Question

Q. Find the vectors t, n, b at the given point to the curve.
γ(t)=(t^2,2/3 t^3,t), (1, 2/3, 1)

Answer 1

Answer on Question 72688, Math / Geometry

Question

Find the vectors $t, n, b$ at the given point to the curve.

$$y(t) = (t^2, 2/3, t^3, 3/t), \quad (1, 2/3, 1).$$

Solution.

We are given the curve

$$\vec{r}(t) = (t^2)\vec{i} + \left(\frac{2}{3}t^3\right)\vec{j} + (t)\vec{k}$$

and the point $(1, 2/3, 1)$

Note that $t = 1$ because $x_0 = (1^2) = 1$, $y_0 = \left(\frac{2}{3} \cdot 1^3\right) = \frac{2}{3}$, $z_0 = (1) = 1$ is the coordinates of the given point.

1) Find the unit tangent vector $\vec{\tau}(t) = \frac{\vec{r}^j(t)}{|\vec{r}^j(t)|}$

$$\vec{r}'(t) = (2t)\vec{i} + (2t^2)\vec{j} + (1)\vec{k}$$ $$|\vec{r}'(t)| = \sqrt{4t^2 + 4t^4 + 1} = \sqrt{(2t^2 + 1)^2} = 2t^2 + 1$$

Thus we get

$$\vec{\tau}(t) = \frac{\vec{r}'(t)}{|\vec{r}'(t)|} = \left(\frac{2t}{2t^2 + 1}\right)\vec{i} + \left(\frac{2t^2}{2t^2 + 1}\right)\vec{j} + \left(\frac{1}{2t^2 + 1}\right)\vec{k}$$

2) Find the unit normal vector $\vec{n}(t) = \frac{\vec{r}^j(t)}{|\vec{r}^j(t)|}$

$$\begin{aligned} \vec{t}'(t) &= \left(\frac{2 \cdot (2t^2 + 1) - 2t \cdot 4t}{(2t^2 + 1)^2}\right)\vec{i} + \left(\frac{4t \cdot (2t^2 + 1) - 2t^2 \cdot 4t}{(2t^2 + 1)^2}\right)\vec{j} + \left(\frac{-4t}{(2t^2 + 1)^2}\right)\vec{k} \\ &= \left(\frac{2 - 4t^2}{(2t^2 + 1)^2}\right)\vec{i} + \left(\frac{4t}{(2t^2 + 1)^2}\right)\vec{j} + \left(\frac{-4t}{(2t^2 + 1)^2}\right)\vec{k} \end{aligned}$$ $$\begin{aligned} |\vec{\tau}'(t)| &= \frac{\sqrt{(2 - 4t^2)^2 + (4t)^2 + (-4t)^2}}{(2t^2 + 1)^2} = \frac{\sqrt{4 - 16t^2 + 16t^4 + 16t^2 + 16t^2}}{(2t^2 + 1)^2} \\ &= \frac{\sqrt{4 + 16t^2 + 16t^4}}{(2t^2 + 1)^2} = \frac{\sqrt{4(1 + 4t^2 + 4t^4)}}{(2t^2 + 1)^2} = \frac{2\sqrt{(1 + 2t^2)^2}}{(2t^2 + 1)^2} = \frac{2(1 + 2t^2)}{(2t^2 + 1)^2} = \frac{2}{(2t^2 + 1)} \end{aligned}$$

Thus we get

$$\begin{aligned} \vec{n}(t) &= \frac{\vec{\tau}'(t)}{|\vec{\tau}'(t)|} = \frac{\left(\frac{2 - 4t^2}{(2t^2 + 1)^2}\right)\vec{i} + \left(\frac{4t}{(2t^2 + 1)^2}\right)\vec{j} - \left(\frac{4t}{(2t^2 + 1)^2}\right)\vec{k}}{\frac{2}{(2t^2 + 1)}} \\ &= \left(\frac{1 - 2t^2}{2t^2 + 1}\right)\vec{i} + \left(\frac{2t}{2t^2 + 1}\right)\vec{j} - \left(\frac{2t}{2t^2 + 1}\right)\vec{k} \end{aligned}$$

3) Find the unit binormal vector $\vec{b}(t) = \vec{\tau} \times \vec{n}$

$$\vec{b}(t) = \vec{\tau} \times \vec{n} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ \dfrac{2t}{2t^2 + 1} & \dfrac{2t^2}{2t^2 + 1} & \dfrac{1}{2t^2 + 1} \\ \dfrac{1 - 2t^2}{2t^2 + 1} & \dfrac{2t}{2t^2 + 1} & -\dfrac{2t}{2t^2 + 1} \end{vmatrix} = \vec{i}\left(-\dfrac{4t^3}{(2t^2 + 1)^2} - \dfrac{2t}{(2t^2 + 1)^2}\right) - \vec{j}\left(-\dfrac{4t^2}{(2t^2 + 1)^2} - \dfrac{1 - 2t^2}{(2t^2 + 1)^2}\right) + \vec{k}\left(\dfrac{4t^2}{(2t^2 + 1)^2} - \dfrac{2t^2 - 4t^4}{(2t^2 + 1)^2}\right) = -\left(\dfrac{2t(2t^2 + 1)}{(2t^2 + 1)^2}\right)\vec{i} + \left(\dfrac{2t^2 + 1}{(2t^2 + 1)^2}\right)\vec{j} + \vec{k}\left(\dfrac{2t^2(2t^2 + 1)}{(2t^2 + 1)^2}\right) = -\left(\dfrac{2t}{2t^2 + 1}\right)\vec{i} + \left(\dfrac{1}{2t^2 + 1}\right)\vec{j} + \left(\dfrac{2t^2}{2t^2 + 1}\right)\vec{k}$$

Thus, for $t = 1$ that is at the given point $(1, 2/3, 1)$ we have:

$$\vec{\tau}(1,2/3,1) = \frac{2}{3}\vec{i} + \frac{2}{3}\vec{j} + \frac{1}{3}\vec{k}$$ $$\vec{n}(1,2/3,1) = -\frac{1}{3}\vec{i} + \frac{2}{3}\vec{j} - \frac{2}{3}\vec{k}$$ $$\vec{b}(1,2/3,1) = -\frac{2}{3}\vec{i} + \frac{1}{3}\vec{j} + \frac{2}{3}\vec{k}$$

Answer: at the given point:

the unit tangent vector is

$$\vec{\tau}(1,2/3,1) = \frac{2}{3}\vec{i} + \frac{2}{3}\vec{j} + \frac{1}{3}\vec{k}$$

the unit normal vector is

$$\vec{n}(1,2/3,1) = -\frac{1}{3}\vec{i} + \frac{2}{3}\vec{j} - \frac{2}{3}\vec{k}$$

the unit binormal vector is

$$\vec{b}(1,2/3,1) = -\frac{2}{3}\vec{i} + \frac{1}{3}\vec{j} + \frac{2}{3}\vec{k}$$

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