Consider the curve given parametrically by $(x,y,z)=(2−t,−1−t^2,−2t−3t^3)$. There is a unique point $P$ on the curve with the property that the tangent line at $P$ passes through the point $(2,8,−162)$. Find the coordinates of the point $P$.
Can anyone give me a hand with this one? I honestly have no idea where to start other than rewriting the curve as,
$\vec{r(t)}$ = $\langle 2-t, -1-t^2, -2t-3t^3\rangle$, and then taking the derivate to obtain,
$\vec{r'(t)}$ = $\langle -1, -2t, -2 -9t^2\rangle$
Best Answer
Let $\vec{r}(t) = \langle 2−t,−1−t^2,−2t−3t^3\rangle$, where $t\in \mathbb{R}$. Then $\vec{r}'(t) = \langle -1, -2t, -2 -9t^2\rangle$.
For any $t_0$ in the domain, a parametric equation of the tangent line at $t_0$ is given by $$ \vec{T}(t) = t \:\vec{r}'(t_0) + \vec{r}(t_0), \mbox{ where } t\in \mathbb{R}. $$ Now, we want to find an equation of the tangent line when it is passing through $(2,8, -162)$.
Thus, setting the following equal to each other, \begin{align*} t\langle -1, -2t, -2 -9t^2\rangle + \langle 2−t_0,−1−t_0^2,−2t_0−3t_0^3\rangle = \langle 2,8, -162 \rangle, \end{align*} we need to solve for $t_0$.
The above vector equation gives us three equations: \begin{align*} -t+2-t_0&=2, \hspace{4mm} (\dagger) \\ -2t_0t-1-t_0^2&=8, \hspace{4mm} (\ddagger) \\ -2t-9t_0^2t-2t_0-3t_0^3&=-162. \hspace{4mm} (\Omega)\\ \end{align*} Use the first equation $(\dagger)$ to solve for $t:$ $$ t=-t_0. $$ Let's substitute this into the second equation $(\ddagger)$: $$ 2t_0^2 -1-t_0^2 = 8, $$ which gives us $t_0^2 = 9$. So $t_0 = \pm 3$ while $t=\mp 3$.
If $t_0=3$, then $t=-3$. We substitute these into the third equation $(\Omega)$ to obtain: $$ 6+9\cdot 9\cdot 3 -2(3)-3(27) = 9(27)-3(27)=6(27) \not= -162. $$ We conclude that $t_0=-3$ while $t=3$, which we can check using the third equation $(\Omega)$: $$ -6-9\cdot 9\cdot 3 +2(3)+3(27) = -9(27)+3(27)=-6(27) = -162. $$ So $P$ has position vector $\vec{r}(-3)=\langle 5,-10,87 \rangle $, which is when its tangent line passes through the point $(2,8,−162)$.