A ladder is 15 feet long and leaning against a wall. The base of the ladder is being pulled away from the wall at a rate of 2 feet per second. If the top of the ladder is sliding down the wall at a rate of 1 foot per second, at what rate is the angle between the ladder and the wall changing when the base of the ladder is 5 feet away from the wall?

To find the rate at which the angle between the ladder and the wall is changing, we can use related rates and trigonometry. Let's define the variables and set up the problem:

- Let \( L \) be the length of the ladder, which is constant at 15 feet.
- Let \( x \) be the distance from the wall to the base of the ladder.
- Let \( y \) be the height of the ladder's top from the ground.
- Let \( \theta \) be the angle between the ladder and the wall.

We know:
- The ladder's length: \( L = 15 \) feet (constant)
- The rate at which the base is moving away from the wall: \( \frac{dx}{dt} = 2 \) feet/second
- The rate at which the top of the ladder is sliding down the wall: \( \frac{dy}{dt} = -1 \) foot/second
- The distance of the base from the wall at the specific moment: \( x = 5 \) feet

We need to find: \( \frac{d\theta}{dt} \) when \( x = 5 \).

### Solution

1. **Relate the sides and the angle using trigonometry:**

   Since the ladder makes a right triangle with the wall and the ground, we have:

   \[
   \cos \theta = \frac{x}{L}
   \]

   Differentiating both sides with respect to \( t \):

   \[
   -\sin \theta \cdot \frac{d\theta}{dt} = \frac{1}{L} \cdot \frac{dx}{dt}
   \]

   Rearrange to solve for \( \frac{d\theta}{dt} \):

   \[
   \frac{d\theta}{dt} = -\frac{\frac{dx}{dt}}{L \sin \theta}
   \]

2. **Find \( \sin \theta \) when \( x = 5 \):**

   We use the Pythagorean theorem to find \( y \):

   \[
   y = \sqrt{L^2 - x^2} = \sqrt{15^2 - 5^2} = \sqrt{225 - 25} = \sqrt{200} = 10\sqrt{2}
   \]

   Now, find \( \sin \theta \):

   \[
   \sin \theta = \frac{y}{L} = \frac{10\sqrt{2}}{15} = \frac{2\sqrt{2}}{3}
   \]

3. **Substitute the known values into the differentiated equation:**

   \[
   \frac{d\theta}{dt} = -\frac{\frac{dx}{dt}}{L \sin \theta} = -\frac{2}{15 \cdot \frac{2\sqrt{2}}{3}}
   \]

   \[
   \frac{d\theta}{dt} = -\frac{2}{15 \cdot \frac{2\sqrt{2}}{3}} = -\frac{2 \cdot 3}{15 \cdot 2\sqrt{2}} = -\frac{6}{30\sqrt{2}}
   \]

   \[
   \frac{d\theta}{dt} = -\frac{1}{5\sqrt{2}}
   \]

   \[
   \frac{d\theta}{dt} = -\frac{\sqrt{2}}{10} \text{ radians/second}
   \]

Therefore, the angle between the ladder and the wall is decreasing at a rate of \( \frac{\sqrt{2}}{10} \) radians per second when the base of the ladder is 5 feet away from the wall.

Would you like a detailed explanation on any part of the solution or any additional questions? Here are some related questions you might find interesting:

1. How does the rate of change of the angle vary with the distance of the base from the wall?
2. What would be the rate of change of the angle if the ladder were shorter or longer?
3. How does the rate at which the top slides down change as the base moves away?
4. What happens to the rate of change of the angle if the ladder length changes?
5. How would the rates change if the ladder were on a slope instead of a flat surface?
6. What if the base was moving at a different rate, like 3 feet per second?
7. How can you calculate the time it takes for the ladder to hit the ground?
8. How does the speed of the ladder's top depend on the angle with the wall?

**Tip:** When dealing with related rates, always differentiate with respect to time and carefully keep track of all changing variables and their rates.

Explore how to calculate the rate at which the angle between a 15-foot ladder and a wall changes as the ladder's base moves away at 2 feet per second and the top slides down at 1 foot per second. Learn about trigonometry and related rates with a step-by-step solution.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation