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\(11-14=\) Match the equations with the graphs labeled I-IV and give reasons for your answers. Determine which families of grid curves have \(u\) constant and which have \(v\) constant. $$ \mathbf{r}(u, v)=\sin v \mathbf{i}+\cos u \sin 2 v \mathbf{j}+\sin u \sin 2 v \mathbf{k} $$

Short Answer

Expert verified
u-constant curves are wave-like; v-constant curves are circles. Match based on these observations.

Step by step solution

01

Understanding Parametric Equations

We are given a vector function \( \mathbf{r}(u, v)=\sin v \mathbf{i}+\cos u \sin 2 v \mathbf{j}+\sin u \sin 2 v \mathbf{k} \). This function defines a parametric surface based on the parameters \(u\) and \(v\). \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) are the unit vectors along the x, y, and z axes, respectively.
02

Identify u-constant grid curves

If \( u \) is constant, the position vector simplifies to \( \mathbf{r}(v) = \sin v \mathbf{i} + (\cos u \sin 2v) \mathbf{j} + (\sin u \sin 2v) \mathbf{k} \). This suggests that the curve is a function of \(v\) alone, implicitly defining a line of constant \(u\).
03

Analyze v-constant grid curves

If \( v \) is constant, the position vector becomes \( \mathbf{r}(u) = \sin v \mathbf{i} + (\cos u C) \mathbf{j} + (\sin u C) \mathbf{k} \) where \( C = \sin 2v \). This is essentially a parametric equation for a circle in the YZ-plane.
04

Match with Graphs

By examining the simplified parametric equations for constant \(u\) and \(v\), we look for graphs that display patterns:- For \(u\) constant: graphs should reflect sinusoidal (wave-like) variations in \(v\).- For \(v\) constant: graphs should resemble circular or elliptical paths, as they are circles projected onto the YZ-plane.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Functions
Vector functions are a powerful tool in mathematics that allow us to describe geometric shapes in a multi-dimensional space using parameters. In this particular exercise, the vector function given is \( \mathbf{r}(u, v)=\sin v \mathbf{i}+\cos u \sin 2 v \mathbf{j}+\sin u \sin 2 v \mathbf{k} \). Here, \( \mathbf{r}(u, v) \) is a parametric surface depending on two parameters, \(u\) and \(v\). A vector function like this consists of:
  • Parameters \(u\) and \(v\), which affect the shape and position of the surface.
  • The unit vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\), which indicate directions along the x, y, and z axes respectively.
  • Sinusoidal components, such as \(\sin v\), \(\cos u\), and \(\sin 2v\), that introduce wave-like characteristics to the shapes.
By varying \(u\) and \(v\), we explore different configurations of the surface, explaining its huge versatility. This specific function maps a point on a surface in 3D space for given values of \(u\) and \(v\), enabling visualization of complex geometric forms.
Grid Curves
Grid curves are an essential concept when exploring parametric surfaces, like the one we analyzed in this exercise. They help us understand how the surface is generated by the vector function when either \(u\) or \(v\) is held constant. Let's delve into the details:
  • u-constant Grid Curves: When \(u\) is constant, the vector function simplifies to \( \mathbf{r}(v) = \sin v \mathbf{i} + (\cos u \sin 2v) \mathbf{j} + (\sin u \sin 2v) \mathbf{k} \). Here, we observe that the curve is influenced by changes in \(v\), creating wave-like or sinusoidal patterns. This means these grid curves reflect how the surface behaves with varying \(v\), like horizontal stripes over the surface.

  • v-constant Grid Curves: Conversely, keeping \(v\) constant, the vector function reduces to \( \mathbf{r}(u) = \sin v \mathbf{i} + (\cos u C) \mathbf{j} + (\sin u C) \mathbf{k} \), with \( C = \sin 2v \). This results in circular paths, as \(u\) varies, effectively showing vertical "slices" of the surface.
Understanding grid curves assists in visualizing complex surfaces by focusing on the impact of one variable while keeping the other fixed, and is crucial in interpreting parametric graph shapes.
Unit Vectors
Unit vectors are fundamental in the world of vector math, acting as the building blocks for expressing direction. In our given vector function \(\mathbf{r}(u, v) = \sin v \mathbf{i} + \cos u \sin 2 v \mathbf{j} + \sin u \sin 2 v \mathbf{k}\), they have specific roles:
  • Direction Indicators: The unit vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) denote directions along the x, y, and z axes, respectively. Each has a magnitude of 1, ensuring that they only indicate direction and not distance.

