Sketch the curve with the given vector equation by finding the following points. r(t) = (2, t, 4-t²) (-4) (x, y, z) = (0) (4) (x, y, z) = (x, y, z) =
Sketch the curve with the given vector equation by finding the following points. r(t) = (2, t, 4-t²) (-4) (x, y, z) = (0) (4) (x, y, z) = (x, y, z) =
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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![### Sketching a Curve Using Vector Equation
To sketch the curve defined by the given vector equation, we need to find specific points on the curve. The vector equation given is:
\[ \mathbf{r}(t) = (2, t, 4 - t^2) \]
We will calculate the coordinates of the curve at specific values of \(t\):
1. **For \(t = -4\):**
\[ \mathbf{r}(-4) = (x, y, z) = \]
- \(x = 2\)
- \(y = -4\)
- \(z = 4 - (-4)^2 = 4 - 16 = -12\)
\[ \mathbf{r}(-4) = (2, -4, -12) \]
2. **For \(t = 0\):**
\[ \mathbf{r}(0) = (x, y, z) = \]
- \(x = 2\)
- \(y = 0\)
- \(z = 4 - 0^2 = 4\)
\[ \mathbf{r}(0) = (2, 0, 4) \]
3. **For \(t = 4\):**
\[ \mathbf{r}(4) = (x, y, z) = \]
- \(x = 2\)
- \(y = 4\)
- \(z = 4 - 4^2 = 4 - 16 = -12\)
\[ \mathbf{r}(4) = (2, 4, -12) \]
### Summary
For the vector equation \(\mathbf{r}(t) = (2, t, 4 - t^2)\), the points on the curve at specific values of \(t\) are:
- At \(t = -4\): \(\mathbf{r}(-4) = (2, -4, -12)\)
- At \(t = 0\): \(\mathbf{r}(0) = (2, 0, 4)\)
- At \(t = 4\): \(\mathbf{r}(4) = (2, 4, -12)\)
These points can be used to sketch the curve defined by the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faf9eba19-7b67-4fcd-8970-a14a27c2cd15%2Fc1f32190-ece0-4975-8b43-8f70b71027c8%2Fzz1etfk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Sketching a Curve Using Vector Equation
To sketch the curve defined by the given vector equation, we need to find specific points on the curve. The vector equation given is:
\[ \mathbf{r}(t) = (2, t, 4 - t^2) \]
We will calculate the coordinates of the curve at specific values of \(t\):
1. **For \(t = -4\):**
\[ \mathbf{r}(-4) = (x, y, z) = \]
- \(x = 2\)
- \(y = -4\)
- \(z = 4 - (-4)^2 = 4 - 16 = -12\)
\[ \mathbf{r}(-4) = (2, -4, -12) \]
2. **For \(t = 0\):**
\[ \mathbf{r}(0) = (x, y, z) = \]
- \(x = 2\)
- \(y = 0\)
- \(z = 4 - 0^2 = 4\)
\[ \mathbf{r}(0) = (2, 0, 4) \]
3. **For \(t = 4\):**
\[ \mathbf{r}(4) = (x, y, z) = \]
- \(x = 2\)
- \(y = 4\)
- \(z = 4 - 4^2 = 4 - 16 = -12\)
\[ \mathbf{r}(4) = (2, 4, -12) \]
### Summary
For the vector equation \(\mathbf{r}(t) = (2, t, 4 - t^2)\), the points on the curve at specific values of \(t\) are:
- At \(t = -4\): \(\mathbf{r}(-4) = (2, -4, -12)\)
- At \(t = 0\): \(\mathbf{r}(0) = (2, 0, 4)\)
- At \(t = 4\): \(\mathbf{r}(4) = (2, 4, -12)\)
These points can be used to sketch the curve defined by the
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