The position function of a particle is given by 
r(t) = 

t^2, 9t, t^2 − 16t

.
 When is the speed a minimum?

To determine when the speed of the particle is at a minimum, we need to follow these steps:

### 1. **Find the Velocity Vector**:
The velocity vector \(\mathbf{v}(t)\) is the derivative of the position vector \(\mathbf{r}(t)\) with respect to time \(t\).

Given:
\[
\mathbf{r}(t) = \langle t^2, 9t, t^2 - 16t \rangle
\]

The velocity vector \(\mathbf{v}(t)\) is:
\[
\mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) = \langle 2t, 9, 2t - 16 \rangle
\]

### 2. **Find the Speed**:
The speed of the particle is the magnitude of the velocity vector \(\mathbf{v}(t)\).

\[
\text{Speed} = \|\mathbf{v}(t)\| = \sqrt{(2t)^2 + 9^2 + (2t - 16)^2}
\]

Simplifying this expression:

\[
\|\mathbf{v}(t)\| = \sqrt{4t^2 + 81 + (2t - 16)^2}
\]

Expanding and simplifying further:

\[
(2t - 16)^2 = 4t^2 - 64t + 256
\]

So, the speed function becomes:

\[
\|\mathbf{v}(t)\| = \sqrt{4t^2 + 81 + 4t^2 - 64t + 256} = \sqrt{8t^2 - 64t + 337}
\]

### 3. **Minimize the Speed**:
To find when the speed is a minimum, we need to minimize the function \( \sqrt{8t^2 - 64t + 337} \).

This is equivalent to minimizing the expression inside the square root:

\[
f(t) = 8t^2 - 64t + 337
\]

### 4. **Find the Critical Points**:
Take the derivative of \(f(t)\) with respect to \(t\) and set it equal to zero:

\[
f'(t) = 16t - 64
\]
\[
16t - 64 = 0
\]

Solve for \(t\):

\[
t = 4
\]

### 5. **Verify Minimum**:
To verify that \(t = 4\) gives a minimum, we check the second derivative:

\[
f''(t) = 16
\]

Since \(f''(t) > 0\), the function \(f(t)\) is concave up at \(t = 4\), confirming a minimum at this point.

### Conclusion:
The speed of the particle is at a minimum when \(t = 4\).

---

Would you like further details on any of these steps, or do you have any questions?

Here are some related questions:
1. How do you find the position vector from a velocity vector?
2. What does the minimum speed of a particle indicate about its motion?
3. How would the process change if the velocity function was given in a different form?
4. Can the speed of the particle be negative?
5. What physical situations correspond to finding a minimum speed?

**Tip:** Minimizing the magnitude of a vector often involves minimizing the square of the magnitude to avoid dealing with the square root directly.

The position function of a particle is given by r(t) = t^2, 9t, t^2 − 16t. When is the speed a minimum?

Find out how to minimize the speed of a particle given its position function r(t) = t^2, 9t, t^2 − 16t. This problem uses calculus concepts such as vector derivatives and quadratic optimization, suitable for advanced high school or early university students.

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