(a) Find $\mathbf{r}(t)$

We are given the following information:

• $\mathbf{r}''(t) = \hat{\mathbf{i}} + 2\hat{\mathbf{k}}$
• $\mathbf{r}(0) = -2\hat{\mathbf{j}} + 5\hat{\mathbf{k}}$
• $\mathbf{r}'(0) = 3\hat{\mathbf{j}} - 4\hat{\mathbf{k}}$

To find $\mathbf{r}(t)$, we first integrate $\mathbf{r}''(t)$ with respect to $t$ to find $\mathbf{r}'(t)$.

$\mathbf{r}''(t) = \hat{\mathbf{i}} + 2\hat{\mathbf{k}}$

Integrating:

$\mathbf{r}'(t) = (\hat{\mathbf{i}} t) + (2\hat{\mathbf{k}} t) + \mathbf{C}$

where $\mathbf{C}$ is a constant vector. To determine $\mathbf{C}$, we use the initial condition $\mathbf{r}'(0) = 3\hat{\mathbf{j}} - 4\hat{\mathbf{k}}$:

$\mathbf{r}'(0) = \mathbf{C} = 3\hat{\mathbf{j}} - 4\hat{\mathbf{k}}$

Thus, the first derivative of $\mathbf{r}(t)$ is:

$\mathbf{r}'(t) = \hat{\mathbf{i}} t + 3\hat{\mathbf{j}} + (2t - 4)\hat{\mathbf{k}}$

Next, integrate $\mathbf{r}'(t)$ to find $\mathbf{r}(t)$:

$\mathbf{r}(t) = \int (\hat{\mathbf{i}} t + 3\hat{\mathbf{j}} + (2t - 4)\hat{\mathbf{k}}) \, dt$

$\mathbf{r}(t) = \frac{1}{2} \hat{\mathbf{i}} t^2 + 3\hat{\mathbf{j}} t + (t^2 - 4t)\hat{\mathbf{k}} + \mathbf{D}$

where $\mathbf{D}$ is another constant vector. To determine $\mathbf{D}$, we use the initial condition $\mathbf{r}(0) = -2\hat{\mathbf{j}} + 5\hat{\mathbf{k}}$:

$\mathbf{r}(0) = \mathbf{D} = -2\hat{\mathbf{j}} + 5\hat{\mathbf{k}}$

Thus, the vector function $\mathbf{r}(t)$ is:

$\mathbf{r}(t) = \frac{1}{2} \hat{\mathbf{i}} t^2 + (3t - 2)\hat{\mathbf{j}} + (t^2 - 4t + 5)\hat{\mathbf{k}}$

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(b) At what $t$ is $\mathbf{r}'(t)$ the shortest?

The length of $\mathbf{r}'(t)$ is given by the magnitude of the vector $\mathbf{r}'(t)$. We want to minimize this magnitude.

$|\mathbf{r}'(t)| = \sqrt{(\hat{\mathbf{i}} t)^2 + (3\hat{\mathbf{j}})^2 + (2t - 4)^2}$

$|\mathbf{r}'(t)| = \sqrt{t^2 + 9 + (2t - 4)^2}$

Simplifying the expression inside the square root:

$|\mathbf{r}'(t)| = \sqrt{t^2 + 9 + (4t^2 - 16t + 16)}$

$|\mathbf{r}'(t)| = \sqrt{5t^2 - 16t + 25}$

To find the minimum of $|\mathbf{r}'(t)|$, we take the derivative of $f(t) = \sqrt{5t^2 - 16t + 25}$ and set it equal to zero:

$f'(t) = \frac{1}{2\sqrt{5t^2 - 16t + 25}}(10t - 16)$

Setting $f'(t) = 0$, we solve:

$10t - 16 = 0$

$t = \frac{8}{5}$

Thus, $\mathbf{r}'(t)$ is shortest when $t = \frac{8}{5}$.

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Summary:

• (a) $\mathbf{r}(t) = \frac{1}{2} \hat{\mathbf{i}} t^2 + (3t - 2)\hat{\mathbf{j}} + (t^2 - 4t + 5)\hat{\mathbf{k}}$
• (b) The velocity $\mathbf{r}'(t)$ is shortest at $t = \frac{8}{5}$.

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Do you have any further questions, or would you like more details?

Here are 5 related questions for you to consider:

1. How do you determine the velocity vector $\mathbf{r}'(t)$ from $\mathbf{r}(t)$?
2. What does the magnitude of $\mathbf{r}'(t)$ represent physically?
3. How do initial conditions affect the general solution to a vector-valued function?
4. What is the geometric significance of minimizing the velocity vector's magnitude?
5. How do you integrate vector-valued functions with respect to time?

Tip: Always use initial conditions to solve for unknown constants after integration, ensuring the solution fits the problem's constraints.