At what point on the curve $x = t ^ { 3 } , y = 9 t , z = t ^ { 4 }$ is the normal plane parallel to the plane $3 x + 9 y - 4 z = 8 ?$
Select the correct answer.
A) $( - 9,1,1 )$
B) $( - 1 , - 9,1 )$
C) $( 9,18 , - 2 )$
D) $( - 18,9,1 )$
E) $( - 1 , - 3,9 )$

Question

Quizplus · Accepted Answer

First, we find the curve's velocity vector by differentiating each component using the chain rule: $\vec{r}'(t) = \langle 3t^2, 9, 4t^3angle$. At the point where the normal plane is parallel to $3x+9y-4z=8$, the normal vector to the curve must be perpendicular to the normal vector of the given plane, which is $\langle 3, 9, -4angle$. Thus, we need to find a point on the curve where $\vec{r}'(t)$ is proportional to $\langle 3, 9, -4angle$. Let $k$ be this proportionality constant, so $\vec{r}'(t) = k\langle 3, 9, -4angle$. Equating the first two components gives $3t^2/k = 3$, so $k = t^2$. Equating the first and third components gives $4t^3/k = -4$, so $k = -t^2$. Thus, $t^2 = -t^2$, which implies $t=0$ or $t=\pm\sqrt{-1}$. Since $t$ must be real, there are no points on the curve that satisfy the given conditions except for $t=0$. Plugging in $t=0$ to the equations for the curve gives the point $(0,0,0)$. To find the normal plane at this point, we need the cross product of the two partial derivatives of $\vec{r}$: $\vec{r}_x = \langle 3t^2, 0, 4t^3angle$ and $\vec{r}_y = \langle 0, 9, 0angle$. Evaluating at $(0,0,0)$ gives $\vec{r}_x(0,0,0) = \langle 0, 0, 0angle$ and $\vec{r}_y(0,0,0) = \langle 0, 9, 0angle$. Thus, the normal vector is $\vec{n} = \vec{r}_x	imes\vec{r}_y = \langle 0, 0, 27angle$. Equating this to a multiple of $\langle 3, 9, -4angle$ gives $\vec{n}/3 = \langle 0, 0, 9angle$ as the normal vector of the desired plane. Thus, the equation of the normal plane at $(0,0,0)$ is $0(x-0)+0(y-0)+9(z-0) = 0$, which simplifies to $z=0$. Since $3x+9y-4z=8$ intersects this plane at $z=8/4=2$, we see that the normal plane and the given plane do not intersect and are thus parallel. The only point on the curve where this occurs is $(0,0,0)$, so the answer is $\boxed{	extbf{(B)}}$.

At What Point on the Curve x=t3,y=9t,z=t4x = t ^ { 3 } , y = 9 t , z = t ^ { 4 }x=t3,y=9t,z=t4

At What Point on the Curve $x = t ^ { 3 } , y = 9 t , z = t ^ { 4 }$