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Chapter 12: Problem 11

Sketch the space curve and find its length over the given interval. Function \(\quad\) Interval \(\mathbf{r}(t)=\langle 3 t, 2 \cos t, 2 \sin t\rangle \quad\left[0, \frac{\pi}{2}\right]\)

Short Answer

Expert verified
The sketch of the given vector function shows a space curve in 3-dimensional space, starting at the origin and spiraling in the direction of increasing t. The length of the space curve in the interval \([0, \frac{\pi}{2}]\) is found to be \(\sqrt{13}\times\frac{\pi}{2}\).

Step by step solution

01

Sketching the space curve

Let's start with sketching the space curve defined by the given function. We can accomplish this by plotting the parametric equations \(x(t) = 3t\), \(y(t) = 2\cos(t)\), and \(z(t) = 2\sin(t)\) over the given interval \([0, \frac{\pi}{2}]\). In a 3D coordinate system, these would represent the coordinates (x, y, z) at a given value of t.
02

Computing the derivative

Next, we need to compute the derivative of the vector function \(\mathbf{r}(t)=\langle 3 t, 2 \cos t, 2 \sin t\rangle\). This gives a new vector function, \(\mathbf{r}^\prime(t)\), that describes how \(\mathbf{r}(t)\) changes with t. Use the standard rules for differentiation: \(\mathbf{r}^\prime(t)=\langle 3, -2 \sin t, 2 \cos t\rangle\).
03

Computing the magnitude of the derivative

The length of the curve can be obtained by taking the integral of the magnitude of the derivative \(\mathbf{r}^\prime(t)\) over the given interval. The magnitude (or length) of \(\mathbf{r}^\prime(t)\) is given by \(\sqrt{(\mathbf{r}^\prime(t))^2} = \sqrt{3^2+(-2 \sin t)^2+(2 \cos t)^2}\). Simplifying, we get \(\sqrt{9+4\sin^2 t + 4\cos^2 t}\).
04

Simplifying and integrating to find the length

The term inside the square root simplifies to \(9 + 4\), as \(\sin^2 t + \cos^2 t = 1\). Which gives us \(\sqrt{13}\) for the magnitude of \(\mathbf{r}^\prime(t)\). We obtain the length of the curve by integrating \(\sqrt{13}\) from 0 to \(\frac{\pi}{2}\), which simplifies to length equal to \(\sqrt{13}\times\frac{\pi}{2}\).

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Key Concepts

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

Space Curve
A space curve is a curve that exists in three-dimensional space. Unlike flat two-dimensional curves, these curves have depth and can twist and turn in complex ways. They are often defined by a set of parametric equations that describe points
  • along the x-axis
  • along the y-axis
  • along the z-axis
For the space curve described in the exercise, the function \( \mathbf{r}(t)=\langle 3t, 2\cos t, 2\sin t\rangle \) describes a path in 3D space.This specific example is part of a helix because of the periodic nature of the cosine and sine equations on the y and z components, while the linear part along the x-axis stretches this movement.
Parametric Equations
Parametric equations enable us to express a geometric shape in terms of one or more parameters. This is particularly useful when defining a path or a trajectory in physics and engineering.For the space curve in question, we have:
  • \( x(t) = 3t \) 鈥� a linear equation that determines the horizontal position along the x-axis.
  • \( y(t) = 2\cos t \) 鈥� causes oscillation along the y-axis, derived from the properties of the cosine function.
  • \( z(t) = 2\sin t \) 鈥� similarly, defines oscillation in the z-axis by using the sine function.
Variations in 't' give us coordinates that plot the path of the curve in the 3D space . This allows for a smooth representation of shapes that can be difficult to express in other forms, such as standard equations.
Vector Function
A vector function uses vectors to represent curves in space. It pairs the notion of vectors, which have both magnitude and direction, with functions.For this curve, the vector function is given as \( \mathbf{r}(t) = \langle 3t, 2\cos t, 2\sin t \rangle \), which aligns three functions along the x, y, and z axes respectively.
This smoothly defines the curve as we input different values for \( t \).The derivative of the vector function, \( \mathbf{r}'(t) = \langle 3, -2\sin t, 2\cos t \rangle \), represents changes along each axis. Understanding how a vector function works offers insights into motion along paths, making this a powerful tool in physics and engineering.
Curve Length
Curve length is the total distance measured along the path of a curve. It's crucial for determining how far you'd travel if you followed that path. To find the length of the space curve, we integrate the magnitude of its derivative.Following these steps:
  • Differentiating gives \( \mathbf{r}'(t) \) with components \( \langle 3, -2\sin t, 2\cos t \rangle \).
  • Magnitude is calculated with:\[\sqrt{3^2 + (-2\sin t)^2 + (2\cos t)^2} = \sqrt{9 + 4\sin^2 t + 4\cos^2 t}\]
  • Simplify the expression: \( \sin^2 t + \cos^2 t = 1 \), therefore magnitude becomes \( \sqrt{13} \).
  • Integrate over the interval \( [0, \frac{\pi}{2}] \) resulting in curve length \( \sqrt{13}\times\frac{\pi}{2} \).
These steps allow computation of the exact path length broken into clear and concise phases.

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Most popular questions from this chapter

(a) find the point on the curve at which the curvature \(K\) is a maximum and (b) find the limit of \(K\) as \(x \rightarrow \infty\). $$ y=x^{2 / 3} $$

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