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Sagot :
To find [tex]\(\left|\frac{d}{dt} \left(a_1 \times a_2\right)\right|_{t=0}\)[/tex], we need to perform several steps. Let's do this step by step.
First, let's write down the vectors [tex]\(a_1\)[/tex] and [tex]\(a_2\)[/tex]:
[tex]\[ a_1 = 5t^2 \mathbf{i} + t \mathbf{j} - t^3 \mathbf{k} \][/tex]
[tex]\[ a_2 = \sin(t) \mathbf{i} - \cos(t) \mathbf{j} \][/tex]
Next, we need to compute the cross product [tex]\(a_1 \times a_2\)[/tex]:
[tex]\[ a_1 \times a_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5t^2 & t & -t^3 \\ \sin(t) & -\cos(t) & 0 \end{vmatrix} \][/tex]
We expand this determinant:
[tex]\[ a_1 \times a_2 = \mathbf{i} \left| \begin{matrix} t & -t^3 \\ -\cos(t) & 0 \end{matrix} \right| - \mathbf{j} \left| \begin{matrix} 5t^2 & -t^3 \\ \sin(t) & 0 \end{matrix} \right| + \mathbf{k} \left| \begin{matrix} 5t^2 & t \\ \sin(t) & -\cos(t) \end{matrix} \right| \][/tex]
Compute each 2x2 determinant:
[tex]\[ \mathbf{i} \left( t \cdot 0 - (-t^3)(-\cos(t)) \right) = -t^3 \cos(t) \mathbf{i} \][/tex]
[tex]\[ -\mathbf{j} \left( 5t^2 \cdot 0 - (-t^3)(\sin(t)) \right) = -(-5t^5 \sin(t)) \mathbf{j} = 5t^5 \sin(t) \mathbf{j} \][/tex]
[tex]\[ \mathbf{k} \left( 5t^2(-\cos(t)) - t \sin(t) \right) = -5t^2 \cos(t) - t \sin(t) \mathbf{k} \][/tex]
So the cross product is:
[tex]\[ a_1 \times a_2 = -t^3 \cos(t) \mathbf{i} + 5t^5 \sin(t) \mathbf{j} + (-5t^2 \cos(t) - t \sin(t)) \mathbf{k} \][/tex]
Now, we need to differentiate this vector with respect to [tex]\(t\)[/tex]:
1. Differentiate [tex]\(-t^3 \cos(t)\mathbf{i}\)[/tex]:
[tex]\[ \frac{d}{dt}[-t^3 \cos(t)] = -3t^2 \cos(t) + t^3 \sin(t) \][/tex]
2. Differentiate [tex]\(5t^5 \sin(t)\mathbf{j}\)[/tex]:
[tex]\[ \frac{d}{dt}[5t^5 \sin(t)] = 25t^4 \sin(t) + 5t^5 \cos(t) \][/tex]
3. Differentiate [tex]\((-5t^2 \cos(t)) - t \sin(t))\mathbf{k}\)[/tex]:
[tex]\[ \frac{d}{dt}[-5t^2 \cos(t) - t \sin(t)] = -10t \cos(t) + 5t^2 \sin(t) - \sin(t) - t \cos(t) \][/tex]
Thus:
[tex]\[ \frac{d}{dt}[a_1 \times a_2] = (-3t^2 \cos(t) + t^3 \sin(t)) \mathbf{i} + (25t^4 \sin(t) + 5t^5 \cos(t)) \mathbf{j} + (-10t \cos(t) + 5t^2 \sin(t) - \sin(t) - t \cos(t)) \mathbf{k} \][/tex]
Evaluate at [tex]\(t = 0\)[/tex]:
For the [tex]\(\mathbf{i}\)[/tex] component:
[tex]\[ -3(0)^2 \cos(0) + (0)^3 \sin(0) = 0 \][/tex]
For the [tex]\(\mathbf{j}\)[/tex] component:
[tex]\[ 25(0)^4 \sin(0) + 5(0)^5 \cos(0) = 0 \][/tex]
For the [tex]\(\mathbf{k}\)[/tex] component:
[tex]\[ -10(0) \cos(0) + 5(0)^2 \sin(0) - \sin(0) - 0 \cos(0) = -\sin(0) = 0 \][/tex]
Finally, the derivative at [tex]\(t = 0\)[/tex] is [tex]\(\mathbf{0} = 0 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k}\)[/tex].
