Orientation & Normal Vectors - Multivariable Calculus

1. A surface $S$ is parameterized by $\mathbf{r}(u,v) = (u^2, v^2, u + v)$ where $u, v \in [0, 1]$. At the point $(1, 1, 2)$ corresponding to $u = 1, v = 1$, which statement about the orientation is correct?
   Answer: The normal vector $\mathbf{r}_u \times \mathbf{r}_v = (-2, -2, 4)$ points in the direction of increasing z When you encounter a surface parameterization problem asking about orientation, you need to find the normal vector using the cross product of the partial derivatives, then determine which direction it points. First, find the partial derivatives of $\mathbf{r}(u,v) = (u^2, v^2, u + v)$: - $\mathbf{r}_u = (2u, 0, 1)$ - $\mathbf{r}_v = (0, 2v, 1)$ At the point where $u = 1, v = 1$: - $\mathbf{r}_u = (2, 0, 1)$ - $\mathbf{r}_v = (0, 2, 1)$ Calculate the cross product $\mathbf{r}_u \times \mathbf{r}_v$: $\mathbf{r}_u \times \mathbf{r}_v = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = (0-2)\mathbf{i} - (2-0)\mathbf{j} + (4-0)\mathbf{k} = (-2, -2, 4)$ Since the z-component is positive (4), this normal vector points in the direction of increasing z. Therefore, answer B is correct. Answer A uses the wrong cross product $\mathbf{r}_v \times \mathbf{r}_u$ (which reverses the orientation) and incorrectly states the direction. Answer C has the correct normal vector but wrongly claims it points toward decreasing z when the positive z-component clearly indicates increasing z. Answer D contains computational errors in both the y and z components of the normal vector. Remember: the cross product $\mathbf{r}_u \times \mathbf{r}_v$ follows the right-hand rule, and the sign of the z-component tells you the vertical orientation—positive means increasing z, negative means decreasing z.
2. Consider the surface $S$ defined by $x^2 + y^2 = z^2$ for $1 \leq z \leq 3$. At the point $(2, 2, 2\sqrt{2})$, if the surface is oriented with normal vectors pointing away from the z-axis, what is the direction of the unit normal vector?
   Answer: $\frac{(1, 1, -\sqrt{2})}{2}$ For the surface $x^2 + y^2 - z^2 = 0$, the normal vector is $\nabla(x^2 + y^2 - z^2) = (2x, 2y, -2z)$. At $(2, 2, 2\sqrt{2})$, this gives $(4, 4, -4\sqrt{2})$. The magnitude is $\sqrt{16 + 16 + 32} = \sqrt{64} = 8$. To check orientation: the vector from the z-axis to point $(2, 2, 2\sqrt{2})$ is $(2, 2, 0)$. The dot product $(4, 4, -4\sqrt{2}) \cdot (2, 2, 0) = 16 > 0$, so this normal points away from the z-axis as desired. The unit normal is $\frac{(4, 4, -4\sqrt{2})}{8} = \frac{(1, 1, -\sqrt{2})}{2}$.
3. Consider the hemisphere $x^2 + y^2 + z^2 = 9$ with $z \geq 0$. If this surface is oriented with the outward-pointing normal vectors, what is the z-component of the unit normal vector at the point $(0, 0, 3)$?
   Answer: $1$ For the sphere $x^2 + y^2 + z^2 = 9$, the gradient $\nabla(x^2 + y^2 + z^2) = (2x, 2y, 2z)$ gives the direction of the normal vector. At $(0, 0, 3)$, this is $(0, 0, 6)$. The unit normal is $(0, 0, 1)$. Since we want the outward-pointing normal and the center of the sphere is at the origin, the vector from center to point $(0, 0, 3)$ is $(0, 0, 3)$, which points in the positive z-direction. Therefore, the outward normal at $(0, 0, 3)$ has z-component $1$.
4. Let $S$ be the part of the plane $2x + 3y + z = 6$ that lies in the first octant. If $S$ is oriented so that the normal vector has a positive z-component, what is the unit normal vector to $S$?
   Answer: $\frac{(2, 3, 1)}{\sqrt{14}}$ For the plane $2x + 3y + z = 6$, the normal vector is the gradient of $2x + 3y + z - 6$, which is $(2, 3, 1)$. The magnitude is $\sqrt{2^2 + 3^2 + 1^2} = \sqrt{14}$. Since we want the normal with positive z-component, we use $(2, 3, 1)$ rather than $(-2, -3, -1)$. Therefore, the unit normal vector is $\frac{(2, 3, 1)}{\sqrt{14}}$.
