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Chapter 14: Problem 31

Find the unit tangent vector at the given value of \(t\) for the following parameterized curves. $$\mathbf{r}(t)=\left\langle 6 t, 6, \frac{3}{t}\right\rangle ; t=1$$

Short Answer

Expert verified
Answer: The unit tangent vector at \(t=1\) is \(\mathbf{T}(t) = \left\langle \frac{6}{\sqrt{45}}, 0, -\frac{3}{\sqrt{45}} \right\rangle\).

Step by step solution

01

Find the derivative of the parameterized curve

To find the derivative of the parameterized curve \(\mathbf{r}(t)=\left\langle 6 t, 6, \frac{3}{t}\right\rangle\) with respect to \(t\), differentiate each component of the vector with respect to \(t\): $$\frac{d\mathbf{r}}{dt} = \left\langle \frac{d(6t)}{dt}, \frac{d(6)}{dt}, \frac{d(3/t)}{dt} \right\rangle$$
02

Calculate the derivative of each component

Differentiate each component of the vector: $$\frac{d(6t)}{dt} = 6$$ $$\frac{d(6)}{dt} = 0$$ $$\frac{d(3/t)}{dt} = -\frac{3}{t^2}$$ Thus, the derivative of the parameterized curve is: $$\frac{d\mathbf{r}}{dt} = \left\langle 6, 0, -\frac{3}{t^2} \right\rangle$$
03

Evaluate the derivative at \(t=1\)

Plug in \(t=1\) into the derivative: $$\frac{d\mathbf{r}}{dt} \Big|_{t=1} = \left\langle 6, 0, -\frac{3}{1^2} \right\rangle = \left\langle 6, 0, -3 \right\rangle$$
04

Find the magnitude of the derivative

To normalize the tangent vector, we need its magnitude. Calculate the magnitude of the vector \(\left\langle 6, 0, -3 \right\rangle\): $$\|\frac{d\mathbf{r}}{dt}\| = \sqrt{(6)^2 + (0)^2 + (-3)^2} = \sqrt{36 + 9} = \sqrt{45}$$
05

Normalize the tangent vector

Divide each component of the tangent vector by its magnitude to obtain the unit tangent vector: $$\mathbf{T}(t) = \frac{\frac{d\mathbf{r}}{dt}}{\|\frac{d\mathbf{r}}{dt}\|} = \frac{\left\langle 6, 0, -3 \right\rangle}{\sqrt{45}} = \left\langle \frac{6}{\sqrt{45}}, 0, -\frac{3}{\sqrt{45}} \right\rangle$$ The unit tangent vector \(\mathbf{T}(t)\) at \(t=1\) for the given parameterized curve is: $$\mathbf{T}(t) = \left\langle \frac{6}{\sqrt{45}}, 0, -\frac{3}{\sqrt{45}} \right\rangle$$

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Key Concepts

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

Parameterized Curves
A parameterized curve in calculus offers a way to describe a path or a curve in space by using a parameter, most commonly denoted as t. Think of it like a set of instructions for how to draw a curve by tracing out points as time passes. Instead of saying 'at x position, the height is y', you describe the position as a function of t, for example, \( \mathbf{r}(t) \).

In our case, the curve is \( \mathbf{r}(t)=\langle 6t, 6, \frac{3}{t}\rangle \), which tells us the x, y, and z coordinates of each point on the curve based on the value of t. This allows for complex curves to be represented that are otherwise hard to express with a single function.

To assist in understanding, you might visualize a parameterized curve as a path traced by a bug crawling in three-dimensional space, where its position at any time t is given by \( \mathbf{r}(t) \).
Vector Differentiation
Vector differentiation is like finding out how a vector-valued function, like our parameterized curve \( \mathbf{r}(t) \), changes as the parameter t changes. It's akin to asking, 'If the bug speeds up, slows down, or changes direction, how does its path change in response?'. To differentiate a vector function, you differentiate each of its components separately with respect to t.