The magnitude of this vector is:
[tex]\[ \left| \mathbf{0} \right| = \sqrt{0^2 + 0^2 + 0^2} = 0 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{0} \][/tex]
First, let's write down the vectors [tex]\(a_1\)[/tex] and [tex]\(a_2\)[/tex]:
[tex]\[ a_1 = 5t^2 \mathbf{i} + t \mathbf{j} - t^3 \mathbf{k} \][/tex]
[tex]\[ a_2 = \sin(t) \mathbf{i} - \cos(t) \mathbf{j} \][/tex]
Next, we need to compute the cross product [tex]\(a_1 \times a_2\)[/tex]:
[tex]\[ a_1 \times a_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 5t^2 & t & -t^3 \\ \sin(t) & -\cos(t) & 0 \end{vmatrix} \][/tex]
We expand this determinant:
[tex]\[ a_1 \times a_2 = \mathbf{i} \left| \begin{matrix} t & -t^3 \\ -\cos(t) & 0 \end{matrix} \right| - \mathbf{j} \left| \begin{matrix} 5t^2 & -t^3 \\ \sin(t) & 0 \end{matrix} \right| + \mathbf{k} \left| \begin{matrix} 5t^2 & t \\ \sin(t) & -\cos(t) \end{matrix} \right| \][/tex]
Compute each 2x2 determinant:
[tex]\[ \mathbf{i} \left( t \cdot 0 - (-t^3)(-\cos(t)) \right) = -t^3 \cos(t) \mathbf{i} \][/tex]
[tex]\[ -\mathbf{j} \left( 5t^2 \cdot 0 - (-t^3)(\sin(t)) \right) = -(-5t^5 \sin(t)) \mathbf{j} = 5t^5 \sin(t) \mathbf{j} \][/tex]
[tex]\[ \mathbf{k} \left( 5t^2(-\cos(t)) - t \sin(t) \right) = -5t^2 \cos(t) - t \sin(t) \mathbf{k} \][/tex]
So the cross product is:
[tex]\[ a_1 \times a_2 = -t^3 \cos(t) \mathbf{i} + 5t^5 \sin(t) \mathbf{j} + (-5t^2 \cos(t) - t \sin(t)) \mathbf{k} \][/tex]
Now, we need to differentiate this vector with respect to [tex]\(t\)[/tex]:
1. Differentiate [tex]\(-t^3 \cos(t)\mathbf{i}\)[/tex]:
[tex]\[ \frac{d}{dt}[-t^3 \cos(t)] = -3t^2 \cos(t) + t^3 \sin(t) \][/tex]
2. Differentiate [tex]\(5t^5 \sin(t)\mathbf{j}\)[/tex]:
[tex]\[ \frac{d}{dt}[5t^5 \sin(t)] = 25t^4 \sin(t) + 5t^5 \cos(t) \][/tex]
3. Differentiate [tex]\((-5t^2 \cos(t)) - t \sin(t))\mathbf{k}\)[/tex]:
[tex]\[ \frac{d}{dt}[-5t^2 \cos(t) - t \sin(t)] = -10t \cos(t) + 5t^2 \sin(t) - \sin(t) - t \cos(t) \][/tex]
Thus:
[tex]\[ \frac{d}{dt}[a_1 \times a_2] = (-3t^2 \cos(t) + t^3 \sin(t)) \mathbf{i} + (25t^4 \sin(t) + 5t^5 \cos(t)) \mathbf{j} + (-10t \cos(t) + 5t^2 \sin(t) - \sin(t) - t \cos(t)) \mathbf{k} \][/tex]
Evaluate at [tex]\(t = 0\)[/tex]:
For the [tex]\(\mathbf{i}\)[/tex] component:
[tex]\[ -3(0)^2 \cos(0) + (0)^3 \sin(0) = 0 \][/tex]
For the [tex]\(\mathbf{j}\)[/tex] component:
[tex]\[ 25(0)^4 \sin(0) + 5(0)^5 \cos(0) = 0 \][/tex]
For the [tex]\(\mathbf{k}\)[/tex] component:
[tex]\[ -10(0) \cos(0) + 5(0)^2 \sin(0) - \sin(0) - 0 \cos(0) = -\sin(0) = 0 \][/tex]
Finally, the derivative at [tex]\(t = 0\)[/tex] is [tex]\(\mathbf{0} = 0 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k}\)[/tex].
The magnitude of this vector is:
[tex]\[ \left| \mathbf{0} \right| = \sqrt{0^2 + 0^2 + 0^2} = 0 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{0} \][/tex]
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