Solve in good English and explain detaily step by step used simple example to show ur point especially the differentiation part

To solve this problem, we need to use differentiation to find velocity and acceleration from the position vector.

• Velocity ($\vec{v}$) is the first derivative of the position vector ($\vec{r}$) with respect to time ($t$): $\vec{v} = \frac{d\vec{r}}{dt}$.
• Acceleration ($\vec{a}$) is the first derivative of the velocity vector with respect to time: $\vec{a} = \frac{d\vec{v}}{dt}$.

We will use two main rules for differentiation:

1. Product Rule: If a function $f(t)$ is a product of two other functions, $u(t)$ and $v(t)$, so $f(t) = u(t)v(t)$, then its derivative is:

   $$\frac{d}{dt}(u(t)v(t)) = u'(t)v(t) + u(t)v'(t)$$

   Example: To differentiate $t \sin(t)$, let $u(t) = t$ and $v(t) = \sin(t)$. Then $u'(t) = 1$ and $v'(t) = \cos(t)$. So, $\frac{d}{dt}(t \sin(t)) = (1)(\sin(t)) + (t)(\cos(t)) = \sin(t) + t\cos(t)$.

2. Chain Rule: If a function $f(t)$ is a function of another function, say $f(t) = g(h(t))$, then its derivative is:

   $$\frac{d}{dt}(g(h(t))) = g'(h(t)) \cdot h'(t)$$

   Example: To differentiate $\sin(2t)$, let $g(x) = \sin(x)$ and $h(t) = 2t$. Then $g'(x) = \cos(x)$ and $h'(t) = 2$. So, $\frac{d}{dt}(\sin(2t)) = \cos(2t) \cdot 2 = 2\cos(2t)$.

The position vector of particle P is given by:

$$\vec{r}(t) = (t \sin(2t))\hat{i} + (\cos(2t))\hat{j}$$

First, let's find the velocity vector $\vec{v}(t)$ by differentiating $\vec{r}(t)$ with respect to $t$:

$$\vec{v}(t) = \frac{d}{dt}(t \sin(2t))\hat{i} + \frac{d}{dt}(\cos(2t))\hat{j}$$

For the $\hat{i}$ component, $t \sin(2t)$:
Using the product rule with $u=t$ and $v=\sin(2t)$:
$u' = \frac{d}{dt}(t) = 1$
$v' = \frac{d}{dt}(\sin(2t)) = 2\cos(2t)$ (using the chain rule)
So, $\frac{d}{dt}(t \sin(2t)) = (1)(\sin(2t)) + (t)(2\cos(2t)) = \sin(2t) + 2t\cos(2t)$.

For the $\hat{j}$ component, $\cos(2t)$:
Using the chain rule:
$\frac{d}{dt}(\cos(2t)) = -2\sin(2t)$.

Thus, the velocity vector is:

$$\vec{v}(t) = (\sin(2t) + 2t\cos(2t))\hat{i} - (2\sin(2t))\hat{j}$$

Next, let's find the acceleration vector $\vec{a}(t)$ by differentiating $\vec{v}(t)$ with respect to $t$:

$$\vec{a}(t) = \frac{d}{dt}(\sin(2t) + 2t\cos(2t))\hat{i} - \frac{d}{dt}(2\sin(2t))\hat{j}$$

For the $\hat{i}$ component, $\sin(2t) + 2t\cos(2t)$:
$\frac{d}{dt}(\sin(2t)) = 2\cos(2t)$.
For $2t\cos(2t)$, using the product rule with $u=2t$ and $v=\cos(2t)$:
$u' = \frac{d}{dt}(2t) = 2$
$v' = \frac{d}{dt}(\cos(2t)) = -2\sin(2t)$ (using the chain rule)
So, $\frac{d}{dt}(2t\cos(2t)) = (2)(\cos(2t)) + (2t)(-2\sin(2t)) = 2\cos(2t) - 4t\sin(2t)$.
Combining these, $\frac{d}{dt}(\sin(2t) + 2t\cos(2t)) = 2\cos(2t) + 2\cos(2t) - 4t\sin(2t) = 4\cos(2t) - 4t\sin(2t)$.

For the $\hat{j}$ component, $2\sin(2t)$:
$\frac{d}{dt}(2\sin(2t)) = 2(2\cos(2t)) = 4\cos(2t)$.

Thus, the acceleration vector is:

$$\vec{a}(t) = (4\cos(2t) - 4t\sin(2t))\hat{i} - (4\cos(2t))\hat{j}$$

Now we can solve each part for $t = \frac{\pi}{4}$.
At $t = \frac{\pi}{4}$, we have $2t = 2 \times \frac{\pi}{4} = \frac{\pi}{2}$.
Recall that $\cos(\frac{\pi}{2}) = 0$ and $\sin(\frac{\pi}{2}) = 1$.