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: $\frac{dy}{dt} = 2 \cdot 1 \cdot \frac{5 \sqrt{2}}{3} = \frac{10 \sqrt{2}}{3} \, \text{cm/min}$

Do you want details or have any questions about this solution?

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 $t$.