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Let's go through each question one by one.

### Given Information:
- The bug moves along the parabola \( y = x^2 \).
- The rate of increase in the distance from the origin (\( D \)) is \( \frac{dD}{dt} = 5 \, \text{cm/min} \).
- We are asked about the rates of change in the \( x \)- and \( y \)-coordinates when the bug is at the point \( (1, 1) \).

### Part (a)
We need to find the rate of change of the \( x \)-coordinate (\( \frac{dx}{dt} \)) when the bug is at \( (1, 1) \).

1. **Distance from the origin** \( D \) is given by:
   \[
   D = \sqrt{x^2 + y^2}
   \]
2. Since \( y = x^2 \), substitute \( y = x^2 \) into the distance formula:
   \[
   D = \sqrt{x^2 + (x^2)^2} = \sqrt{x^2 + x^4} = x \sqrt{1 + x^2}
   \]
3. Differentiate \( D \) with respect to \( t \) to find the relation between \( \frac{dD}{dt} \) and \( \frac{dx}{dt} \):
   \[
   \frac{dD}{dt} = \frac{d}{dt} \left( x \sqrt{1 + x^2} \right)
   \]
4. Apply the product rule:
   \[
   \frac{dD}{dt} = \frac{dx}{dt} \cdot \sqrt{1 + x^2} + x \cdot \frac{d}{dt} \left( \sqrt{1 + x^2} \right)
   \]
5. To differentiate \( \sqrt{1 + x^2} \):
   \[
   \frac{d}{dt} \left( \sqrt{1 + x^2} \right) = \frac{1}{2\sqrt{1 + x^2}} \cdot 2x \cdot \frac{dx}{dt} = \frac{x \frac{dx}{dt}}{\sqrt{1 + x^2}}
   \]
6. Substitute back:
   \[
   \frac{dD}{dt} = \frac{dx}{dt} \sqrt{1 + x^2} + x \cdot \frac{x \frac{dx}{dt}}{\sqrt{1 + x^2}}
   \]
7. Simplify:
   \[
   \frac{dD}{dt} = \frac{dx}{dt} \left( \sqrt{1 + x^2} + \frac{x^2}{\sqrt{1 + x^2}} \right)
   \]
   \[
   \frac{dD}{dt} = \frac{dx}{dt} \cdot \frac{1 + 2x^2}{\sqrt{1 + x^2}}
   \]
8. Solve for \( \frac{dx}{dt} \):
   \[
   \frac{dx}{dt} = \frac{\frac{dD}{dt} \cdot \sqrt{1 + x^2}}{1 + 2x^2}
   \]
9. Substitute \( x = 1 \), \( \frac{dD}{dt} = 5 \):
   \[
   \frac{dx}{dt} = \frac{5 \cdot \sqrt{1 + 1^2}}{1 + 2 \cdot 1^2} = \frac{5 \cdot \sqrt{2}}{3}
   \]

### Part (b)
Use the equation \( y = x^2 \) to relate \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \):
1. Differentiate both sides of \( y = x^2 \) with respect to \( t \):
   \[
   \frac{dy}{dt} = 2x \frac{dx}{dt}
   \]

### Part (c)
To find \( \frac{dy}{dt} \) at \( (1,1) \):
1. Substitute \( x = 1 \) and \( \frac{dx}{dt} = \frac{5 \sqrt{2}}{3} \) into the equation:
   \[
   \frac{dy}{dt} = 2 \cdot 1 \cdot \frac{5 \sqrt{2}}{3} = \frac{10 \sqrt{2}}{3} \, \text{cm/min}
   \]

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#### Follow-up Questions:
1. How would the calculations change if the bug's speed was given in terms of its \( y \)-coordinate?
2. What if the parabola had a different shape, such as \( y = x^3 \)? How would that affect \( \frac{dy}{dt} \)?
3. How does the rate of change in distance relate to the rates of change in both \( x \) and \( y \) coordinates?
4. Can you derive the rate of change formula if the bug moved along a circle instead?
5. How would you interpret these rates of change in a real-world context?

#### Tip:
When dealing with related rates, always ensure each variable’s derivative is expressed with respect to the same independent variable, typically time \( t \).

A bug is moving along the right side of the parabola y = x^2 at a rate such that its distance from the origin is increasing at 5 cm/min.

a. At what rate is the x-coordinate of the bug increasing when the bug is at the point (1, 1)?
b. Use the equation y = x^2 to find an equation relating dy/dt to dx/dt.
c. At what rate is the y-coordinate of the bug increasing when the bug is at the point (1, 1)?

Learn how to determine the rates of change in x- and y-coordinates for a bug moving along y = x^2, with a constant increase in distance from the origin. Explore related rates, derivatives, and the chain rule in this calculus problem suitable for Grades 11-12 or college-level calculus.

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