In the given exercise, the derivative found after differentiation, \( \frac{d\mathbf{r}}{dt} = \langle 6, 0, -\frac{3}{t^2} \rangle \), tells us the velocity of the bug at any given moment t. At t=1, we find that the velocity vector is \( \langle 6, 0, -3 \rangle \), meaning the bug moves quickly along the x-axis and downwards along the z-axis, with no movement in the y direction.

Vector differentiation is vital as it provides a snapshot of the motion's dynamics at any given instant, illustrating not just the path's shape but how it is being traversed.
Magnitude of a Vector
Determining the magnitude of a vector is like measuring the length of a straight line from the starting point to the end point of the vector's arrow. In essence, it quantifies how far the bug has traveled from the origin, irrespective of its path. The magnitude of a vector is always non-negative and is calculated by using the Pythagorean theorem for each component of the vector.

For a vector \( \mathbf{v} = \langle x, y, z \rangle \), its magnitude is \( \|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2} \). From our example, the magnitude of the derivative vector \( \|\frac{d\mathbf{r}}{dt}\| \) ends up being \( \sqrt{45} \), which is essential for finding the unit tangent vector, as this will tell us the direction of the path without considering how quickly the bug is moving.

Magnitude is an intrinsic property of vectors and is fundamental when normalizing a vector to find a unit vector in the same direction. This attribute plays a pivotal role in diverse fields, from physics to engineering, whenever direction and length are crucial factors.

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Most popular questions from this chapter

A projectile (such as a bascball or a cannonball) launched from the origin with an initial horizontal velocity \(u_{0}\) and an initial vertical velocity \(v_{0}\) moves in a parabolic trajectory given by $$\mathbf{r}(t)=\left\langle u_{0} t_{1}-\frac{1}{2} g t^{2}+v_{0} t\right\rangle, \quad \text { for } t \geq 0$$ where air resistance is neglected and \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity (see Section 14.3 ). a. Let \(u_{0}=20 \mathrm{m} / \mathrm{s}\) and \(v_{0}=25 \mathrm{m} / \mathrm{s} .\) Assuming the projectile is launched over horizontal ground, at what time does it return to Earth? b. Find the integral that gives the length of the trajectory from launch to landing. c. Evaluate the integral in part (b) by first making the change of variables \(u=-g t+v_{0} .\) The resulting integral is evaluated either by making a second change of variables or by using a calculator. What is the length of the trajectory? d. How far does the projectile land from its launch site?

Evaluate the following definite integrals. $$\int_{0}^{\pi / 4}\left(\sec ^{2} t \mathbf{i}-2 \cos t \mathbf{j}-\mathbf{k}\right) d t$$

Compute \(\mathbf{r}^{\prime \prime}(t)\) and \(\mathbf{r}^{\prime \prime \prime}(t)\) for the following functions. $$\mathbf{r}(t)=\tan t \mathbf{i}+\left(t+\frac{1}{t}\right) \mathbf{j}-\ln (t+1) \mathbf{k}$$

Relationship between \(\mathbf{r}\) and \(\mathbf{r}^{\prime}.\) Consider the helix \(\mathbf{r}(t)=\langle\cos t, \sin t, t\rangle,\) for \(-\infty

Cusps and noncusps a. Graph the curve \(\mathbf{r}(t)=\left\langle t^{3}, t^{3}\right\rangle .\) Show that \(\mathbf{r}^{\prime}(0)=0\) and the curve does not have a cusp at \(t=0 .\) Explain. b. Graph the curve \(\mathbf{r}(t)=\left\langle t^{3}, t^{2}\right\rangle .\) Show that \(\mathbf{r}^{\prime}(0)=\mathbf{0}\) and the curve has a cusp at \(t=0 .\) Explain. c. The functions \(\mathbf{r}(t)=\left\langle t, t^{2}\right\rangle\) and \(\mathbf{p}(t)=\left\langle t^{2}, t^{4}\right\rangle\) both satisfy \(y=x^{2} .\) Explain how the curves they parameterize are different. d. Consider the curve \(\mathbf{r}(t)=\left\langle t^{m}, t^{n}\right\rangle,\) where \(m>1\) and \(n>1\) are integers with no common factors. Is it true that the curve has a cusp at \(t=0\) if one (not both) of \(m\) and \(n\) is even? Explain.
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