Question

A baseball diamond is a square with each side 90 feet in length. A player take an aggressive turn around second and realizes he must return. Therefore, he is running from third base back to second base at a rate of 10 f/s. At what rate is angle a decreasing when D is 18 feet? Round your answer to 3 decimal places or answer as a reduced fraction.

Answer 1

Rounded to three decimal places,
`\( \frac{{da}}{{dt}} \approx 0.147 \)`. Therefore, the rate at which angle a is decreasing when D is 18 feet is approximately 0.147 radians per second.



To solve this problem, we can use the related rates formula, which relates the rates of change of different variables. In this case, we'll use the formula for the tangent of an angle in a right triangle:

`\[ \tan(\theta) = \frac{{\text{opposite side}}}{{\text{adjacent side}}} \]`

Now, let's define the variables:

- Let D be the distance from the player to the base (which is changing over time).

- Let x be the distance along the baseline from third base to the player.

- Let a be the angle between the baseline and the line from the player to third base.

We know that x and D are related by the Pythagorean theorem:

`\[ x^2 + D^2 = 90^2 \]`

Differentiate both sides of this equation with respect to time t :

`\[ 2x \frac{{dx}}{{dt}} + 2D \frac{{dD}}{{dt}} = 0 \]`

Now, we want to find the rate at which the angle a is decreasing, which is
`\( \frac{{da}}{{dt}} \)`. We can use the tangent of a in terms of x and D :

`\[ \tan(a) = (D)/(x) \]`

Differentiate both sides of this equation with respect to time t :

`\[ \sec^2(a) \frac{{da}}{{dt}} = \frac{{\frac{{dx}}{{dt}}x - \frac{{dD}}{{dt}}D}}{{x^2}} \]`

Now, we can substitute
`\( \frac{{dx}}{{dt}} \)` and
`\( \frac{{dD}}{{dt}} \)` using the relations we found earlier. We know
`\( \frac{{dx}}{{dt}} = -\frac{D}{{x}} \)`, and
`\( x^2 + D^2 = 90^2 \)`.

After substituting and simplifying, we get:

`\[ \frac{{da}}{{dt}} = -\frac{{D^2}}{{x^2}} \cdot \frac{{dx}}{{dt}} \]`

Now, plug in the values D = 18 feet, x can be found from the Pythagorean theorem, and
`\( \frac{{dx}}{{dt}} = -\frac{D}{{x}} \)`:

`\[ \frac{{da}}{{dt}} = -\frac{{18^2}}{{x^2}} \cdot \left(-\frac{D}{{x}}\right) \]\[ \frac{{da}}{{dt}} = \frac{{18^2}}{{x^3}} \cdot D \]`

Now, find x using the Pythagorean theorem:

`\[ x^2 + D^2 = 90^2 \]\[ x^2 = 90^2 - D^2 \]\[ x = √(90^2 - D^2) \]`

Now, plug in the values and calculate
`\( \frac{{da}}{{dt}} \)`:

`\[ \frac{{da}}{{dt}} = \frac{{18^2}}{{\left(√(90^2 - 18^2)\right)^3}} \cdot 18 \]`

Let's calculate the numerical value for
`\( \frac{{da}}{{dt}} \)`:

`\[ x = √(90^2 - 18^2) \]\[ x = √(8100 - 324) \]\[ x = √(7776) \]\[ x = 88.132 \]`

Now, plug this value into the expression for
`\( \frac{{da}}{{dt}} \)`:

`\frac{{da}}{{dt}} = \frac{{18^2}}{{(88.132)^3}} \cdot 18 \]\[ \frac{{da}}{{dt}} = \frac{{324}}{{(88.132)^3}} \cdot 18 \]\[ \frac{{da}}{{dt}} \approx \frac{{324}}{{693,711.193}} \cdot 18 \]\[ \frac{{da}}{{dt}} \approx 0.008172 \cdot 18 \]\[ \frac{{da}}{{dt}} \approx 0.147 \]`

Rounded to three decimal places,
`\( \frac{{da}}{{dt}} \approx 0.147 \)`. Therefore, the rate at which angle a is decreasing when D is 18 feet is approximately 0.147 radians per second.