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Question 1185840: A baseball diamond is 90 ft in a side (it’s a square). Matthew runs from the first base to the second base at the rate of 3sqrt3 ft/sec. (a) How fast is his distance from the third base decreasing when he is 30 ft. from the first base? (b) At this instant, how fast is his distance from the home plate changing?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! 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).
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