}+\frac{2t}{2}\cdot\overrightarrow{j}\\[0.3cm] \boxed{\overrightarrow{r}'(t)=t^2\cdot\overrightarrow{i}+t\cdot\overrightarrow{j}}\\[0.3cm] \left|\overrightarrow{r}'(t)\right|=\sqrt{(t^2)^2+t^2}=\sqrt{t^2\cdot\left(t^2+1\right)}=|t|\cdot\sqrt{t^2+1}\\[0.3cm] \left|\overrightarrow{r}'(t)\right|=[t>0]=t\cdot\sqrt{t^2+1}\\[0.3cm] \overrightarrow{T}(t)=\frac{\overrightarrow{r}'(t)}{\left|\overrightarrow{r}'(t)\right|}=\frac{t^2\cdot\overrightarrow{i}+t\cdot\overrightarrow{j}}{t\cdot\sqrt{t^2+1}}=\frac{t^2\cdot\overrightarrow{i}}{t\cdot\sqrt{t^2+1}}+\frac{t\cdot\overrightarrow{j}}{t\cdot\sqrt{t^2+1}}\\[0.3cm] \boxed{\overrightarrow{T}(t)=\frac{t\cdot\overrightarrow{i}}{\sqrt{t^2+1}}+\frac{\overrightarrow{j}}{\sqrt{t^2+1}}}\\[0.3cm] \overrightarrow{T}'(t)=\frac{d}{dt}\left(\frac{t\cdot\overrightarrow{i}}{\sqrt{t^2+1}}+\frac{\overrightarrow{j}}{\sqrt{t^2+1}}\right)=\\[0.3cm] =\left(\frac{1}{\sqrt{t^2+1}}+t\cdot\left(-\frac{1}{2}\cdot\frac{2t}{\left(t^2+1\right)^{3/2}}\right)\right)\cdot\overrightarrow{i}+\left(-\frac{1}{2}\cdot\frac{2t}{\left(t^2+1\right)^{3/2}}\right)\cdot\overrightarrow{j}=\\[0.3cm] =\left(\frac{t^2+1-t^2}{\left(t^2+1\right)^{3/2}}\right)\cdot\overrightarrow{i}+\left(-\frac{t}{\left(t^2+1\right)^{3/2}}\right)\cdot\overrightarrow{j}=\\[0.3cm] =\left(\frac{1}{\left(t^2+1\right)^{3/2}}\right)\cdot\overrightarrow{i}+\left(-\frac{t}{\left(t^2+1\right)^{3/2}}\right)\cdot\overrightarrow{j}\\[0.3cm] \left|\overrightarrow{T}(t)\right|=\sqrt{\left(\frac{1}{\left(t^2+1\right)^{3/2}}\right)^2+\left(\frac{t}{\left(t^2+1\right)^{3/2}}\right)^2}=\\[0.3cm] =\sqrt{\frac{t^2+1}{\left(t^2+1\right)^3}}=\sqrt{\frac{1}{\left(t^2+1\right)^2}}=\frac{1}{t^2+1}\longrightarrow\frac{1}{\left|\overrightarrow{T}'(t)\right|}=(t^2+1)\\[0.3cm] \overrightarrow{N}(t)=\frac{\overrightarrow{T}'(t)}{\left|\overrightarrow{T}'(t)\right|}=(t^2+1)\cdot\left(\left(\frac{1}{\left(t^2+1\right)^{3/2}}\right)\cdot\overrightarrow{i}+\left(-\frac{t}{\left(t^2+1\right)^{3/2}}\right)\cdot\overrightarrow{j}\right)\\[0.3cm] =\left(\frac{1}{\left(t^2+1\right)^{1/2}}\right)\cdot\overrightarrow{i}+\left(-\frac{t}{\left(t^2+1\right)^{1/2}}\right)\cdot\overrightarrow{j}\\[0.3cm] \boxed{\overrightarrow{N}(t)=\frac{\overrightarrow{i}}{\sqrt{t^2+1}}-\frac{t\cdot\overrightarrow{j}}{\sqrt{t^2+1}}}\\[0.3cm] \overrightarrow{B}=\overrightarrow{T}\times\overrightarrow{N}=\left|\begin{array}{ccc} \overrightarrow{i}&\overrightarrow{j}&\overrightarrow{k}\\ T_x&T_y&T_z=0\\ N_x&N_y&N_z=0 \end{array}\right|=\overrightarrow{k}\cdot(T_xN_y-T_yN_x)=\\[0.3cm] =\overrightarrow{k}\cdot\left(\left(\frac{t}{\sqrt{t^2+1}}\right)\cdot\left(\frac{-t}{\sqrt{t^2+1}}\right)-\left(\frac{1}{\sqrt{t^2+1}}\right)\cdot\left(\frac{1}{\sqrt{t^2+1}}\right)\right)=\\[0.3cm] =\overrightarrow{k}\cdot\left(\frac{-(t^2+1)}{t^2+1}\right)=-1\cdot\overrightarrow{k}\\[0.3cm] \boxed{\overrightarrow{B}=-1\cdot\overrightarrow{k}}$$

