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Chapter 12: Problem 44

Compute \(\mathbf{r}^{\prime \prime}(t)\) and \(\mathbf{r}^{\prime \prime \prime}(t)\) for the following functions. $$\mathbf{r}(t)=\left\langle e^{4 t}, 2 e^{-4 t}+1,2 e^{-t}\right\rangle$$

Short Answer

Expert verified
**Question**: Compute the second and third derivative of the vector-valued function \(\mathbf{r}(t) = \left\langle e^{4t}, 2e^{-4t} + 1, 2e^{-t} \right\rangle\). **Answer**: The second and third derivatives of the vector-valued function are: $$\mathbf{r}''(t) = \left\langle 16e^{4t}, 32e^{-4t}, 2e^{-t} \right\rangle$$ $$\mathbf{r}'''(t) = \left\langle 64e^{4t}, -128e^{-4t}, -2e^{-t} \right\rangle$$

Step by step solution

01

Find the first derivative of the vector function components

Differentiate each component of the vector function with respect to \(t\). \(\frac{d}{dt} e^{4t} = 4e^{4t}\) \(\frac{d}{dt} (2e^{-4t} + 1) = -8e^{-4t}\) \(\frac{d}{dt} 2e^{-t} = -2e^{-t}\)
02

Write the first derivative in vector form

Combine the derivatives of the components from Step 1 to create the first derivative of the vector function: $$\mathbf{r}'(t) = \left\langle 4e^{4t}, -8e^{-4t}, -2e^{-t} \right\rangle$$
03

Find the second derivative of the vector function components

Next, differentiate each component of the first derivative with respect to \(t\). \(\frac{d}{dt} 4e^{4t} = 16e^{4t}\) \(\frac{d}{dt} (-8e^{-4t}) = 32e^{-4t}\) \(\frac{d}{dt} (-2e^{-t}) = 2e^{-t}\)
04

Write the second derivative in vector form

Combine the derivatives of the components from Step 3 to create the second derivative of the vector function: $$\mathbf{r}''(t) = \left\langle 16e^{4t}, 32e^{-4t}, 2e^{-t} \right\rangle$$
05

Find the third derivative of the vector function components

Differentiate each component of the second derivative with respect to \(t\). \(\frac{d}{dt} 16e^{4t} = 64e^{4t}\) \(\frac{d}{dt} 32e^{-4t} = -128e^{-4t}\) \(\frac{d}{dt} 2e^{-t} = -2e^{-t}\)
06

Write the third derivative in vector form

Combine the derivatives of the components from Step 5 to create the third derivative of the vector function: $$\mathbf{r}'''(t) = \left\langle 64e^{4t}, -128e^{-4t}, -2e^{-t} \right\rangle$$
07

Final results

The computed second and third derivatives of the given vector-valued function are: $$\mathbf{r}''(t) = \left\langle 16e^{4t}, 32e^{-4t}, 2e^{-t} \right\rangle$$ $$\mathbf{r}'''(t) = \left\langle 64e^{4t}, -128e^{-4t}, -2e^{-t} \right\rangle$$

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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 fundamentally vectors where each component is a function of a single variable, typically denoted as \( t \). These functions are often expressed as \( \mathbf{r}(t) = \langle f(t), g(t), h(t) \rangle \), and they are pivotal in describing curves and paths in three-dimensional space.

They essentially allow us to map a real number, the parameter \( t \), to a curve in space, revealing the path traced by a point as \( t \) varies.

  • Expression: Represented by component functions like \( f(t) \), \( g(t) \), and \( h(t) \).
  • Dimensionality: Though the example here uses three dimensions, vector functions can have as many dimensions as needed, depending on the model or problem.
  • Applications: Widely used in physics, engineering, and computer graphics for modeling physical phenomena and geometric paths.
Understanding vector functions forms the basis of interpreting spatial changes over time. Each derivative of the vector function provides additional insight into such changes.
Derivatives
A derivative represents the rate of change of a function with respect to a variable. In vector calculus, the derivative of a vector function involves differentiating each component of the vector. This process tells us how the vector function behaves as the parameter changes.

