A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 0.7 ft/s, how fast (in rad/s) is the angle (in radians) between the ladder and the ground changing when the bottom of the ladder is 8 ft from the wall? (That is, find the angle's rate of change when the bottom of the ladder is 8 ft from the wall.)

Given:
- Ladder length \( L = 10 \) ft (constant)
- Rate at which the bottom of the ladder slides away from the wall \( \frac{dx}{dt} = 0.7 \) ft/s
- Distance of the bottom of the ladder from the wall at the specific moment \( x = 8 \) ft

We need to find:
- The rate at which the angle \( \theta \) between the ladder and the ground is changing, \( \frac{d\theta}{dt} \), in radians per second.

First, we apply the Pythagorean theorem to relate the height \( y \) of the ladder on the wall to the distance \( x \) from the wall:

\[
x^2 + y^2 = L^2
\]

Given \( L = 10 \) ft,

\[
8^2 + y^2 = 10^2 \implies 64 + y^2 = 100 \implies y^2 = 36 \implies y = 6 \text{ ft}
\]

To find the rate at which \( \theta \) is changing, we use trigonometric relationships. Specifically, \( \cos(\theta) \) is given by:

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

Taking the derivative with respect to time \( t \):

\[
\frac{d}{dt} [ \cos(\theta) ] = \frac{d}{dt} \left[ \frac{x}{L} \right]
\]

Since \( L \) is constant,

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

Solving for \( \frac{d\theta}{dt} \),

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

We already know \( L = 10 \) ft and \( \frac{dx}{dt} = 0.7 \) ft/s. To find \( \sin(\theta) \), use:

\[
\sin(\theta) = \frac{y}{L} = \frac{6}{10} = 0.6
\]

Substitute the values into the equation:

\[
\frac{d\theta}{dt} = -\frac{1}{10 \times 0.6} \times 0.7 = -\frac{0.7}{6} = -\frac{7}{60} \text{ rad/s}
\]

Therefore, the rate at which the angle is changing is:

\[
\frac{d\theta}{dt} = -\frac{7}{60} \text{ rad/s}
\]

This negative sign indicates that the angle is decreasing.

Would you like more details or have any questions?

Here are some related questions:
1. How is the Pythagorean theorem used to solve related rates problems?
2. What is the relationship between the rates of change of the sides of a right triangle and its angles?
3. How can trigonometric identities help in solving related rates problems?
4. What is the significance of the negative sign in the rate of change of the angle?
5. How does the rate of change of the distance of the ladder's bottom from the wall affect the rate of change of the angle?
6. How do you differentiate trigonometric functions with respect to time?
7. What are some other common real-life applications of related rates problems?
8. How does the length of the ladder influence the rate at which the angle changes?

**Tip:** Always draw a diagram when solving related rates problems to visualize the relationship between the different quantities and their rates of change.

Learn how to solve a related rates problem involving a ladder sliding against a wall. Calculate the rate at which the angle between the ladder and the ground changes using trigonometry and the Pythagorean theorem. Suitable for high school students and those interested in calculus and trigonometry applications.

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