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

Short Answer

Expert verified
Question: Determine the second and third derivatives of the vector-valued function: \(\mathbf{r}(t) = \langle \tan t, t + \frac{1}{t}, -\ln(t+1) \rangle\). Answer: The second and third derivatives of the function are as follows: \(\mathbf{r}^{\prime \prime}(t) = \left\langle 4 \tan t \sec^2 t, \frac{2}{t^3}, \frac{1}{(t+1)^2} \right\rangle\), and \(\mathbf{r}^{\prime \prime \prime}(t) = \left\langle 8 \sec^4 t, -\frac{6}{t^4}, -\frac{2}{(t+1)^3} \right\rangle\).

Step by step solution

01

Compute \(\mathbf{r}^{\prime}(t)\)

To compute the first derivative of \(\mathbf{r}(t)\), we need to find the first derivative of each component function with respect to \(t\). So, we have: $$\frac{d}{dt}(\tan t) = \sec^2 t \quad , \quad \frac{d}{dt}\left(t + \frac{1}{t}\right) = 1 - \frac{1}{t^2} \quad \mathrm{and} \quad \frac{d}{dt}(-\ln(t+1)) = -\frac{1}{t+1}$$ Thus, we get: $$\mathbf{r}^{\prime}(t) = \langle \sec^2 t, 1 - \frac{1}{t^2}, -\frac{1}{t+1} \rangle$$
02

Compute \(\mathbf{r}^{\prime \prime}(t)\)

Next, we need to compute the second derivative of each component function. So we have: $$\frac{d^2}{dt^2}(\sec^2 t) = 4 \tan t \sec^2 t \quad , \quad \frac{d^2}{dt^2}\left(1 - \frac{1}{t^2}\right) = \frac{2}{t^3} \quad \mathrm{and} \quad \frac{d^2}{dt^2}\left(-\frac{1}{t+1}\right) = \frac{1}{(t+1)^2}$$ Therefore, we get: $$\mathbf{r}^{\prime \prime}(t) = \left\langle 4 \tan t \sec^2 t, \frac{2}{t^3}, \frac{1}{(t+1)^2} \right\rangle$$
03

Compute \(\mathbf{r}^{\prime \prime \prime}(t)\)

Finally, we need to compute the third derivative of each component function. We have: $$\frac{d^3}{dt^3}(\sec^2 t) = 8 \sec^4 t \quad , \quad \frac{d^3}{dt^3}\left(\frac{2}{t^3}\right) = -\frac{6}{t^4} \quad \mathrm{and} \quad \frac{d^3}{dt^3}\left(\frac{1}{(t+1)^2}\right) = -\frac{2}{(t+1)^3}$$ Thus, we obtain: $$\mathbf{r}^{\prime \prime \prime}(t) = \left\langle 8 \sec^4 t, -\frac{6}{t^4}, -\frac{2}{(t+1)^3} \right\rangle$$

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Key Concepts

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

Higher Order Derivatives
Derivatives provide crucial information about the behavior of functions. Higher order derivatives go beyond the first derivative, allowing us to explore the rate of change of the rate of change, or even further. In calculus, the first derivative is often used to find velocity if the function represents position. The second derivative can help describe acceleration, giving insights into how the velocity itself is changing over time. As we take more derivatives, like the third derivative, we get into the rate of change of acceleration, which can reveal even deeper layers of how a system evolves.

In the provided exercise, we are asked to find the second and third derivatives of a vector-valued function \(\mathbf{r}(t)\). This requires differentiating each component function multiple times, which involves applying rules like the power rule, product rule, or chain rule depending on the function's form. For example, differentiating \(\sec^2 t\) involves the chain rule due to its composition of functions.
  • The second derivative, \(\mathbf{r}''(t)\), provides information about the curvature of the vector function in space.
  • The third derivative, \(\mathbf{r}'''(t)\), can indicate the changing rate of the curvature, which can be important in fields like physics and engineering.
Vector Functions
Vector functions are mathematical expressions that assign a vector to each input from a domain, typically real numbers. These functions are often used to describe paths or trajectories in multi-dimensional space. A vector function like \(\mathbf{r}(t)\) may consist of several components which themselves are functions of \(t\). Each component represents a dimension in the space, such as \(\mathbf{i}, \mathbf{j}, \mathbf{k}\) in the three-dimensional setting.

