To compute part (C), we need to differentiate the expression $(|v| |u|^2)'$ at $t = \frac{\pi}{3}$, where $v(t)$ and $u(t)$ are vector-valued functions.

Step 1: Break down the expression

We want to differentiate: $(|v| |u|^2)' = \frac{d}{dt} \left( |v| \cdot |u|^2 \right)$ We can apply the product rule of differentiation here: $(|v| |u|^2)' = \frac{d}{dt} |v| \cdot |u|^2 + |v| \cdot \frac{d}{dt} |u|^2$

Step 2: Compute $|v|$

The magnitude of $v(t)$ is given by: $|v| = \sqrt{v_1^2 + v_2^2 + v_3^2}$ At $t = \frac{\pi}{3}$, we know $v\left(\frac{\pi}{3}\right) = (1, -2, 2)$. Therefore, $|v| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

Step 3: Compute $\frac{d}{dt} |v|$

We can use the formula for differentiating the magnitude of a vector: $\frac{d}{dt} |v| = \frac{v \cdot v'}{|v|}$ At $t = \frac{\pi}{3}$, we know $v\left(\frac{\pi}{3}\right) = (1, -2, 2)$ and $v'\left(\frac{\pi}{3}\right) = (1.1, 0.6, 0.2)$. Therefore, $v \cdot v' = 1(1.1) + (-2)(0.6) + 2(0.2) = 1.1 - 1.2 + 0.4 = 0.3$ Thus, $\frac{d}{dt} |v| = \frac{0.3}{3} = 0.1$

Step 4: Compute $|u|$ and $|u|^2$

The magnitude of $u(t) = (\cos t, \sin t, t)$ is: $|u| = \sqrt{\cos^2 t + \sin^2 t + t^2}$ Since $\cos^2 t + \sin^2 t = 1$, this simplifies to: $|u| = \sqrt{1 + t^2}$ At $t = \frac{\pi}{3}$, $|u| = \sqrt{1 + \left(\frac{\pi}{3}\right)^2} = \sqrt{1 + \frac{\pi^2}{9}} = \sqrt{\frac{9 + \pi^2}{9}} = \frac{\sqrt{9 + \pi^2}}{3}$ Therefore, $|u|^2 = 1 + \frac{\pi^2}{9}$

Step 5: Compute $\frac{d}{dt} |u|^2$

Differentiating $|u|^2 = 1 + t^2$, we get: $\frac{d}{dt} |u|^2 = 2t$ At $t = \frac{\pi}{3}$, $\frac{d}{dt} |u|^2 = 2 \times \frac{\pi}{3} = \frac{2\pi}{3}$

Step 6: Combine everything

We now combine all the components: $(|v| |u|^2)' = \frac{d}{dt} |v| \cdot |u|^2 + |v| \cdot \frac{d}{dt} |u|^2$ At $t = \frac{\pi}{3}$, this becomes: $(|v| |u|^2)' = (0.1)(1 + \frac{\pi^2}{9}) + 3 \cdot \frac{2\pi}{3}$ Simplifying: $(|v| |u|^2)' = 0.1 \left(1 + \frac{\pi^2}{9}\right) + 2\pi$ Substitute $\pi \approx 3.1416$: $(|v| |u|^2)' = 0.1 \left(1 + \frac{(3.1416)^2}{9}\right) + 2 \times 3.1416$ $(|v| |u|^2)' = 0.1 \left(1 + \frac{9.8696}{9}\right) + 6.2832$ $(|v| |u|^2)' = 0.1 \times 2.0966 + 6.2832 = 0.2097 + 6.2832 = 6.4929$

Thus, the value of $(|v| |u|^2)'$ at $t = \frac{\pi}{3}$ is approximately 6.493 (to four significant figures).

Would you like further details or have any questions?

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Related Questions:

1. How do you compute the derivative of vector-valued functions?
2. What are the steps to find the magnitude of a vector?
3. How does the product rule apply to vector magnitudes?
4. Can you explain the formula for differentiating the magnitude of a vector?
5. How does the chain rule apply to functions involving vectors?

Tip: Always verify that the dot products and magnitudes are computed accurately, as small errors can lead to significant deviations in final answers.