To solve the problem, let’s find both the derivative of the given vector $\vec{r}$ with respect to $t$ and the magnitude of the derivative at $t = \theta$.

Step 1: Given vector function

$\vec{r}(t) = t \hat{i} - t^2 \hat{j} + \sin t \hat{k}.$

Step 2: Differentiate $\vec{r}(t)$ with respect to $t$

$\frac{d\vec{r}}{dt} = \frac{d}{dt}(t) \hat{i} - \frac{d}{dt}(t^2) \hat{j} + \frac{d}{dt}(\sin t) \hat{k}.$ Computing each term:

• $\frac{d}{dt}(t) = 1$,
• $\frac{d}{dt}(t^2) = 2t$,
• $\frac{d}{dt}(\sin t) = \cos t$.

Thus: $\frac{d\vec{r}}{dt} = \hat{i} - 2t \hat{j} + \cos t \hat{k}.$

Step 3: Find the magnitude of $\frac{d\vec{r}}{dt}$ at $t = \theta$

The magnitude of $\frac{d\vec{r}}{dt}$ is: $\left|\frac{d\vec{r}}{dt}\right| = \sqrt{\left(1\right)^2 + \left(-2t\right)^2 + \left(\cos t\right)^2}.$

Substitute $t = \theta$: $\left|\frac{d\vec{r}}{dt}\right| = \sqrt{1 + (-2\theta)^2 + (\cos \theta)^2}.$

Simplify: $\left|\frac{d\vec{r}}{dt}\right| = \sqrt{1 + 4\theta^2 + \cos^2 \theta}.$

Final Results:

1. Derivative of $\vec{r}$ with respect to $t$: $\frac{d\vec{r}}{dt} = \hat{i} - 2t \hat{j} + \cos t \hat{k}.$

2. Magnitude of $\frac{d\vec{r}}{dt}$ at $t = \theta$: $\left|\frac{d\vec{r}}{dt}\right| = \sqrt{1 + 4\theta^2 + \cos^2 \theta}.$

Would you like more details or examples on this?

---

5 Related Questions:

1. How is the magnitude of a vector derivative interpreted geometrically?
2. What happens to $|\frac{d\vec{r}}{dt}|$ when $\theta = 0$?
3. How do you compute the unit tangent vector to $\vec{r}(t)$?
4. What are the physical applications of $\frac{d\vec{r}}{dt}$ in mechanics?
5. How does the magnitude of $\frac{d\vec{r}}{dt}$ change as $\theta$ increases?

Tip: For parametric vector functions, the derivative $\frac{d\vec{r}}{dt}$ represents the velocity vector, and its magnitude gives the speed of the motion.