IDNLearn.com is the perfect place to get detailed and accurate answers to your questions. Ask your questions and receive reliable, detailed answers from our dedicated community of experts.
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]
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com has the solutions to your questions. Thanks for stopping by, and come back for more insightful information.
