A particle moves along the curve `y=x^(3/2)` in the first quadrant in such a way that its distance from the origin increases at the rate of `11` units per second. Then the value of `dx/dt` when `x=3` is given by

Options

A. 4
B. `(9)/(2)`
C. `(3sqrt(3))/(2)`
D. `12`

Answer

To solve the problem step by step, we will follow the given information and apply the necessary calculus concepts.

### Step 1: Understand the Problem
We have a particle moving along the curve \( y = x^{3/2} \) in the first quadrant. The distance \( D \) from the origin to the point \( (x, y) \) increases at a rate of \( 11 \) units per second. We need to find \( \frac{dx}{dt} \) when \( x = 3 \).

### Step 2: Express the Distance
The distance \( D \) from the origin to the point \( (x, y) \) can be expressed as:
\[ D = \sqrt{x^2 + y^2} \]
Substituting \( y = x^{3/2} \):
\[ D = \sqrt{x^2 + (x^{3/2})^2} = \sqrt{x^2 + x^3} = \sqrt{x^2 + x^3} \]

### Step 3: Differentiate the Distance with Respect to Time
To find how the distance changes with time, we differentiate both sides with respect to \( t \):
\[ \frac{dD}{dt} = \frac{1}{2\sqrt{x^2 + x^3}} \cdot \left( 2x \frac{dx}{dt} + 3x^2 \frac{dx}{dt} \right) \]
This simplifies to:
\[ \frac{dD}{dt} = \frac{(2x + 3x^2)}{2\sqrt{x^2 + x^3}} \cdot \frac{dx}{dt} \]

### Step 4: Substitute Known Values
We know that \( \frac{dD}{dt} = 11 \) units per second. Therefore, we can set up the equation:
\[ 11 = \frac{(2x + 3x^2)}{2\sqrt{x^2 + x^3}} \cdot \frac{dx}{dt} \]

### Step 5: Evaluate at \( x = 3 \)
Now, we substitute \( x = 3 \):
1. Calculate \( D \):
\[ D = \sqrt{3^2 + 3^3} = \sqrt{9 + 27} = \sqrt{36} = 6 \]
2. Substitute \( x = 3 \) into the equation:
\[ 11 = \frac{(2(3) + 3(3^2))}{2\sqrt{36}} \cdot \frac{dx}{dt} \]
Simplifying:
\[ 11 = \frac{(6 + 27)}{2 \cdot 6} \cdot \frac{dx}{dt} \]
\[ 11 = \frac{33}{12} \cdot \frac{dx}{dt} \]

### Step 6: Solve for \( \frac{dx}{dt} \)
Now, we can solve for \( \frac{dx}{dt} \):
\[ \frac{dx}{dt} = \frac{11 \cdot 12}{33} = \frac{132}{33} = 4 \]

### Final Answer
Thus, the value of \( \frac{dx}{dt} \) when \( x = 3 \) is:
\[ \frac{dx}{dt} = 4 \]