If [tex]$a_1=5i t^2 + jt - t^3 k$[/tex] and [tex]$a_2=i \sin t - j \cos t$[/tex], then [tex]\left|\frac{d}{dt}\left(a_1 \times a_2\right)\right|_{t=0}[/tex] is:

A. 0
B. 1
C. [tex]\sqrt{136}[/tex]
D. [tex]\sqrt{12}[/tex]
E. None of the above



Answer :

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]