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Sagot :
To find the arc length [tex]\( s \)[/tex] of the vector function [tex]\(\mathbf{r}(t) = 2 \sin(2t) \mathbf{i} - 5 \cos(t) \mathbf{j}\)[/tex] from [tex]\(t = 0\)[/tex] to [tex]\(t = \frac{\pi}{2}\)[/tex], we can follow these steps:
1. Find the derivative of [tex]\(\mathbf{r}(t)\)[/tex] with respect to [tex]\(t\)[/tex]:
[tex]\[ \mathbf{r}(t) = \begin{pmatrix} 2 \sin(2t) \\ -5 \cos(t) \end{pmatrix} \][/tex]
To find [tex]\(\mathbf{r}'(t)\)[/tex]:
[tex]\[ \mathbf{r}'(t) = \begin{pmatrix} 2 \frac{d}{dt} \sin(2t) \\ -5 \frac{d}{dt} \cos(t) \end{pmatrix} \][/tex]
Using the chain rule and standard derivatives:
[tex]\[ \frac{d}{dt} \sin(2t) = 2 \cos(2t) \][/tex]
And:
[tex]\[ \frac{d}{dt} \cos(t) = - \sin(t) \][/tex]
Therefore:
[tex]\[ \mathbf{r}'(t) = \begin{pmatrix} 2 \cdot 2 \cos(2t) \\ -5 \cdot (-\sin(t)) \end{pmatrix} = \begin{pmatrix} 4 \cos(2t) \\ 5 \sin(t) \end{pmatrix} \][/tex]
2. Find the magnitude of [tex]\(\mathbf{r}'(t)\)[/tex]:
The magnitude of [tex]\(\mathbf{r}'(t)\)[/tex] is given by:
[tex]\[ \|\mathbf{r}'(t)\| = \sqrt{\left(4 \cos(2t)\right)^2 + \left(5 \sin(t)\right)^2} \][/tex]
Simplifying inside the square root:
[tex]\[ \|\mathbf{r}'(t)\| = \sqrt{16 \cos^2(2t) + 25 \sin^2(t)} \][/tex]
3. Set up the integral for the arc length [tex]\(s\)[/tex]:
The arc length [tex]\(s\)[/tex] is the integral of the magnitude of [tex]\(\mathbf{r}'(t)\)[/tex] with respect to [tex]\(t\)[/tex] from [tex]\(t = 0\)[/tex] to [tex]\(t = \frac{\pi}{2}\)[/tex]:
[tex]\[ s = \int_{0}^{\frac{\pi}{2}} \|\mathbf{r}'(t)\| \, dt = \int_{0}^{\frac{\pi}{2}} \sqrt{16 \cos^2(2t) + 25 \sin^2(t)} \, dt \][/tex]
Hence, the solution gives us the derivative of the vector function, the magnitude of the derivative, and the integral representing the arc length. Specifically, the results are:
1. The derivative of [tex]\(\mathbf{r}(t)\)[/tex]:
[tex]\[ \mathbf{r}'(t) = \begin{pmatrix} 4 \cos(2t) \\ 5 \sin(t) \end{pmatrix} \][/tex]
2. The magnitude of [tex]\(\mathbf{r}'(t)\)[/tex]:
[tex]\[ \sqrt{16 \cos^2(2t) + 25 \sin^2(t)} \][/tex]
3. The integral representing the arc length [tex]\(s\)[/tex]:
[tex]\[ \int_{0}^{\frac{\pi}{2}} \sqrt{16 \cos^2(2t) + 25 \sin^2(t)} \, dt \][/tex]
1. Find the derivative of [tex]\(\mathbf{r}(t)\)[/tex] with respect to [tex]\(t\)[/tex]:
[tex]\[ \mathbf{r}(t) = \begin{pmatrix} 2 \sin(2t) \\ -5 \cos(t) \end{pmatrix} \][/tex]
To find [tex]\(\mathbf{r}'(t)\)[/tex]:
[tex]\[ \mathbf{r}'(t) = \begin{pmatrix} 2 \frac{d}{dt} \sin(2t) \\ -5 \frac{d}{dt} \cos(t) \end{pmatrix} \][/tex]
Using the chain rule and standard derivatives:
[tex]\[ \frac{d}{dt} \sin(2t) = 2 \cos(2t) \][/tex]
And:
[tex]\[ \frac{d}{dt} \cos(t) = - \sin(t) \][/tex]
Therefore:
[tex]\[ \mathbf{r}'(t) = \begin{pmatrix} 2 \cdot 2 \cos(2t) \\ -5 \cdot (-\sin(t)) \end{pmatrix} = \begin{pmatrix} 4 \cos(2t) \\ 5 \sin(t) \end{pmatrix} \][/tex]
2. Find the magnitude of [tex]\(\mathbf{r}'(t)\)[/tex]:
The magnitude of [tex]\(\mathbf{r}'(t)\)[/tex] is given by:
[tex]\[ \|\mathbf{r}'(t)\| = \sqrt{\left(4 \cos(2t)\right)^2 + \left(5 \sin(t)\right)^2} \][/tex]
Simplifying inside the square root:
[tex]\[ \|\mathbf{r}'(t)\| = \sqrt{16 \cos^2(2t) + 25 \sin^2(t)} \][/tex]
3. Set up the integral for the arc length [tex]\(s\)[/tex]:
The arc length [tex]\(s\)[/tex] is the integral of the magnitude of [tex]\(\mathbf{r}'(t)\)[/tex] with respect to [tex]\(t\)[/tex] from [tex]\(t = 0\)[/tex] to [tex]\(t = \frac{\pi}{2}\)[/tex]:
[tex]\[ s = \int_{0}^{\frac{\pi}{2}} \|\mathbf{r}'(t)\| \, dt = \int_{0}^{\frac{\pi}{2}} \sqrt{16 \cos^2(2t) + 25 \sin^2(t)} \, dt \][/tex]
Hence, the solution gives us the derivative of the vector function, the magnitude of the derivative, and the integral representing the arc length. Specifically, the results are:
1. The derivative of [tex]\(\mathbf{r}(t)\)[/tex]:
[tex]\[ \mathbf{r}'(t) = \begin{pmatrix} 4 \cos(2t) \\ 5 \sin(t) \end{pmatrix} \][/tex]
2. The magnitude of [tex]\(\mathbf{r}'(t)\)[/tex]:
[tex]\[ \sqrt{16 \cos^2(2t) + 25 \sin^2(t)} \][/tex]
3. The integral representing the arc length [tex]\(s\)[/tex]:
[tex]\[ \int_{0}^{\frac{\pi}{2}} \sqrt{16 \cos^2(2t) + 25 \sin^2(t)} \, dt \][/tex]
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