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Chapter 2: Problem 61

Find the derivatives of u(t), u?(t), u?(t)脳u(t), u(t)脳u?(t), and u(t)?u?(t). Find the unit tangent vector. $$ \mathbf{u}(t)=\left\langle e^{t}, e^{-t}\right\rangle $$

Short Answer

Expert verified
The unit tangent vector is \( \langle \frac{e^t}{\sqrt{e^{2t} + e^{-2t}}}, \frac{-e^{-t}}{\sqrt{e^{2t} + e^{-2t}}} \rangle \).

Step by step solution

01

Differentiate u(t)

To find the derivative of \( \mathbf{u}(t) = \langle e^t, e^{-t} \rangle \), we differentiate each component with respect to \( t \). \( e^t \) differentiates to \( e^t \) and \( e^{-t} \) differentiates to \( -e^{-t} \). Thus, \( \mathbf{u}'(t) = \langle e^t, -e^{-t} \rangle \).
02

Differentiate u?(t)

Since \( \mathbf{u}?(t) \) simply refers to \( \mathbf{u}'(t) \), we've already computed this in Step 1. So, \( \mathbf{u}?(t) = \langle e^t, -e^{-t} \rangle \).
03

Compute u?(t)脳u(t)

In two dimensions, the cross product \( \mathbf{u}'(t) \times \mathbf{u}(t) \) is scalar, calculated as \( e^t \, e^{-t} - (-e^{-t}) \, e^t = 1 + e^{2t} \).
04

Compute u(t)脳u?(t)

The expression \( \mathbf{u}(t) \times \mathbf{u}'(t) \) is the negative of \( \mathbf{u}'(t) \times \mathbf{u}(t) \) in two dimensions, thus \( \mathbf{u}(t) \times \mathbf{u}'(t) = -\left( 1 + e^{2t} \right) \).
05

Compute u(t)?u?(t)

The dot product \( \mathbf{u}(t) \, \cdot \, \mathbf{u}'(t) \) is calculated as \( e^t \, e^t + e^{-t} \, (-e^{-t}) = \left( e^{2t} - 1 \right) \).
06

Normalize the Tangent Vector

The unit tangent vector is found by normalizing \( \mathbf{u}'(t) = \langle e^t, -e^{-t} \rangle \). Compute the magnitude: \( \|\mathbf{u}'(t)\| = \sqrt{(e^t)^2 + (-e^{-t})^2} = \sqrt{e^{2t} + e^{-2t}} \). Thus, the unit tangent vector is \( \langle \frac{e^t}{\sqrt{e^{2t} + e^{-2t}}}, \frac{-e^{-t}}{\sqrt{e^{2t} + e^{-2t}}} \rangle \).

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Key Concepts

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

Derivative
The derivative of a vector function, like the one in our exercise, is found by differentiating each of its components individually. This results in a new vector function, made up of the derivatives of the original components. For example, when finding the derivative of \( \mathbf{u}(t) = \langle e^t, e^{-t} \rangle \), you treat \( e^t \) and \( e^{-t} \) as separate entities.
  • To differentiate \( e^t \), recognize that it remains \( e^t \), because the derivative of an exponential function with base \( e \) is the function itself.
  • \( e^{-t} \) differentiates according to the chain rule, resulting in \( -e^{-t} \).
Thus, the derivative \( \mathbf{u}'(t) \) of the vector function \( \langle e^t, e^{-t} \rangle \) is \( \langle e^t, -e^{-t} \rangle \). Differentiation helps us understand the rate at which each component of a vector changes with respect to the parameter \( t \).
Cross Product
The cross product is an operation commonly used to calculate a vector that is orthogonal to two given vectors in three-dimensional space. However, when dealing with two-dimensional vectors, the cross product is scalar. The unique property holds significance in physics and engineering, where the computation of rotational effects is important.
  • For two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \) in the plane, their cross product \( \mathbf{a} \times \mathbf{b} \) simplifies to a scalar: \( a_1b_2 - a_2b_1 \).
In the context of this exercise, calculating the cross products \( \mathbf{u}'(t) \times \mathbf{u}(t) \) and \( \mathbf{u}(t) \times \mathbf{u}'(t) \) simplifies to scalars. Specifically, \( \mathbf{u}'(t) \times \mathbf{u}(t) = 1 + e^{2t} \) and \( \mathbf{u}(t) \times \mathbf{u}'(t) \) is the negative of the former: \( -(1 + e^{2t}) \).
Dot Product
The dot product is an operation that takes two equal-length sequences of numbers and returns a single number. This operation is particularly significant in measuring the angle between two vectors, as it can reveal orthogonality when the product is zero.
  • For two vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), their dot product is computed as: \( a_1b_1 + a_2b_2 \).
In our exercise, the dot product of \( \mathbf{u}(t) \) and \( \mathbf{u}'(t) \) results in \( e^t \cdot e^t + e^{-t} \cdot (-e^{-t}) = e^{2t} - 1 \). This operation provides insight into how aligned the function and its derivative are with each other at any particular point \( t \).
Unit Tangent Vector
To understand the direction in which a vector is pointing, we often normalize it. This results in a unit vector, a vector with a magnitude of 1, pointing in the same direction as the original vector. This is particularly useful in physics, helping us focus solely on direction while disregarding magnitude.
  • To find the unit tangent vector of \( \mathbf{u}'(t) \), first determine its magnitude: \( \|\mathbf{u}'(t)\| = \sqrt{(e^t)^2 + (-e^{-t})^2} = \sqrt{e^{2t} + e^{-2t}} \).
  • Scale \( \mathbf{u}'(t) \) by dividing each of its components by this magnitude.
Therefore, the unit tangent vector is \( \langle \frac{e^t}{\sqrt{e^{2t} + e^{-2t}}}, \frac{-e^{-t}}{\sqrt{e^{2t} + e^{-2t}}} \rangle \). This vector maintains the direction of \( \mathbf{u}'(t) \), allowing us to analyze directional behavior effectively.

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Most popular questions from this chapter

The force on a particle is given by \(\mathbf{f}(t)=(\cos t) \mathbf{i}+(\sin t) \mathbf{j}\). The particle is located at point \((c, 0)\) at \(t=0\). The initial velocity of the particle is given by \(\mathbf{v}(0)=v_{0} \mathbf{j}\). Find the path of the particle of mass \(\mathbf{m} \cdot(\operatorname{Recall}, \mathbf{F}=m \cdot \mathbf{a}\).)

Find the domains of the vector-valued functions. $$ \mathbf{r}(t)=\langle\sin (t), \ln (t), \sqrt{t}\rangle $$

The position of a particle is given by \(\mathbf{r}(t)=\left\langle t^{2}, \ln (t), \sin (\pi t)\right\rangle\), where \(t\) is measured in seconds and \(\mathbf{r}\) is measured in meters. Find the velocity, acceleration, and speed functions. What are the position, velocity, speed, and acceleration of the particle at \(1 \mathrm{sec} ?\)

Find the unit tangent vector for the following parameterized curves.\(\mathbf{r}(t)=6 \mathbf{i}+\cos (3 t) \mathbf{j}+3 \sin (4 t) \mathbf{k}, 0 \leq t<2 \pi\)

For the following problems, find a tangent vector at the indicated value of \(t$$\mathbf{r}(t)=3 e^{t} \mathbf{i}+2 e^{-3 t} \mathbf{j}+4 e^{2 t} \mathbf{k} ; t=\ln (2)\)
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