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 $$ \frac{3 \sqrt{37}}{73} \mathrm{~cm} / \mathrm{min} $$. (Type an exact answer, using radicals as needed.) b. Use the equation $$ y=x^{2} $$ to find an equation relating $$ \frac{d y}{d t} $$ to $$ \frac{d x}{d t} $$.

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To solve the problem, we will follow the steps outlined in the prompt.

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

1. **Extract known conditions:**
   - The bug is moving along the parabola $$ y = x^2 $$.
   - The distance from the origin is increasing at a rate of $$ \frac{dD}{dt} = 3 $$ cm/min.
   - The point of interest is $$ (6, 36) $$.

2. **Distance from the origin:**
   The distance $$ D $$ from the origin to the point $$ (x, y) $$ is given by:
   $$
   D = \sqrt{x^2 + y^2}
   $$
   Substituting $$ y = x^2 $$:
   $$
   D = \sqrt{x^2 + (x^2)^2} = \sqrt{x^2 + x^4} = \sqrt{x^2(1 + x^2)} = x\sqrt{1 + x^2}
   $$

3. **Differentiate with respect to time $$ t $$:**
   Using the product rule:
   $$
   \frac{dD}{dt} = \frac{d}{dt}(x\sqrt{1 + x^2}) = \frac{dx}{dt}\sqrt{1 + x^2} + x \cdot \frac{1}{2\sqrt{1 + x^2}} \cdot 2x \cdot \frac{dx}{dt}
   $$
   Simplifying:
   $$
   \frac{dD}{dt} = \frac{dx}{dt}\sqrt{1 + x^2} + \frac{x^2}{\sqrt{1 + x^2}} \cdot \frac{dx}{dt} = \frac{dx}{dt} \left( \sqrt{1 + x^2} + \frac{x^2}{\sqrt{1 + x^2}} \right)
   $$
   $$
   = \frac{dx}{dt} \cdot \frac{1 + x^2 + x^2}{\sqrt{1 + x^2}} = \frac{dx}{dt} \cdot \frac{1 + 2x^2}{\sqrt{1 + x^2}}
   $$

4. **Substituting known values:**
   At the point $$ (6, 36) $$:
   - $$ x = 6 $$
   - $$ y = 36 $$
   - $$ D = 3 $$ cm/min

We need to find $$ \frac{dx}{dt} $$:
   $$
   3 = \frac{dx}{dt} \cdot \frac{1 + 2(6^2)}{\sqrt{1 + 6^2}} = \frac{dx}{dt} \cdot \frac{1 + 72}{\sqrt{37}} = \frac{dx}{dt} \cdot \frac{73}{\sqrt{37}}
   $$

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

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

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

This equation relates the rate of change of $$ y $$ with respect to time to the rate of change of $$ x $$ with respect to time.

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

1. **Substituting known values:**
   At the point $$ (6, 36) $$:
   $$
   \frac{dy}{dt} = 2(6) \cdot \frac{dx}{dt} = 12 \cdot \frac{3\sqrt{37}}{73}
   $$

2. **Calculating $$ \frac{dy}{dt} $$:**
   $$
   \frac{dy}{dt} = \frac{36\sqrt{37}}{73}
   $$

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

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