Question 1185840
Here's how to solve this related rates problem:

**(a) Distance from Third Base:**

1. **Diagram:** Draw a square representing the baseball diamond. Label the bases 1st, 2nd, 3rd, and Home. Let *x* be Matthew's distance from first base, and *y* be his distance from third base.

2. **Relationship between x and y:** By the Pythagorean theorem, we have:

   ```
   x^2 + 90^2 = y^2
   ```

3. **Given Information:**
   * dx/dt = 3√3 ft/sec (Matthew's speed from 1st to 2nd)
   * x = 30 ft

4. **Find dy/dt:** We want to find how fast *y* is changing (dy/dt) when x = 30.

5. **Differentiate the equation:** Differentiate both sides of the equation with respect to time (t):

   ```
   2x(dx/dt) + 0 = 2y(dy/dt)
   ```

6. **Solve for y:** When x = 30:

   ```
   30^2 + 90^2 = y^2
   900 + 8100 = y^2
   y^2 = 9000
   y = 30√10 ft
   ```

7. **Substitute and solve for dy/dt:**

   ```
   2(30)(3√3) = 2(30√10)(dy/dt)
   180√3 = 60√10(dy/dt)
   dy/dt = (180√3) / (60√10)
   dy/dt = (3√3) / √10
   dy/dt = (3√30) / 10 ft/sec
   ```

   Since the distance from third base is *decreasing*, the rate is negative:

   ```
   dy/dt = -(3√30) / 10 ft/sec
   ```

**(b) Distance from Home Plate:**

1. **Let z be the distance from home plate:** When Matthew is at distance *x* from first base, the distance *z* from home plate can be found by the Pythagorean theorem: z^2 = 90^2 + (90-x)^2

2. **Find dz/dt:** We want to find how fast *z* is changing (dz/dt) when x = 30.

3. **Differentiate the equation with respect to t:**

   ```
   2z(dz/dt) = 0 + 2(90-x)(-dx/dt)
   z(dz/dt) = (90-x)(-dx/dt)
   ```

4. **Solve for z when x = 30:**

   ```
   z^2 = 90^2 + (90-30)^2
   z^2 = 8100 + 3600
   z^2 = 11700
   z = 30√13 ft
   ```

5. **Substitute and solve for dz/dt:**

   ```
   (30√13)(dz/dt) = (90-30)(-3√3)
   (30√13)(dz/dt) = (60)(-3√3)
   dz/dt = (-180√3) / (30√13)
   dz/dt = (-6√3) / √13
   dz/dt = (-6√39) / 13 ft/sec
   ```

Therefore, the distance from home plate is decreasing at a rate of (6√39)/13 ft/sec.

**Final Answers:**

* (a) The distance from the third base is decreasing at a rate of (3√30)/10 ft/sec.
* (b) The distance from home plate is changing at a rate of -(6√39)/13 ft/sec (decreasing).