Find r'(t), r"(t), r'(t) · r"(t), and r'(t) × r"(t). P(C) = 21²1 - 48 + ²P²H 2³k r(t) 3 (a) r(t) 3it + 2t²k - 4j (b) r"(t) 4kt + 3i (c) r'(t) r"(t) (d) r(t) x r"(t) X Your answer cannot be understood or graded. More Information

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
icon
Related questions
Question
---

### Vector Calculus Problem Set

#### Task: Compute derivatives and perform vector operations

Given the vector function:
\[ \mathbf{r}(t) = \frac{3}{2}t^2\mathbf{i} - 4t\mathbf{j} + \frac{2}{3}t^3\mathbf{k} \]

Find the following:
1. \(\mathbf{r}'(t)\)
2. \(\mathbf{r}''(t)\)
3. \(\mathbf{r}'(t) \cdot \mathbf{r}''(t)\)
4. \(\mathbf{r}'(t) \times \mathbf{r}''(t)\)

#### Solutions:

**(a) \(\mathbf{r}'(t)\)**
\[ \boxed{3t\mathbf{i} + 2t^2\mathbf{k} - 4\mathbf{j}} \]
This solution is correct.

**(b) \(\mathbf{r}''(t)\)**
\[ \boxed{4t\mathbf{k} + 3\mathbf{i}} \]
This solution is correct.

**(c) \(\mathbf{r}'(t) \cdot \mathbf{r}''(t)\)**
\[ \boxed{} \]
\[ \textcolor{red}{\text{Your answer cannot be understood or graded.}} \]

**(d) \(\mathbf{r}'(t) \times \mathbf{r}''(t)\)**
\[ \boxed{} \]
\[ \textcolor{red}{\text{Your answer cannot be understood or graded.}} \]

---

### Explanation of Graphs and Calculations

- **Part (a):** 
  - Compute the first derivative \(\mathbf{r}'(t)\) by differentiating each component of \(\mathbf{r}(t)\):
    \[ \mathbf{r}'(t) = \frac{d}{dt} \left( \frac{3}{2}t^2 \mathbf{i} - 4t\mathbf{j} + \frac{2}{3}t^3\mathbf{k} \right) \]
    \[ = 3t\mathbf{i} - 4\mathbf{j} + 2t^2\mathbf{k} \]

- **Part (b):**
Transcribed Image Text:--- ### Vector Calculus Problem Set #### Task: Compute derivatives and perform vector operations Given the vector function: \[ \mathbf{r}(t) = \frac{3}{2}t^2\mathbf{i} - 4t\mathbf{j} + \frac{2}{3}t^3\mathbf{k} \] Find the following: 1. \(\mathbf{r}'(t)\) 2. \(\mathbf{r}''(t)\) 3. \(\mathbf{r}'(t) \cdot \mathbf{r}''(t)\) 4. \(\mathbf{r}'(t) \times \mathbf{r}''(t)\) #### Solutions: **(a) \(\mathbf{r}'(t)\)** \[ \boxed{3t\mathbf{i} + 2t^2\mathbf{k} - 4\mathbf{j}} \] This solution is correct. **(b) \(\mathbf{r}''(t)\)** \[ \boxed{4t\mathbf{k} + 3\mathbf{i}} \] This solution is correct. **(c) \(\mathbf{r}'(t) \cdot \mathbf{r}''(t)\)** \[ \boxed{} \] \[ \textcolor{red}{\text{Your answer cannot be understood or graded.}} \] **(d) \(\mathbf{r}'(t) \times \mathbf{r}''(t)\)** \[ \boxed{} \] \[ \textcolor{red}{\text{Your answer cannot be understood or graded.}} \] --- ### Explanation of Graphs and Calculations - **Part (a):** - Compute the first derivative \(\mathbf{r}'(t)\) by differentiating each component of \(\mathbf{r}(t)\): \[ \mathbf{r}'(t) = \frac{d}{dt} \left( \frac{3}{2}t^2 \mathbf{i} - 4t\mathbf{j} + \frac{2}{3}t^3\mathbf{k} \right) \] \[ = 3t\mathbf{i} - 4\mathbf{j} + 2t^2\mathbf{k} \] - **Part (b):**
Expert Solution
steps

Step by step

Solved in 2 steps with 2 images

Blurred answer
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning