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Find \((a) r^{\prime \prime}(t)\) and \((b) r^{\prime}(t) \cdot r^{\prime \prime}(t)\). $$ \mathbf{r}(t)=\langle\cos t+t \sin t, \sin t-t \cos t, t\rangle $$

Short Answer

Expert verified
The second derivative \(r''(t)\) is \(\langle -t\sin(t), -2\sin(t) - 2t\cos(t), 0 \rangle\) and the dot product \( \mathbf{r}'(t) \cdot \mathbf{r}''(t) \) is \(-2t^{2}\cos(t)\sin(t) - 4\cos(t)\sin(t) + 2t\cos^{2}(t)\).

Step by step solution

01

Calculate \( \mathbf{r}'(t) \)

To obtain the first derivative of \( \mathbf{r}(t) \), use the basic rules of differentiation applied to each coordinate of the function. So we have: \n \\( \mathbf{r}'(t) = \langle-\sin(t) + \sin(t) + t\cos(t), \cos(t) + \cos(t) - t\sin(t), 1\rangle \) = \langle t\cos(t), 2\cos(t) - t\sin(t), 1\rangle
02

Calculate \( \mathbf{r}''(t) \)

To get the second derivative, \( \mathbf{r}''(t) \), take the derivative once again of \( \mathbf{r}'(t) \). We get: \n \ \( \mathbf{r}''(t) = \langle -t\sin(t), 2(-\sin(t) - t\cos(t)), 0\rangle = \langle -t\sin(t), -2\sin(t) - 2t\cos(t), 0 \rangle
03

Calculate \( \mathbf{r}'(t) \cdot \mathbf{r}''(t) \)

The dot product of these two vectors \( \mathbf{r}' \) and \( \mathbf{r}'' \) is calculated as follows: \n \ \( \mathbf{r}'(t) \cdot \mathbf{r}''(t) = t\cos(t)(-t\sin(t)) + [2\cos(t) - t\sin(t)][-2\sin(t) - 2t\cos(t)] + 1 \cdot 0 = -t^{2}\cos(t)\sin(t) - 4\cos(t)\sin(t) + 2t\cos^{2}(t) - 2t^{2}\cos(t)\sin(t) = -2t^{2}\cos(t)\sin(t) - 4\cos(t)\sin(t) + 2t\cos^{2}(t)

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Key Concepts

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

Derivative Computation
Understanding how to compute the derivative of a vector function is essential in vector calculus. A vector function, such as \( \mathbf{r}(t) \), comprises individual component functions, each with its own variables. To find the derivative \( \mathbf{r}'(t) \), apply differentiation rules to each component separately.
- For example, consider a vector function \( \mathbf{r}(t) = \langle \cos t + t \sin t, \sin t - t \cos t, t \rangle \).
- Each component requires differentiation with respect to \( t \).
- The first component \( \cos t + t \sin t \) differentiates using the product rule and basic trigonometric derivative rules.
- The resulting derivative is \( -\sin(t) + \sin(t) + t\cos(t) = t\cos(t) \) for the first component.

Repeat this process for the remaining components. Derivatives of vector functions not only transform the vector function but showcase changes in vector direction and magnitude over time.
When calculating the second derivative \( \mathbf{r}''(t) \), repeat the differentiation process on \( \mathbf{r}'(t) \). This operation reveals the acceleration vector, describing the rate of change of the velocity vector.
Dot Product
The dot product, also known as the scalar product, is a crucial operation in vector calculus. It measures how much two vectors align with each other. In mathematical terms, the dot product between two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \) is calculated as:
  • \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)
When applied to vector functions like \( \mathbf{r}'(t) \cdot \mathbf{r}''(t) \), computing the dot product can reveal essential properties of their interaction.
- If the dot product is zero, the vectors are orthogonal, meaning they meet at a right angle.
- A positive dot product indicates the vectors are pointing in roughly the same direction.
- A negative result suggests they're pointing oppositely.

In practical exercises, evaluating the dot product between a velocity and acceleration vector can help you understand dynamic systems like motion direction changes over time.
Vector Functions
Vector functions offer a dynamic way to describe mathematical spaces as collections of vectors. Each vector function assigns a vector to every point in a space, like time or distance, making them particularly powerful for modeling physical systems.
For example, a vector function \( \mathbf{r}(t) = \langle \cos t + t \sin t, \sin t - t \cos t, t \rangle \) can represent a path in space where \( t \) could signify time. Each vector in the function delineates position coordinates at a specific moment.
Key characteristics of vector functions include:
  • Representation of motion or force along multiple axes simultaneously.
  • Ability to describe curves and surfaces in 3D space by mapping a parameter \( t \) to points in a vector field.
  • They can be differentiated like scalar functions, providing velocity and acceleration vectors.
Vector functions serve as a foundation for modeling real-world phenomena like aerodynamic flows, electromagnetic fields, and natural forces. Grasping how they operate opens doors to analyzing complex systems with precision.

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Most popular questions from this chapter

Use the model for projectile motion, assuming there is no air resistance. Rogers Centre in Toronto, Ontario has a center field fence that is 10 feet high and 400 feet from home plate. A ball is hit 3 feet above the ground and leaves the bat at a speed of 100 miles per hour. (a) The ball leaves the bat at an angle of \(\theta=\theta_{0}\) with the horizontal. Write the vector-valued function for the path of the ball. (b) Use a graphing utility to graph the vector-valued function for \(\theta_{0}=10^{\circ}, \theta_{0}=15^{\circ}, \theta_{0}=20^{\circ},\) and \(\theta_{0}=25^{\circ} .\) Use the graphs to approximate the minimum angle required for the hit to be a home run. (c) Determine analytically the minimum angle required for the hit to be a home run.

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