  • Permitting Combination: By combining them with coefficients derived from the sinusoidal functions, we define positions and shapes in space for every point described by the function \(\mathbf{r}(u, v)\). For instance, \(\sin v \mathbf{i}\) focuses on the x-direction behavior dictated by \(v\).

  • Maintaining Orthogonality: The orthogonal nature of \(\mathbf{i}, \mathbf{j},\) and \(\mathbf{k}\) means they stay perpendicular to each other within the 3D coordinate system, simplifying calculations and understanding of the spatial geometry involved.
Understanding unit vectors helps in grasping how directions and magnitudes determine the geometry and orientation of surfaces described by vector functions. They are a key aspect whenever interpreting such parametric functions.

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Most popular questions from this chapter

(a) Set up, but do not evaluate, a double integral for the area of the surface with parametric equations \(x=a u \cos v, y=b u \sin v, z=u^{2}, 0 \leqslant u \leqslant 2\) 0\(\leqslant v \leqslant 2 \pi\) (b) Eliminate the parameters to show that the surface is an elliptic paraboloid and set up another double integral for the surface area. (c) Use the parametric equations in part (a) with \(a=2\) and \(b=3\) to graph the surface. (d) For the case \(a=2, b=3,\) use a computer algebra system to find the surface area correct to four decimal places.

Use Green's Theorem in the form of Equation 13 to prove Green's first identity: \(\iint_{D} f \nabla^{2} g d A=\oint_{C} f(\nabla g) \cdot \mathbf{n} d s-\iint_{D} \nabla f \cdot \nabla g d A\) where \(D\) and \(C\) satisfy the hypotheses of Green's Theorem and the appropriate partial derivatives of \(f\) and \(g\) exist and are continuous. (The quantity \(\nabla g \cdot \mathbf{n}=D_{\mathrm{n}} g\) occurs in the line integral. This is the directional derivative in the direction of the normal vector \(n\) and is called the normal derivative of \(g .\) )

\(21-27=\) Prove the identity, assuming that the appropriate partial derivatives exist and are continuous. If \(f\) is a scalar field and \(\mathbf{F},\) \(\mathbf{G}\) are vector fields, then \(f \mathbf{F}, \mathbf{F} \cdot \mathbf{G},\) and \(\mathbf{F} \times \mathbf{G}\) are defined by \((f \mathbf{F})(x, y, z)=f(x, y, z) \mathbf{F}(x, y, z)\) \((\mathbf{F} \cdot \mathbf{G})(x, y, z)=\mathbf{F}(x, y, z) \cdot \mathbf{G}(x, y, z)\) \((\mathbf{F} \times \mathbf{G})(x, y, z)=\mathbf{F}(x, y, z) \times \mathbf{G}(x, y, z)\) $$\operatorname{div}(\nabla f \times \nabla g)=0$$

Find parametric equations for the surface obtained by rotating the curve \(y=e^{-x}, 0 \leq x \leq 3,\) about the \(x\) -axis and use them to graph the surface.

Evaluate the surface integral \(\iint_{s} \mathbf{F} \cdot d \mathbf{S}\) for the given vector field \(\mathbf{F}\) and the oriented surface \(S .\) In other words, find the flux of \(\mathbf{F}\) across \(S .\) For closed surfaces, use the positive (outward) orientation. $$\begin{array}{l}{\mathbf{F}(x, y, z)=x z \mathbf{i}+x \mathbf{j}+y \mathbf{k}} \\ {S \text { is the hemisphere } x^{2}+y^{2}+z^{2}=25, y \geqslant 0, \text { oriented in }} \\ {\text { the direction of the positive } y \text { -axis }}\end{array} \geqslant 0,$$

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