Solve the problem.
-$$\text { Find parametric equations for the normal line to the surface } z = \ln \left( 7 x ^ { 2 } + 4 y ^ { 2 } + 1 \right) \text { at the point } ( 0,0,0 ) \text {. }$$
A) x = 0, y = 0, z = t
B) x = t , y = t, z = 0
C) x = t, y = t, z = -1
D) x = 1, y = 1, z = -t

Question

Quizplus · Accepted Answer

To find the normal line to the surface, we first need to find the gradient vector evaluated at the point $(0,0,0)$. The surface equation $z = \ln \left( 7 x ^ { 2 } + 4 y ^ { 2 } + 1 ight)$ can be rewritten as $F(x,y,z) = \ln \left( 7 x ^ { 2 } + 4 y ^ { 2 } + 1 ight) - z = 0$. Then, the gradient vector of $F$ is given by $ abla F = \left< \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z} ight>$. Taking partial derivatives, we have $\frac{\partial F}{\partial x} = \frac{14x}{7 x ^ { 2 } + 4 y ^ { 2 } + 1}$, $\frac{\partial F}{\partial y} = \frac{8y}{7 x ^ { 2 } + 4 y ^ { 2 } + 1}$, and $\frac{\partial F}{\partial z} = -1$. Evaluating at $(0,0,0)$, we have $ abla F(0,0,0) = \left< \frac{0}{1}, \frac{0}{1}, -1 ight> = \left< 0, 0, -1 ight>$. Since the normal line passes through $(0,0,0)$, it has the parameterization $\vec{r}(t) = <0,0,0> + t \vec{N}$, where $\vec{N}$ is the unit normal vector. Therefore, the parametric equations for the normal line are $x=0$, $y=0$, and $z = -t$ (since $\vec{N}$ is in the direction of $\left< 0, 0, -1 ight>$). Thus, the best choice is $\boxed{ extbf{(A)}}$.

Solve the Problem A) X = 0, Y = 0, Z = T