The first derivative \( \mathbf{r}'(t) \) gives the velocity of the curve, indicating direction and speed. For the components of a vector function \( \mathbf{r}(t) = \langle e^{4t}, 2e^{-4t}+1, 2e^{-t} \rangle \), the first derivatives are:
  • \( \frac{d}{dt} e^{4t} = 4e^{4t} \)
  • \( \frac{d}{dt} (2e^{-4t} + 1) = -8e^{-4t} \)
  • \( \frac{d}{dt} 2e^{-t} = -2e^{-t} \)
The derivative is computed by applying basic differentiation rules to each component individually.
Vector Differentiation
Vector differentiation extends the concept of differentiating a single function to differentiating each component of a vector function. This is a crucial tool in vector calculus, enabling the analysis of complex paths and motions.

  • Process: Differentiate each component separately. As shown in the steps, performing differentiation on \( 4e^{4t} \), \( -8e^{-4t} \), and \(-2e^{-t} \) leads to subsequent derivatives that provide deeper insights into the vector's behavior.
  • Interpretation: Each derivative represents:
    • First derivative \( \mathbf{r}'(t) \) - Velocity or rate of change
    • Second derivative \( \mathbf{r}''(t) \) - Acceleration
    • Third derivative \( \mathbf{r}'''(t) \) - Rate of change of acceleration
This clear differentiation procedure enhances understanding of how a vector changes over time, crucial for applications like motion analysis.
Exponential Functions
Exponential functions feature a constant base raised to a variable exponent and play a significant role in vector calculus. In our original exercise, functions like \( e^{4t} \) and \( e^{-t} \) are exponential. Here's why they're important:

  • Behavior: Exponential functions grow or decay at an increasing rate. For \( e^{4t} \), the function grows quickly as \( t \) increases, whereas \( e^{-4t} \) decays rapidly.
  • Derivatives: Derivative of \( e^{kt} \) is \( ke^{kt} \). This property allows us to easily compute derivatives like \( \frac{d}{dt} e^{4t} = 4e^{4t} \).
  • Applications: These functions are key in modeling exponential growth/decay processes, such as population dynamics or radioactive decay.
Grasping exponential functions empowers you to tackle dynamic models effectively, integrating seamlessly into the larger framework of calculus.

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Most popular questions from this chapter

The definition \(\mathbf{u} \cdot \mathbf{v}=|\mathbf{u}||\mathbf{v}| \cos \theta\) implies that \(|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}|(\text {because}|\cos \theta| \leq 1) .\) This inequality, known as the Cauchy-Schwarz Inequality, holds in any number of dimensions and has many consequences. Verify that the Cauchy-Schwarz Inequality holds for \(\mathbf{u}=\langle 3,-5,6\rangle\) and \(\mathbf{v}=\langle-8,3,1\rangle\).

A golfer launches a tee shot down a horizontal fairway and it follows a path given by \(\mathbf{r}(t)=\left\langle a t,(75-0.1 a) t,-5 t^{2}+80 t\right\rangle,\) where \(t \geq 0\) measures time in seconds and \(\mathbf{r}\) has units of feet. The \(y\) -axis points straight down the fairway and the z-axis points vertically upward. The parameter \(a\) is the slice factor that determines how much the shot deviates from a straight path down the fairway. a. With no slice \((a=0),\) sketch and describe the shot. How far does the ball travel horizontally (the distance between the point the ball leaves the ground and the point where it first strikes the ground)? b. With a slice \((a=0.2),\) sketch and describe the shot. How far does the ball travel horizontally? c. How far does the ball travel horizontally with \(a=2.5 ?\)

Determine the equation of the line that is perpendicular to the lines \(\mathbf{r}(t)=\langle 4 t, 1+2 t, 3 t\rangle\) and \(\mathbf{R}(s)=\langle-1+s,-7+2 s,-12+3 s\rangle\) and passes through the point of intersection of the lines \(\mathbf{r}\) and \(\mathbf{R}\).

A pair of lines in \(\mathbb{R}^{3}\) are said to be skew if they are neither parallel nor intersecting. Determine whether the following pairs of lines are parallel, intersecting, or skew. If the lines intersect. determine the point(s) of intersection. $$\begin{aligned} &\mathbf{r}(t)=\langle 1+2 t, 7-3 t, 6+t\rangle;\\\ &\mathbf{R}(s)=\langle-9+6 s, 22-9 s, 1+3 s\rangle \end{aligned}$$

Evaluate the following limits. $$\lim _{t \rightarrow \pi / 2}\left(\cos 2 t \mathbf{i}-4 \sin t \mathbf{j}+\frac{2 t}{\pi} \mathbf{k}\right)$$
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