5. Consider the surface $S$ given by $z = \sqrt{4 - x^2 - y^2}$ for $x^2 + y^2 \leq 4$. This surface can be oriented in two ways. If we choose the orientation such that the normal vectors point away from the origin, what is the correct expression for the unit normal vector?
   Answer: $\frac{(x, y, z)}{2}$ When finding unit normal vectors to surfaces, you need to determine both the direction and magnitude of the normal vector, then consider which orientation matches the given conditions. The surface $z = \sqrt{4 - x^2 - y^2}$ represents the upper half of a sphere with radius 2 centered at the origin. To find the normal vector, rewrite this as $F(x,y,z) = x^2 + y^2 + z^2 - 4 = 0$. The gradient $\nabla F = (2x, 2y, 2z)$ gives a normal vector, which simplifies to $(x, y, z)$ after factoring out the constant. Since we want the normal vectors to point away from the origin, we need the outward normal. For any point $(x, y, z)$ on this sphere, the vector from the origin to that point is exactly $(x, y, z)$, which points outward. The magnitude of this vector is $\sqrt{x^2 + y^2 + z^2} = \sqrt{4} = 2$ (since all points satisfy $x^2 + y^2 + z^2 = 4$). Therefore, the unit normal vector is $\frac{(x, y, z)}{2}$, making B correct. Answer A, $\frac{(-x, -y, z)}{2}$, points inward in the $x$ and $y$ directions, which would be incorrect for the outward orientation. Answer C, $\frac{(x, y, -z)}{2}$, has the wrong $z$-component and would point toward the lower hemisphere. Answer D, $\frac{(-x, -y, -z)}{2}$, points completely inward toward the origin. Remember: for spheres, the outward normal at any point is simply the position vector from the center to that point, normalized by the radius.
6. Consider the surface $S$ parameterized by $\mathbf{r}(s,t) = (s\cos t, s\sin t, s^2)$ for $s > 0$. At which of the following points does the normal vector point in the direction that makes the smallest angle with the positive z-axis?
   Answer: $(\frac{1}{2}, 0, \frac{1}{4})$ corresponding to $s = \frac{1}{2}, t = 0$ The normal vector is $\mathbf{r}_s \times \mathbf{r}_t$ where $\mathbf{r}_s = (\cos t, \sin t, 2s)$ and $\mathbf{r}_t = (-s\sin t, s\cos t, 0)$. Computing the cross product: $\mathbf{r}_s \times \mathbf{r}_t = (-2s^2\cos t, -2s^2\sin t, s)$. The angle with the positive z-axis depends on the z-component relative to the magnitude. The unit normal has z-component $\frac{s}{\sqrt{4s^4 + s^2}} = \frac{1}{\sqrt{4s^2 + 1}}$. This is maximized when $s$ is smallest, which occurs at $s = \frac{1}{2}$, giving the smallest angle with the z-axis.
7. Let $S$ be the surface parameterized by $\mathbf{r}(\theta, \phi) = (\sin\phi\cos\theta, \sin\phi\sin\theta, \cos\phi)$ for $0 \leq \theta \leq 2\pi$ and $0 \leq \phi \leq \frac{\pi}{2}$. This parameterization naturally induces an orientation on $S$. At the point corresponding to $\theta = \frac{\pi}{4}, \phi = \frac{\pi}{4}$, does the normal vector point toward or away from the origin?
   Answer: Toward the origin, because $\mathbf{r}_{\theta} \times \mathbf{r}_{\phi}$ points inward for this parameterization For the spherical parameterization, $\mathbf{r}_{\theta} = (-\sin\phi\sin\theta, \sin\phi\cos\theta, 0)$ and $\mathbf{r}_{\phi} = (\cos\phi\cos\theta, \cos\phi\sin\theta, -\sin\phi)$. The normal vector $\mathbf{r}_{\theta} \times \mathbf{r}_{\phi}$ can be computed, but more directly: this parameterization gives $\mathbf{r}_{\theta} \times \mathbf{r}_{\phi} = -\sin\phi \cdot \mathbf{r}$, which points toward the origin (opposite to the position vector). This is the inward normal orientation for this particular parameterization order.

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