Find the parametric equation for the line that is tangent to the curve

Question

Find the parametric equation for the line that is tangent to the curve $ \ \ \vec{r}(t)=(\frac{8}{t}, -\frac{1}{2}t^{2}, \frac{1}{8}t^{3}) $ and parallel to the plane $ x=y$. $$ $$ My approach- $ \frac{dr}{dt}=(-\frac{8}{t^{2}}, -t, \frac{3}{8}t^{2} ) $ . Now the plane is $ \ x-y=0 \ $. The normal to the plane is $ \ \vec{n}=(1,-1,0) $. Therefore, $ \frac{dr}{dt} \cdot \vec{n}=0 $. This gives $ t=2$. Then the tangent vector is $ (-\frac{8}{t^{2}}, -t, \frac{3}{8}t^{2})_{t=1} $= $(-2-2,\frac{3}{2}) $. This is the tangent vector. But I need the parametric form . Any help finding the parametric equation of the tangent according the question ?

Answer 1

The coordinates of the point on the curve where $t=2$ are $(4,-2,1)$ so the equation of the line is $$\underline{r}=\left(\begin{matrix}4\\-2\\1\end{matrix}\right)+\lambda\left(\begin{matrix}-2\\-2\\ \frac 32\end{matrix}\right)$$

Answer 2

For the tangent straight line: $\vec s(2)=(4,-2,1)$ and $\vec s'(t)=(-2,-2,3/2)$ constant.

$\vec s(t)=(-2t+4,-2t-2,3t/2+1)$

Answer 3

The point of tangency is common to both the curve and the plane, $x=y$. We find the value of $t$ when this happen. $$\frac{dr}{dt} \bot n \mbox { hence } \frac{dr}{dt}\cdot n = 0 $$. From your calculation,it is right that $ t=2$. That is the parameter at the point of intersection of $\frac{dr}{dt}$ and the normal to the plane, $n$. The coordinates of the point will be $(\frac{8}{2},-\frac{1}{2}\times 2^2,\frac{1}{8}\times 2^3)=(4,-2,1)$

The slope of the line is $\frac{dr}{dt}=-\frac{8}{t^2}\hat{i} -t\hat{j}+\frac{3}{8}t^2\hat{k}$. At $t=2$,this will be $\frac{dr}{dt}=-2\hat{i}-2\hat{j}+\frac{3}{2}\hat{k}$.

The parametric equations of the line are $x(s)=x_0 + v_xs, y(s)=y_0+v_y s , z(s)=z_0+v_z s $ where $s$ is a parameter and $v=\langle v_x, v_y, v_z \rangle$ the vector in the direction of the line.

Thus, the parametric equations are $\begin{cases}x(s)&=4 -2s\\ y(s)&=-2-2s \\ z(s)&=1+\frac{3}{2}s \end{cases}$