3. Suppose that a vector-valued function ~r(t) satisfies ~r 00(t) = ~ı + 2~k (so its second derivative is
constant) along with ~r(0) = −2~ + 5~k and ~r 0
(0) = 3~ − 4
~k.
(a) Find ~r(t).
(b) At what t is ~r 0
the shortest (as we’ll see, this is where ~r is moving most slowly)?


### (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:

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

---

### (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 \(t = \frac{8}{5}\).

---

### 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 \(t = \frac{8}{5}\).

---

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.

Suppose that a vector-valued function r(t) satisfies r''(t) = i + 2k (so its second derivative is constant) along with r(0) = -2j + 5k and r'(0) = 3j - 4k. (a) Find r(t). (b) At what t is r'(t) the shortest (this is where r is moving most slowly)?

Explore how to solve a vector-valued function problem where r''(t) = i + 2k, and initial conditions r(0) = -2j + 5k, r'(0) = 3j - 4k are provided. The solution involves integration, finding r(t), and minimizing the magnitude of r'(t) to determine the point where the motion is slowest. Suitable for Calculus II students.

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