A particle is moving along the curve below.
y = sqrt x
As the particle passes through the point (4, 2), its x-coordinate increases at a rate of 7 cm/s. How fast is the distance from the particle to the origin changing at this instant? (Round your answer to three decimal places.)

We are asked to find the rate at which the distance from the particle to the origin is changing as it moves along the curve \( y = \sqrt{x} \) at the point \( (4, 2) \), given that the x-coordinate increases at a rate of 7 cm/s.

### Step 1: Define the variables
Let:
- \( x \) be the x-coordinate of the particle.
- \( y \) be the y-coordinate of the particle, which is given by \( y = \sqrt{x} \).
- \( D \) be the distance from the particle to the origin. Using the distance formula, we have:
  \[
  D = \sqrt{x^2 + y^2}
  \]

### Step 2: Implicit differentiation
We are looking for \( \frac{dD}{dt} \), the rate of change of the distance \( D \). First, substitute \( y = \sqrt{x} \) into the distance formula:
\[
D = \sqrt{x^2 + (\sqrt{x})^2} = \sqrt{x^2 + x}
\]

Now, differentiate both sides of this equation with respect to time \( t \), using the chain rule:
\[
\frac{dD}{dt} = \frac{1}{2\sqrt{x^2 + x}} \cdot \left(2x \frac{dx}{dt} + \frac{dx}{dt} \right)
\]

Simplify:
\[
\frac{dD}{dt} = \frac{1}{2\sqrt{x^2 + x}} \cdot \left( (2x + 1) \frac{dx}{dt} \right)
\]

### Step 3: Plug in the known values
At the point \( (4, 2) \), we know:
- \( x = 4 \)
- \( \frac{dx}{dt} = 7 \) cm/s (the rate at which the x-coordinate increases)

Substitute these values into the equation:
\[
\frac{dD}{dt} = \frac{1}{2\sqrt{4^2 + 4}} \cdot \left( (2(4) + 1) \cdot 7 \right)
\]

Simplify the terms inside the square root:
\[
\frac{dD}{dt} = \frac{1}{2\sqrt{16 + 4}} \cdot \left( (8 + 1) \cdot 7 \right) = \frac{1}{2\sqrt{20}} \cdot 63
\]

Since \( \sqrt{20} = 2\sqrt{5} \), we have:
\[
\frac{dD}{dt} = \frac{63}{2 \times 2\sqrt{5}} = \frac{63}{4\sqrt{5}} = \frac{63\sqrt{5}}{20}
\]

### Step 4: Calculate the numerical value
Using \( \sqrt{5} \approx 2.236 \):
\[
\frac{63\sqrt{5}}{20} \approx \frac{63 \times 2.236}{20} = \frac{140.868}{20} = 7.043 \, \text{cm/s}
\]

### Final Answer:
The rate at which the distance from the particle to the origin is changing at this instant is approximately \( \boxed{7.043} \) cm/s.

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Do you want further details or have any questions?

Here are five related questions you might want to explore:
1. What is the significance of the chain rule in implicit differentiation?
2. How does the curve \( y = \sqrt{x} \) affect the particle's movement?
3. How would the rate of change of distance vary if the x-coordinate increased at a different rate?
4. What happens if the particle's y-coordinate decreases instead of increases?
5. How does the distance formula change in three-dimensional space?

**Tip:** Always keep track of units when solving real-world rate of change problems; they help ensure accuracy in the final answer.

A particle is moving along the curve y = sqrt(x). As the particle passes through the point (4, 2), its x-coordinate increases at a rate of 7 cm/s. How fast is the distance from the particle to the origin changing at this instant? (Round your answer to three decimal places.)

Explore how to solve a related rates problem involving a particle moving along the curve y = sqrt(x). Given the x-coordinate increases at 7 cm/s, this problem uses calculus concepts like implicit differentiation and the chain rule to find the rate of distance change from the origin. Ideal for Grades 11-12 calculus students.

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