Find r'(t), r(t), and r'(to) for the given value of to. r(t) = (et, e24), to = 0 r'(t) = r(t) = r'(to) = (No Response) Sketch the curve represented by the vector-valued function, and sketch the vectors r(t) and r'(to). (No Response) (No Response) -1 r (1, 0) 2 -1 (0, 1) r 1 2 لات -1 1 (1, 1) 2 -1 (1, 1) 1 2
Find r'(t), r(t), and r'(to) for the given value of to. r(t) = (et, e24), to = 0 r'(t) = r(t) = r'(to) = (No Response) Sketch the curve represented by the vector-valued function, and sketch the vectors r(t) and r'(to). (No Response) (No Response) -1 r (1, 0) 2 -1 (0, 1) r 1 2 لات -1 1 (1, 1) 2 -1 (1, 1) 1 2
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![### Vector-Valued Functions and Their Derivatives
**Problem Statement:**
Find \(r'(t)\), \(r(t_0)\), and \(r'(t_0)\) for the given value of \(t_0\).
Given the vector-valued function:
\[ r(t) = \left(e^t, e^{2t}\right), \quad t_0 = 0 \]
**Derivatives:**
\[ r'(t) = \left(e^t, 2e^{2t}\right) \]
\[ r(t_0) = \left(e^0, e^{2 \cdot 0}\right) = (1, 1) \]
\[ r'(t_0) = \left(e^0, 2e^{2 \cdot 0}\right) = (1, 2) \]
### Visualizations:
**Graphical Representations:**
The following graphs and vectors visually represent the curve and its derivatives.
1. **First Plot:**
- The curve \(r(t)\) is drawn in black.
- The position vector \(r(t_0) = (1, 0)\) is shown originating from the origin to the point (1,0).
- The tangent vector \(r'(t_0)\) at \(t_0 = 0\) is shown in gray, indicating the derivative at this point.
2. **Second Plot:**
- The curve \(r(t)\) in black with the position vector \(r(t_0) = (0, 1)\) drawn from the origin to the point (0, 1).
- The derivative vector \(r'(t_0)\) originating from point (0, 1), showing the direction of the tangent to the curve at this point.
3. **Third Plot:**
- The graph of \(r(t)\) is in black.
- The position vector \(r(t_0) = (1, 1)\) illustrating the point (1, 1).
- The tangent vector \(r'(t_0)\) at this point displayed in gray, showing the slope or direction of the curve at \(t_0 = 0\).
4. **Fourth Plot:**
- The plotted curve of \(r(t)\) in black with the position vector \(r(t_0) = (1,1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7ac392e2-4675-4f88-9d78-8c7af6559e48%2F5fb72c49-5b35-4dc4-b7d2-5516c6a5a88f%2F2ix18of_processed.png&w=3840&q=75)
Transcribed Image Text:### Vector-Valued Functions and Their Derivatives
**Problem Statement:**
Find \(r'(t)\), \(r(t_0)\), and \(r'(t_0)\) for the given value of \(t_0\).
Given the vector-valued function:
\[ r(t) = \left(e^t, e^{2t}\right), \quad t_0 = 0 \]
**Derivatives:**
\[ r'(t) = \left(e^t, 2e^{2t}\right) \]
\[ r(t_0) = \left(e^0, e^{2 \cdot 0}\right) = (1, 1) \]
\[ r'(t_0) = \left(e^0, 2e^{2 \cdot 0}\right) = (1, 2) \]
### Visualizations:
**Graphical Representations:**
The following graphs and vectors visually represent the curve and its derivatives.
1. **First Plot:**
- The curve \(r(t)\) is drawn in black.
- The position vector \(r(t_0) = (1, 0)\) is shown originating from the origin to the point (1,0).
- The tangent vector \(r'(t_0)\) at \(t_0 = 0\) is shown in gray, indicating the derivative at this point.
2. **Second Plot:**
- The curve \(r(t)\) in black with the position vector \(r(t_0) = (0, 1)\) drawn from the origin to the point (0, 1).
- The derivative vector \(r'(t_0)\) originating from point (0, 1), showing the direction of the tangent to the curve at this point.
3. **Third Plot:**
- The graph of \(r(t)\) is in black.
- The position vector \(r(t_0) = (1, 1)\) illustrating the point (1, 1).
- The tangent vector \(r'(t_0)\) at this point displayed in gray, showing the slope or direction of the curve at \(t_0 = 0\).
4. **Fourth Plot:**
- The plotted curve of \(r(t)\) in black with the position vector \(r(t_0) = (1,1
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