Consider the curve r(t) = (t²- 4t, -3t + 1, −2t² + 5t+4). This is called a planar curve, which means it lies entirely on a single plane. Find an equation for the plane on which this curve lies.
Consider the curve r(t) = (t²- 4t, -3t + 1, −2t² + 5t+4). This is called a planar curve, which means it lies entirely on a single plane. Find an equation for the plane on which this curve lies.
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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![**Understanding Planar Curves and Their Equations**
Consider the curve \(\mathbf{r}(t) = \langle t^2 - 4t, -3t + 1, -2t^2 + 5t + 4 \rangle\). This is called a **planar curve**, which means it lies entirely on a single plane.
To find an equation for the plane on which this curve lies, we can follow a few mathematical steps involving vector calculus and algebra. Below, we provide a step-by-step solution to demonstrate this process.
1. **Identify the Curve Components**:
The function \(\mathbf{r}(t)\) represents a 3-dimensional curve where:
- The x-component is \(x = t^2 - 4t\),
- The y-component is \(y = -3t + 1\),
- The z-component is \(z = -2t^2 + 5t + 4\).
2. **Determine the Parametric Points**:
Select values for \(t\) to generate points on the curve:
- If \(t = 0\), then \( \mathbf{r}(0) = \langle 0, 1, 4 \rangle \),
- If \(t = 1\), then \( \mathbf{r}(1) = \langle -3, -2, 7 \rangle \),
- If \(t = 2\), then \( \mathbf{r}(2) = \langle 0, -5, 4 \rangle \).
3. **Find Two Direction Vectors**:
Using the points from the curve, calculate two directional vectors:
- Vector \(\mathbf{v_1}\) from \(\mathbf{r}(0)\) to \(\mathbf{r}(1)\)
\[
\mathbf{v_1} = \langle -3 - 0, -2 - 1, 7 - 4 \rangle = \langle -3, -3, 3 \rangle
\]
- Vector \(\mathbf{v_2}\) from \(\mathbf{r}(0)\) to \(\mathbf{r}(2)\)
\[
\mathbf{v_2} = \langle 0 - 0](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3d7a07fd-0eaf-4cca-88de-0657a73b7fc8%2F59b5d20b-6f2f-4b79-b0f8-6addc2aed844%2Fys79us_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Understanding Planar Curves and Their Equations**
Consider the curve \(\mathbf{r}(t) = \langle t^2 - 4t, -3t + 1, -2t^2 + 5t + 4 \rangle\). This is called a **planar curve**, which means it lies entirely on a single plane.
To find an equation for the plane on which this curve lies, we can follow a few mathematical steps involving vector calculus and algebra. Below, we provide a step-by-step solution to demonstrate this process.
1. **Identify the Curve Components**:
The function \(\mathbf{r}(t)\) represents a 3-dimensional curve where:
- The x-component is \(x = t^2 - 4t\),
- The y-component is \(y = -3t + 1\),
- The z-component is \(z = -2t^2 + 5t + 4\).
2. **Determine the Parametric Points**:
Select values for \(t\) to generate points on the curve:
- If \(t = 0\), then \( \mathbf{r}(0) = \langle 0, 1, 4 \rangle \),
- If \(t = 1\), then \( \mathbf{r}(1) = \langle -3, -2, 7 \rangle \),
- If \(t = 2\), then \( \mathbf{r}(2) = \langle 0, -5, 4 \rangle \).
3. **Find Two Direction Vectors**:
Using the points from the curve, calculate two directional vectors:
- Vector \(\mathbf{v_1}\) from \(\mathbf{r}(0)\) to \(\mathbf{r}(1)\)
\[
\mathbf{v_1} = \langle -3 - 0, -2 - 1, 7 - 4 \rangle = \langle -3, -3, 3 \rangle
\]
- Vector \(\mathbf{v_2}\) from \(\mathbf{r}(0)\) to \(\mathbf{r}(2)\)
\[
\mathbf{v_2} = \langle 0 - 0
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