Find the tangential and normal components of acceleration at the given time t for the space curve 
r(t).
 (If an answer is undefined, enter UNDEFINED.)
r(t) = ti + 3tj − 6tk,	    t = 3


To find the tangential and normal components of acceleration for the space curve \( \mathbf{r}(t) \), we need to follow these steps:

### 1. Tangential Component of Acceleration ( \( a_T \) ):
The tangential component is the projection of the acceleration vector \( \mathbf{a}(t) \) along the direction of the velocity vector \( \mathbf{v}(t) \). It is given by the formula:

\[
a_T = \frac{\mathbf{v}(t) \cdot \mathbf{a}(t)}{|\mathbf{v}(t)|}
\]

### 2. Normal Component of Acceleration ( \( a_N \) ):
The normal component is the part of the acceleration vector that is perpendicular to the velocity vector. It is calculated by:

\[
a_N = \frac{|\mathbf{v}(t) \times \mathbf{a}(t)|}{|\mathbf{v}(t)|}
\]

### Given:
The position vector is \( \mathbf{r}(t) = t\mathbf{i} + 3t\mathbf{j} - 6t\mathbf{k} \).

Let’s compute the tangential and normal components step by step.

---

### Step 1: Velocity Vector \( \mathbf{v}(t) \)
\[
\mathbf{v}(t) = \frac{d}{dt} \mathbf{r}(t) = \frac{d}{dt} (t\mathbf{i} + 3t\mathbf{j} - 6t\mathbf{k})
\]
\[
\mathbf{v}(t) = \mathbf{i} + 3\mathbf{j} - 6\mathbf{k}
\]

At \( t = 3 \), the velocity vector is:
\[
\mathbf{v}(3) = \mathbf{i} + 3\mathbf{j} - 6\mathbf{k}
\]

### Step 2: Acceleration Vector \( \mathbf{a}(t) \)
\[
\mathbf{a}(t) = \frac{d}{dt} \mathbf{v}(t)
\]
Since \( \mathbf{v}(t) \) is constant, \( \mathbf{a}(t) = 0 \).

At \( t = 3 \), the acceleration vector is:
\[
\mathbf{a}(3) = 0
\]

---

### Step 3: Tangential Component of Acceleration \( a_T \)
\[
a_T = \frac{\mathbf{v}(3) \cdot \mathbf{a}(3)}{|\mathbf{v}(3)|}
\]
Since \( \mathbf{a}(3) = 0 \), we have:
\[
a_T = 0
\]

### Step 4: Normal Component of Acceleration \( a_N \)
\[
a_N = \frac{|\mathbf{v}(3) \times \mathbf{a}(3)|}{|\mathbf{v}(3)|}
\]
Again, since \( \mathbf{a}(3) = 0 \), the cross product \( \mathbf{v}(3) \times \mathbf{a}(3) = 0 \). Therefore:
\[
a_N = 0
\]

---

### Final Answer:
- **Tangential Component \( a_T \) = 0**
- **Normal Component \( a_N \) = 0**

Let me know if you need more details or have any questions!

Here are 5 related questions:
1. What is the magnitude of the velocity vector at \( t = 3 \)?
2. How do we compute curvature for a space curve like \( \mathbf{r}(t) \)?
3. What is the direction of the velocity vector at \( t = 3 \)?
4. How would acceleration behave for a non-constant velocity function?
5. What does it mean when the normal component of acceleration is zero?

**Tip:** The tangential acceleration reflects the change in speed, while the normal acceleration reflects the change in direction!

Find the tangential and normal components of acceleration at the given time t for the space curve r(t). r(t) = ti + 3tj − 6tk, t = 3

Learn how to calculate the tangential and normal components of acceleration for the space curve r(t) = ti + 3tj - 6tk at t = 3. This problem involves vector calculus concepts such as dot products, cross products, and 3D kinematics. Suitable for Grades 11-12 and early college students.

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