A particle motion on a shape curve is governed by x=2sint, y = 3cost and `z = sqrt5` sint. What is the magnitude of velocity of the particle at any timet? 

Options

A. `3sqrt2 sint `
B. 3
C. `3sqrt 2 cot t `
D. `3sqrt2`

Explanation

To find the magnitude of the velocity of the particle at any time \( t \), we will follow these steps:

### Step 1: Define the position vector
The position of the particle is given by the equations:
- \( x = 2 \sin t \)
- \( y = 3 \cos t \)
- \( z = \sqrt{5} \sin t \)

We can express the position vector \( \mathbf{r} \) as:
\[ \mathbf{r} = 2 \sin t \, \hat{i} + 3 \cos t \, \hat{j} + \sqrt{5} \sin t \, \hat{k} \]

### Step 2: Differentiate to find the velocity vector
The velocity vector \( \mathbf{v} \) is the derivative of the position vector with respect to time \( t \):
\[ \mathbf{v} = \frac{d\mathbf{r}}{dt} = \frac{dx}{dt} \hat{i} + \frac{dy}{dt} \hat{j} + \frac{dz}{dt} \hat{k} \]

Calculating each component:
- \( \frac{dx}{dt} = \frac{d}{dt}(2 \sin t) = 2 \cos t \)
- \( \frac{dy}{dt} = \frac{d}{dt}(3 \cos t) = -3 \sin t \)
- \( \frac{dz}{dt} = \frac{d}{dt}(\sqrt{5} \sin t) = \sqrt{5} \cos t \)

Thus, the velocity vector is:
\[ \mathbf{v} = 2 \cos t \, \hat{i} - 3 \sin t \, \hat{j} + \sqrt{5} \cos t \, \hat{k} \]

### Step 3: Calculate the magnitude of the velocity
The magnitude of the velocity \( |\mathbf{v}| \) is given by:
\[ |\mathbf{v}| = \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 + \left( \frac{dz}{dt} \right)^2 } \]

Substituting the components:
\[ |\mathbf{v}| = \sqrt{(2 \cos t)^2 + (-3 \sin t)^2 + (\sqrt{5} \cos t)^2} \]

Calculating each term:
- \( (2 \cos t)^2 = 4 \cos^2 t \)
- \( (-3 \sin t)^2 = 9 \sin^2 t \)
- \( (\sqrt{5} \cos t)^2 = 5 \cos^2 t \)

Combining these:
\[ |\mathbf{v}| = \sqrt{4 \cos^2 t + 9 \sin^2 t + 5 \cos^2 t} \]
\[ |\mathbf{v}| = \sqrt{(4 + 5) \cos^2 t + 9 \sin^2 t} \]
\[ |\mathbf{v}| = \sqrt{9 \cos^2 t + 9 \sin^2 t} \]

Using the Pythagorean identity \( \cos^2 t + \sin^2 t = 1 \):
\[ |\mathbf{v}| = \sqrt{9(\cos^2 t + \sin^2 t)} = \sqrt{9} = 3 \]

### Final Answer
The magnitude of the velocity of the particle at any time \( t \) is \( 3 \).