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Chapter 11: Problem 34

Graph the curves described by the following functions, indicating the positive orientation. $$\mathbf{r}(t)=4 \sin t \mathbf{i}+4 \cos t \mathbf{j}+e^{-t / 10} \mathbf{k}, \text { for } 0 \leq t<\infty$$

Short Answer

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Question: Provide a brief description of the shape and orientation of the space curve represented by the given vector function $$\mathbf{r}(t) = 4 \sin t \mathbf{i} + 4 \cos t \mathbf{j} + e^{-t/10} \mathbf{k}, \text { for } 0 \leq t<\infty$$. Answer: The space curve is a gradually descending spiral shape, starting at a circle with radius 4 in the xy-plane at a height of z(0) = 1. As t increases, the curve spirals counterclockwise in the xy-plane and downward in the z-axis, getting closer to the xy-plane but never quite reaching it.

Step by step solution

01

Understanding the vector components

The given vector function is $$\mathbf{r}(t) = 4 \sin t \mathbf{i} + 4 \cos t \mathbf{j} + e^{-t/10} \mathbf{k}, \text { for } 0 \leq t<\infty$$ This represents a space curve, with positions in 3D space specified by the components \(x(t) = 4 \sin t\), \(y(t) = 4 \cos t\), and \(z(t) = e^{-t/10}\). **Step 2: Plot the 2D projection**
02

Visualize in the xy-plane

As \(0 \leq t <\infty\), first, let's consider the projection of the curve on the xy-plane by plotting \(x(t) = 4 \sin t\) and \(y(t) = 4 \cos t\) ignoring the z-component which is \(z(t) = e^{-t/10}\). It forms a circle with radius 4 in xy-plane (see polar coordinates: \(r = 4\) and \(\theta = t\)). **Step 3: Analyze the z-component**
03

Understanding the z-component

The z-component of the curve is given by \(z(t) = e^{-t/10}\). Here, as \(t\) increases from 0 to infinity, the z-component will decrease from its maximum value at \(t = 0\) (which is \(z(0) = 1\)) to 0 when \(t\) approaches infinity (since \(e^{-t/10}\) approaches 0 for large \(t\)). **Step 4: Combine 2D projection with the z-component**
04

3D visualization of the curve

Now, combining the 2D projection (a circle in the xy-plane) with the z-component, imagine it as a circle that starts at \((0, 4, 1)\) and gradually spirals down in the positive \(z\) direction, approaching the xy-plane as the z-component becomes closer to 0. The positive orientation is determined by the direction of the space curve as \(t\) increases, so it goes counterclockwise in the xy-plane and downwards in the z-axis. **Step 5: Graph the curve**
05

Plot the space curve

With the 2D projection and z-components combined, plot the space curve as follows: 1. Start with a circle of radius 4 centered at the origin of the xy-plane. 2. Project the circle upwards so that it starts at a height of \(z(0) = 1\). 3. Gradually spiral the curve downwards, following the orientation given by the increasing values of \(t\) (counterclockwise in the xy-plane and downward in the z-axis). 4. As \(t\) approaches infinity, the curve gets closer and closer to the xy-plane, but never quite touches it ('\(z\) approaches 0). The resulting graph represents the given space curve with the indicated positive orientation.

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Most popular questions from this chapter

Let $$\mathbf{u}(t)=2 t^{3} \mathbf{i}+\left(t^{2}-1\right) \mathbf{j}-8 \mathbf{k} \text { and } \mathbf{v}(t)=e^{t} \mathbf{i}+2 e^{-t} \mathbf{j}-e^{2 t} \mathbf{k}$$ Compute the derivative of the following functions. $$\mathbf{v}(\sqrt{t})$$

Direction angles and cosines Let \(\mathbf{v}=\langle a, b, c\rangle\) and let \(\alpha, \beta\) and \(\gamma\) be the angles between \(\mathbf{v}\) and the positive \(x\) -axis, the positive \(y\) -axis, and the positive \(z\) -axis, respectively (see figure). a. Prove that \(\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\) b. Find a vector that makes a \(45^{\circ}\) angle with \(\mathbf{i}\) and \(\mathbf{j}\). What angle does it make with \(\mathbf{k} ?\) c. Find a vector that makes a \(60^{\circ}\) angle with i and \(\mathbf{j}\). What angle does it make with k? d. Is there a vector that makes a \(30^{\circ}\) angle with \(\mathbf{i}\) and \(\mathbf{j} ?\) Explain. e. Find a vector \(\mathbf{v}\) such that \(\alpha=\beta=\gamma .\) What is the angle?

Consider the curve \(\mathbf{r}(t)=(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j}+(e \cos t+f \sin t) \mathbf{k}\) where \(a, b, c, d, e,\) and \(f\) are real numbers. It can be shown that this curve lies in a plane. Find a general expression for a nonzero vector orthogonal to the plane containing the curve. $$\begin{aligned}\mathbf{r}(t)=&(a \cos t+b \sin t) \mathbf{i}+(c \cos t+d \sin t) \mathbf{j} \\ &+(e \cos t+f \sin t) \mathbf{k},\end{aligned}$$ where \(\langle a, c, e\rangle \times\langle b, d, f\rangle \neq \mathbf{0}\).

Compute the following derivatives. $$\frac{d}{d t}\left(\left(t^{3} \mathbf{i}-2 t \mathbf{j}-2 \mathbf{k}\right) \times\left(t \mathbf{i}-t^{2} \mathbf{j}-t^{3} \mathbf{k}\right)\right)$$

Jack pulls east on a rope attached to a camel with a force of 40 ib. Jill pulls north on a rope attached to the same camel with a force of 30 Ib. What is the magnitude and direction of the force on the camel? Assume the vectors lie in a horizontal plane.
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