The shortest distance between origin and a point on the space curve `x=2sint, y=2cost, z=3t` is….

Answer

To find the shortest distance between the origin and a point on the space curve defined by the equations \( x = 2 \sin t \), \( y = 2 \cos t \), and \( z = 3t \), we can follow these steps:

### Step 1: Define the distance formula
The distance \( D \) from the origin \( O(0, 0, 0) \) to a point \( P(x, y, z) \) is given by the distance formula:
\[ D = \sqrt{x^2 + y^2 + z^2} \]

### Step 2: Substitute the parametric equations into the distance formula
Substituting the expressions for \( x \), \( y \), and \( z \) into the distance formula:
\[ D = \sqrt{(2 \sin t)^2 + (2 \cos t)^2 + (3t)^2} \]

### Step 3: Simplify the expression
Now, we simplify the expression:
\[ D = \sqrt{4 \sin^2 t + 4 \cos^2 t + 9t^2} \]
Using the Pythagorean identity \( \sin^2 t + \cos^2 t = 1 \):
\[ D = \sqrt{4(\sin^2 t + \cos^2 t) + 9t^2} = \sqrt{4 + 9t^2} \]

### Step 4: Minimize the distance
To find the shortest distance, we need to minimize \( D \). The expression \( D = \sqrt{4 + 9t^2} \) is minimized when \( t^2 \) is minimized. Since \( t^2 \) is always non-negative, the minimum value occurs at \( t = 0 \).

### Step 5: Calculate the minimum distance
Substituting \( t = 0 \) into the distance formula:
\[ D_{\text{min}} = \sqrt{4 + 9(0)^2} = \sqrt{4} = 2 \]

### Conclusion
Thus, the shortest distance between the origin and a point on the space curve is:
\[ \text{Shortest distance} = 2 \text{ units} \]

---