2 case :

$$\overrightarrow{r}(t)=\frac{t^3}{3}\cdot\overrightarrow{i}+\frac{t^2}{2}\cdot\overrightarrow{j},\,\,\,t\ge0\\[0.3cm] \overrightarrow{r}'(t)=\frac{d}{dt}\left(\frac{t^3}{3}\cdot\overrightarrow{i}+\frac{t^2}{2}\cdot\overrightarrow{j}\right)=\frac{3t^2}{3}\cdot\overrightarrow{i}+\frac{2t}{2}\cdot\overrightarrow{j}\\[0.3cm] \boxed{\overrightarrow{r}'(t)=t^2\cdot\overrightarrow{i}+t\cdot\overrightarrow{j}}\\[0.3cm] \left|\overrightarrow{r}'(t)\right|=\sqrt{(t^2)^2+t^2}=\sqrt{t^2\cdot\left(t^2+1\right)}=|t|\cdot\sqrt{t^2+1}\\[0.3cm] \left|\overrightarrow{r}'(t)\right|=[t\ge0]=t\cdot\sqrt{t^2+1}\\[0.3cm]$$

There is a problem with the parameter value $t=0$ , since

$$\overrightarrow{r}'(0)=0^2\cdot\overrightarrow{i}+0\cdot\overrightarrow{j}\\[0.3cm] \left|\overrightarrow{r}'(0)\right|=0\cdot\sqrt{0^2+1}=0\\[0.3cm] \overrightarrow{T}(0)=\frac{0}{0}\cdot\overrightarrow{i}+\frac{0}{0}\cdot\overrightarrow{j}-\text{formal entry}$$

It means that $\nexists\overrightarrow{T}(0)$ and as a consequence $\nexists\overrightarrow{N}(0)$ and $\nexists\overrightarrow{B}(0)$ .

ANSWER

$$\overrightarrow{r}(t)=\frac{t^3}{3}\cdot\overrightarrow{i}+\frac{t^2}{2}\cdot\overrightarrow{j},\,\,\,t>0\\[0.3cm] \overrightarrow{T}(t)=\frac{t\cdot\overrightarrow{i}}{\sqrt{t^2+1}}+\frac{\overrightarrow{j}}{\sqrt{t^2+1}}\\[0.3cm] \overrightarrow{N}(t)=\frac{\overrightarrow{i}}{\sqrt{t^2+1}}-\frac{t\cdot\overrightarrow{j}}{\sqrt{t^2+1}}\\[0.3cm] \overrightarrow{B}=-1\cdot\overrightarrow{k}\\[0.3cm] \overrightarrow{r}(t)=\frac{t^3}{3}\cdot\overrightarrow{i}+\frac{t^2}{2}\cdot\overrightarrow{j},\,\,\,t\ge0\\[0.3cm] \overrightarrow{T}(t)=\frac{t\cdot\overrightarrow{i}}{\sqrt{t^2+1}}+\frac{\overrightarrow{j}}{\sqrt{t^2+1}},\,\,\,t>0\,\,\,\text{and}\,\,\,\nexists\overrightarrow{T}(0)\\[0.3cm] \overrightarrow{N}(t)=\frac{\overrightarrow{i}}{\sqrt{t^2+1}}-\frac{t\cdot\overrightarrow{j}}{\sqrt{t^2+1}},\,\,\,t>0\,\,\,\text{and}\,\,\,\nexists\overrightarrow{N}(0)\\[0.3cm] \overrightarrow{B}=-1\cdot\overrightarrow{k},\,\,\,t>0\,\,\,\text{and}\,\,\,\nexists\overrightarrow{B}(0)$$