In the exercise, \(\mathbf{r}(t)\) is given by the components \(\tan t \mathbf{i}\), \(\left(t + \frac{1}{t}\right) \mathbf{j}\) and \(-\ln(t+1) \mathbf{k}\). Each of these components has its dynamics governed by \(t\), and differentiating the entire vector function involves treating each component separately. This allows the study of changes and behavior in each direction defined by the basis vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\).
  • It's essential to handle each component according to its defining scalar function.
  • This enables us to analyze complex motions that combine different types of behavior along different axes.
Parametric Equations
Parametric equations are a set of equations where a set of quantities are expressed as functions of one or more independent parameters. They are particularly useful in describing complex curves and surfaces. In the context of vector functions, the parameter often represents time, providing a way to trace out a path as time progresses.

For the vector function \(\mathbf{r}(t)\), \(t\) acts as the parameter, dictating the position along the path at any given moment. With parametric equations, we have the flexibility to describe intricate shapes and motions that are not easily captured by traditional Cartesian coordinates. As such, solving derivatives of vector functions with parametric equations involves:
  • Identifying how each component of the vector function changes with the parameter.
  • Using these derivatives to study motion and dynamic systems in physics and engineering.
Derivatives like \(\mathbf{r}'(t)\) and \(\mathbf{r}''(t)\) give deeper insight into the nature of these paths, revealing velocity and curvature, respectively.

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Most popular questions from this chapter

Torsion formula Show that the formula defining the torsion, \(\tau=-\frac{d \mathbf{B}}{d s} \cdot \mathbf{N},\) is equivalent to \(\tau=-\frac{1}{|\mathbf{v}|} \frac{d \mathbf{B}}{d t} \cdot \mathbf{N} .\) The second formula is generally easier to use.

Note that two lines \(y=m x+b\) and \(y=n x+c\) are orthogonal provided \(m n=-1\) (the slopes are negative reciprocals of each other). Prove that the condition \(m n=-1\) is equivalent to the orthogonality condition \(\mathbf{u} \cdot \mathbf{v}=0\) where \(\mathbf{u}\) points in the direction of one line and \(\mathbf{v}\) points in the direction of the other line.

\(\mathbb{R}^{2}\) Consider the vectors \(\mathbf{I}=\langle 1 / \sqrt{2}, 1 / \sqrt{2}\rangle\) and \(\mathbf{J}=\langle-1 / \sqrt{2}, 1 / \sqrt{2}\rangle\). Write the vector \langle 2,-6\rangle in terms of \(\mathbf{I}\) and \(\mathbf{J}\).

An object moves along an ellipse given by the function \(\mathbf{r}(t)=\langle a \cos t, b \sin t\rangle,\) for \(0 \leq t \leq 2 \pi,\) where \(a > 0\) and \(b > 0\) a. Find the velocity and speed of the object in terms of \(a\) and \(b\) for \(0 \leq t \leq 2 \pi\) b. With \(a=1\) and \(b=6,\) graph the speed function, for \(0 \leq t \leq 2 \pi .\) Mark the points on the trajectory at which the speed is a minimum and a maximum. c. Is it true that the object speeds up along the flattest (straightest) parts of the trajectory and slows down where the curves are sharpest? d. For general \(a\) and \(b\), find the ratio of the maximum speed to the minimum speed on the ellipse (in terms of \(a\) and \(b\) ).

Consider the curve described by the vector function \(\mathbf{r}(t)=\left(50 e^{-t} \cos t\right) \mathbf{i}+\left(50 e^{-t} \sin t\right) \mathbf{j}+\left(5-5 e^{-t}\right) \mathbf{k},\) for \(t \geq 0\). a. What is the initial point of the path corresponding to \(\mathbf{r}(0) ?\) b. What is \(\lim _{t \rightarrow \infty} \mathbf{r}(t) ?\) c. Sketch the curve. d. Eliminate the parameter \(t\) to show that \(z=5-r / 10\), where \(r^{2}=x^{2}+y^{2}\).

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