The position co-ordinates of a particle moving in a 3-D coordinate system is given
by `x = a" cos"wt`
`y = a" sin"wt `
and `z=awt`
The speed of the particle is :

Explanation

To find the speed of the particle given its position coordinates in a 3-D coordinate system, we will follow these steps:

### Step 1: Write the position coordinates
The position coordinates of the particle are given as:
- \( x = a \cos(\omega t) \)
- \( y = a \sin(\omega t) \)
- \( z = a \omega t \)

### Step 2: Write the displacement vector
The displacement vector \( \mathbf{S} \) can be expressed in terms of its components:
\[ \mathbf{S} = x \hat{i} + y \hat{j} + z \hat{k} = a \cos(\omega t) \hat{i} + a \sin(\omega t) \hat{j} + a \omega t \hat{k} \]

### Step 3: Differentiate the displacement vector to find velocity
To find the velocity \( \mathbf{v} \), we differentiate the displacement vector \( \mathbf{S} \) with respect to time \( t \):
\[ \mathbf{v} = \frac{d\mathbf{S}}{dt} = \frac{d}{dt}(a \cos(\omega t) \hat{i} + a \sin(\omega t) \hat{j} + a \omega t \hat{k}) \]

### Step 4: Apply differentiation
Using the chain rule, we differentiate each component:
- For \( x \):
\[ \frac{dx}{dt} = -a \omega \sin(\omega t) \]
- For \( y \):
\[ \frac{dy}{dt} = a \omega \cos(\omega t) \]
- For \( z \):
\[ \frac{dz}{dt} = a \omega \]

Thus, the velocity vector becomes:
\[ \mathbf{v} = (-a \omega \sin(\omega t)) \hat{i} + (a \omega \cos(\omega t)) \hat{j} + (a \omega) \hat{k} \]

### Step 5: Calculate the speed
The speed of the particle is the magnitude of the velocity vector:
\[ \text{Speed} = |\mathbf{v}| = \sqrt{(-a \omega \sin(\omega t))^2 + (a \omega \cos(\omega t))^2 + (a \omega)^2} \]

### Step 6: Simplify the expression
Calculating each term:
\[ = \sqrt{(a^2 \omega^2 \sin^2(\omega t)) + (a^2 \omega^2 \cos^2(\omega t)) + (a^2 \omega^2)} \]
Factoring out \( a^2 \omega^2 \):
\[ = \sqrt{a^2 \omega^2 (\sin^2(\omega t) + \cos^2(\omega t) + 1)} \]
Using the identity \( \sin^2(\theta) + \cos^2(\theta) = 1 \):
\[ = \sqrt{a^2 \omega^2 (1 + 1)} = \sqrt{2a^2 \omega^2} = a \omega \sqrt{2} \]

### Final Answer
The speed of the particle is:
\[ \text{Speed} = a \omega \sqrt{2} \]

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