We are tasked with finding the magnitude of the second derivative of the position vector $\mathbf{r}(t)$ at $t = \frac{\pi}{2}$.

Given the position vector:

$\mathbf{r}(t) = 2 \cos(t) \hat{i} - 3 \sin(t) \hat{j} + 5 \hat{k}$

Step 1: First derivative of $\mathbf{r}(t)$

The first derivative, which represents the velocity vector, is:

$\frac{d\mathbf{r}}{dt} = \frac{d}{dt} \left( 2 \cos(t) \hat{i} - 3 \sin(t) \hat{j} + 5 \hat{k} \right)$

Taking the derivative of each component:

$\frac{d}{dt} \left( 2 \cos(t) \right) = -2 \sin(t)$ $\frac{d}{dt} \left( -3 \sin(t) \right) = -3 \cos(t)$ $\frac{d}{dt} \left( 5 \right) = 0$

Thus, the velocity vector is:

$\frac{d\mathbf{r}}{dt} = -2 \sin(t) \hat{i} - 3 \cos(t) \hat{j}$

Step 2: Second derivative of $\mathbf{r}(t)$

Now, we find the second derivative, which represents the acceleration vector:

$\frac{d^2\mathbf{r}}{dt^2} = \frac{d}{dt} \left( -2 \sin(t) \hat{i} - 3 \cos(t) \hat{j} \right)$

Taking the derivative of each component:

$\frac{d}{dt} \left( -2 \sin(t) \right) = -2 \cos(t)$ $\frac{d}{dt} \left( -3 \cos(t) \right) = 3 \sin(t)$

Thus, the acceleration vector is:

$\frac{d^2\mathbf{r}}{dt^2} = -2 \cos(t) \hat{i} + 3 \sin(t) \hat{j}$

Step 3: Magnitude of $\frac{d^2\mathbf{r}}{dt^2}$ at $t = \frac{\pi}{2}$

To find the magnitude of the acceleration vector at $t = \frac{\pi}{2}$, substitute $t = \frac{\pi}{2}$ into the acceleration vector components:

$\mathbf{a}\left( \frac{\pi}{2} \right) = -2 \cos\left( \frac{\pi}{2} \right) \hat{i} + 3 \sin\left( \frac{\pi}{2} \right) \hat{j}$

Since $\cos\left( \frac{\pi}{2} \right) = 0$ and $\sin\left( \frac{\pi}{2} \right) = 1$, we get:

$\mathbf{a}\left( \frac{\pi}{2} \right) = -2(0) \hat{i} + 3(1) \hat{j} = 3 \hat{j}$

The magnitude of the acceleration vector is:

$|\mathbf{a}\left( \frac{\pi}{2} \right)| = \sqrt{0^2 + 3^2} = 3$

Final Answer:

The magnitude of $\frac{d^2\mathbf{r}}{dt^2}$ at $t = \frac{\pi}{2}$ is $\boxed{3}$.

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Would you like more details on any of the steps or have any questions?

Here are 5 related questions for you to explore further:

1. How do you compute the first and second derivatives of a position vector in general?
2. What is the physical interpretation of the first and second derivatives of a vector function?
3. How does the magnitude of acceleration relate to the motion of an object?
4. How can you apply this to motion in three dimensions?
5. How would you compute the velocity vector and acceleration vector if the position vector had a different functional form?

Tip: Remember that to find the magnitude of a vector, you take the square root of the sum of the squares of its components.