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 $$ 3 \mathrm{~cm} / \mathrm{min} $$. a. At what rate is the $$ x $$-coordinate of the bug increasing when the bug is at the point $$ (6,36) $$ ? b. Use the equation $$ y=x^{2} $$ to find an equation relating $$ \frac{d y}{d t} $$ to $$ \frac{d x}{d t} $$. c. At what rate is the $$ y $$-coordinate of the bug increasing when the bug is at the point $$ (6,36) $$ ? a. At what rate is the $$ x $$-coordinate of the bug increasing when the bug is at the point $$ (6,36) $$ ? The $$ x $$-coordinate of the bug is increasing at a rate of (Type an exact answer, using radicals as needed.)

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### Known Conditions
1. The bug is moving along the parabola $$ y = x^2 $$.
2. The distance from the origin is increasing at a rate of $$ \frac{dD}{dt} = 3 \, \text{cm/min} $$.
3. We need to find the rates of change of the $$ x $$-coordinate and $$ y $$-coordinate when the bug is at the point $$ (6, 36) $$.

### Part a: Rate of Change of the $$ x $$-coordinate

1. **Distance from the origin**: The distance $$ D $$ from the origin to the point $$ (x, y) $$ is given by:
   $$
   D = \sqrt{x^2 + y^2}
   $$
   Since $$ y = x^2 $$, we can substitute this into the distance formula:
   $$
   D = \sqrt{x^2 + (x^2)^2} = \sqrt{x^2 + x^4} = \sqrt{x^2(1 + x^2)} = x\sqrt{1 + x^2}
   $$

2. **Differentiate with respect to time $$ t $$**:
   Using the chain rule, we differentiate $$ D $$:
   $$
   \frac{dD}{dt} = \frac{d}{dt}(x\sqrt{1 + x^2}) = \sqrt{1 + x^2} \frac{dx}{dt} + x \cdot \frac{1}{2\sqrt{1 + x^2}} \cdot 2x \frac{dx}{dt}
   $$
   Simplifying this gives:
   $$
   \frac{dD}{dt} = \sqrt{1 + x^2} \frac{dx}{dt} + \frac{x^2}{\sqrt{1 + x^2}} \frac{dx}{dt} = \left( \sqrt{1 + x^2} + \frac{x^2}{\sqrt{1 + x^2}} \right) \frac{dx}{dt}
   $$
   $$
   \frac{dD}{dt} = \frac{(1 + x^2) + x^2}{\sqrt{1 + x^2}} \frac{dx}{dt} = \frac{1 + 2x^2}{\sqrt{1 + x^2}} \frac{dx}{dt}
   $$

3. **Substituting known values**:
   At the point $$ (6, 36) $$, we have $$ x = 6 $$:
   $$
   \frac{dD}{dt} = 3 = \frac{1 + 2(6^2)}{\sqrt{1 + 6^2}} \frac{dx}{dt}
   $$
   Calculating $$ 6^2 = 36 $$:
   $$
   \frac{dD}{dt} = 3 = \frac{1 + 72}{\sqrt{1 + 36}} \frac{dx}{dt} = \frac{73}{\sqrt{37}} \frac{dx}{dt}
   $$

4. **Solving for $$ \frac{dx}{dt} $$**:
   $$
   3 = \frac{73}{\sqrt{37}} \frac{dx}{dt}
   $$
   $$
   \frac{dx}{dt} = \frac{3\sqrt{37}}{73}
   $$

### Part b: Relating $$ \frac{dy}{dt} $$ to $$ \frac{dx}{dt} $$

1. **Differentiate $$ y = x^2 $$** with respect to $$ t $$:
   $$
   \frac{dy}{dt} = 2x \frac{dx}{dt}
   $$

### Part c: Rate of Change of the $$ y $$-coordinate

1. **Substituting $$ x = 6 $$** into the equation from part b:
   $$
   \frac{dy}{dt} = 2(6) \frac{dx}{dt} = 12 \frac{dx}{dt}
   $$
   Now substituting $$ \frac{dx}{dt} = \frac{3\sqrt{37}}{73} $$:
   $$
   \frac{dy}{dt} = 12 \cdot \frac{3\sqrt{37}}{73} = \frac{36\sqrt{37}}{73}
   $$

### Final Answers
- a. The $$ x $$-coordinate of the bug is increasing at a rate of $$ \frac{3\sqrt{37}}{73} \, \text{cm/min} $$.
- b. The equation relating $$ \frac{dy}{dt} $$ to $$ \frac{dx}{dt} $$ is $$ \frac{dy}{dt} = 2x \frac{dx}{dt} $$.
- c. The $$ y $$-coordinate of the bug is increasing at a rate of $$ \frac{36\sqrt{37}}{73} \, \text{cm/min} $$.
- a. The $$ x $$-coordinate is increasing at $$ \frac{3\sqrt{37}}{73} \, \text{cm/min} $$. - b. $$ \frac{dy}{dt} = 2x \frac{dx}{dt} $$. - c. The $$ y $$-coordinate is increasing at $$ \frac{36\sqrt{37}}{73} \, \text{cm/min} $$.

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