Calculate the acceleration of a particle at `t=(pi)/(6)` , if the position vectors of the particle
`vecr=(5 cos t ) hati +(2 sin t) hatj+(6t^(2) sin 2t) hatkm`

Answer

To solve the problem of calculating the acceleration of a particle at \( t = \frac{\pi}{6} \), given the position vector

\[ \vec{r} = 5 \cos t \hat{i} + 2 \sin t \hat{j} + 6t^2 \sin 2t \hat{k}, \]

we will follow these steps:

### Step 1: Differentiate the Position Vector to Find Velocity

The velocity \( \vec{v} \) is given by the derivative of the position vector \( \vec{r} \) with respect to time \( t \):

\[ \vec{v} = \frac{d\vec{r}}{dt} = \frac{d}{dt}(5 \cos t) \hat{i} + \frac{d}{dt}(2 \sin t) \hat{j} + \frac{d}{dt}(6t^2 \sin 2t) \hat{k}. \]

Calculating each component:

1. For the \( \hat{i} \) component:
\[ \frac{d}{dt}(5 \cos t) = -5 \sin t. \]

2. For the \( \hat{j} \) component:
\[ \frac{d}{dt}(2 \sin t) = 2 \cos t. \]

3. For the \( \hat{k} \) component, we use the product rule:
\[ \frac{d}{dt}(6t^2 \sin 2t) = 6 \left(2t \sin 2t + t^2 \cdot \frac{d}{dt}(\sin 2t)\right) = 6 \left(2t \sin 2t + t^2 \cdot 2 \cos 2t\right) = 12t \sin 2t + 12t^2 \cos 2t. \]

Thus, the velocity vector is:

\[ \vec{v} = -5 \sin t \hat{i} + 2 \cos t \hat{j} + (12t \sin 2t + 12t^2 \cos 2t) \hat{k}. \]

### Step 2: Differentiate the Velocity Vector to Find Acceleration

The acceleration \( \vec{a} \) is given by the derivative of the velocity vector \( \vec{v} \):

\[ \vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(-5 \sin t) \hat{i} + \frac{d}{dt}(2 \cos t) \hat{j} + \frac{d}{dt}(12t \sin 2t + 12t^2 \cos 2t) \hat{k}. \]

Calculating each component:

1. For the \( \hat{i} \) component:
\[ \frac{d}{dt}(-5 \sin t) = -5 \cos t. \]

2. For the \( \hat{j} \) component:
\[ \frac{d}{dt}(2 \cos t) = -2 \sin t. \]

3. For the \( \hat{k} \) component, we again use the product rule:
\[ \frac{d}{dt}(12t \sin 2t + 12t^2 \cos 2t) = 12 \left(\sin 2t + 2t \cos 2t\right) + 12 \left(2t \cos 2t - 2t^2 \sin 2t\right). \]
Simplifying gives:
\[ 12 \left(\sin 2t + 2t \cos 2t + 2t \cos 2t - 2t^2 \sin 2t\right) = 12 \left(\sin 2t + 4t \cos 2t - 2t^2 \sin 2t\right). \]

Thus, the acceleration vector is:

\[ \vec{a} = -5 \cos t \hat{i} - 2 \sin t \hat{j} + 12 \left(\sin 2t + 4t \cos 2t - 2t^2 \sin 2t\right) \hat{k}. \]

### Step 3: Evaluate the Acceleration at \( t = \frac{\pi}{6} \)

Substituting \( t = \frac{\pi}{6} \):

1. For the \( \hat{i} \) component:
\[ -5 \cos\left(\frac{\pi}{6}\right) = -5 \cdot \frac{\sqrt{3}}{2} = -\frac{5\sqrt{3}}{2}. \]

2. For the \( \hat{j} \) component:
\[ -2 \sin\left(\frac{\pi}{6}\right) = -2 \cdot \frac{1}{2} = -1. \]

3. For the \( \hat{k} \) component:
\[ 12 \left(\sin\left(\frac{\pi}{3}\right) + 4\left(\frac{\pi}{6}\right) \cos\left(\frac{\pi}{3}\right) - 2\left(\frac{\pi}{6}\right)^2 \sin\left(\frac{\pi}{3}\right)\right). \]
Calculating each term:
- \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \),
- \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \),
- \( \left(\frac{\pi}{6}\right)^2 = \frac{\pi^2}{36} \).

Thus, the \( \hat{k} \) component becomes:
\[ 12 \left(\frac{\sqrt{3}}{2} + 4 \cdot \frac{\pi}{6} \cdot \frac{1}{2} - 2 \cdot \frac{\pi^2}{36} \cdot \frac{\sqrt{3}}{2}\right). \]
Simplifying gives:
\[ 12 \left(\frac{\sqrt{3}}{2} + \frac{2\pi}{6} - \frac{\pi^2 \sqrt{3}}{36}\right) = 12 \left(\frac{\sqrt{3}}{2} + \frac{\pi}{3} - \frac{\pi^2 \sqrt{3}}{36}\right). \]

### Final Result

Thus, the acceleration at \( t = \frac{\pi}{6} \) is:

\[ \vec{a} = -\frac{5\sqrt{3}}{2} \hat{i} - 1 \hat{j} + 12 \left(\frac{\sqrt{3}}{2} + \frac{\pi}{3} - \frac{\pi^2 \sqrt{3}}{36}\right) \hat{k}. \]