Let r(t) = (sin(t), cos(t), 6 sin(t) + 9 cos(2t)). Find the projection of r(t) onto the xz-plane for-1 ≤ x (Use symbolic notation and fractions where needed.)
Let r(t) = (sin(t), cos(t), 6 sin(t) + 9 cos(2t)). Find the projection of r(t) onto the xz-plane for-1 ≤ x (Use symbolic notation and fractions where needed.)
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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![### Problem Statement
Let \(\mathbf{r}(t) = \langle \sin(t), \cos(t), 6 \sin(t) + 9 \cos(2t) \rangle\).
Find the projection of \(\mathbf{r}(t)\) onto the \(xz\)-plane for \(-1 \leq x \leq 1\).
(Use symbolic notation and fractions where needed.)
### Solution
To project the vector function \(\mathbf{r}(t)\) onto the \(xz\)-plane, we essentially ignore the \(y\)-component. The \(xz\)-plane projection of \(\mathbf{r}(t)\) is given by:
\[ \mathbf{r}_{xz}(t) = \langle \sin(t), 6 \sin(t) + 9 \cos(2t) \rangle \]
Expressing \(z\) as a function of \(x\):
Let \( x = \sin(t) \).
We need to find the corresponding \(z\)-component in terms of \(x\):
\[ z = 6 \sin(t) + 9 \cos(2t) \]
Since \( \cos(2t) = 1 - 2 \sin^2(t) \),
we substitute \( \sin(t) = x \):
\[ z(x) = 6x + 9 (1 - 2x^2) \]
\[ z(x) = 6x + 9 - 18x^2 \]
\[ z(x) = -18x^2 + 6x + 9 \]
Therefore, the projection of \(\mathbf{r}(t)\) onto the \(xz\)-plane can be written as:
\[ z(x) = -18x^2 + 6x + 9 \]
This equation represents the relationship between \(z\) and \(x\) on the \(xz\)-plane for \(-1 \leq x \leq 1\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F657f96d0-e702-4515-9e22-2987647efb37%2F36790409-b4ca-451f-8da6-b1f593926ad3%2Fzwfs7dcj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Let \(\mathbf{r}(t) = \langle \sin(t), \cos(t), 6 \sin(t) + 9 \cos(2t) \rangle\).
Find the projection of \(\mathbf{r}(t)\) onto the \(xz\)-plane for \(-1 \leq x \leq 1\).
(Use symbolic notation and fractions where needed.)
### Solution
To project the vector function \(\mathbf{r}(t)\) onto the \(xz\)-plane, we essentially ignore the \(y\)-component. The \(xz\)-plane projection of \(\mathbf{r}(t)\) is given by:
\[ \mathbf{r}_{xz}(t) = \langle \sin(t), 6 \sin(t) + 9 \cos(2t) \rangle \]
Expressing \(z\) as a function of \(x\):
Let \( x = \sin(t) \).
We need to find the corresponding \(z\)-component in terms of \(x\):
\[ z = 6 \sin(t) + 9 \cos(2t) \]
Since \( \cos(2t) = 1 - 2 \sin^2(t) \),
we substitute \( \sin(t) = x \):
\[ z(x) = 6x + 9 (1 - 2x^2) \]
\[ z(x) = 6x + 9 - 18x^2 \]
\[ z(x) = -18x^2 + 6x + 9 \]
Therefore, the projection of \(\mathbf{r}(t)\) onto the \(xz\)-plane can be written as:
\[ z(x) = -18x^2 + 6x + 9 \]
This equation represents the relationship between \(z\) and \(x\) on the \(xz\)-plane for \(-1 \leq x \leq 1\).
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