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.