This problem involves related rates and trigonometry. Let's break it down step by step.

Problem Breakdown:

• The ladder is 5 meters long.
• The top of the ladder is sliding downward at a rate of $\frac{dy}{dt} = -10 \, \text{m/s}$.
• We need to find the rate at which the angle $\theta$ between the floor and the ladder is decreasing when the bottom of the ladder is 2 meters away from the wall.

Notation:

• Let $x$ be the distance from the bottom of the ladder to the wall.
• Let $y$ be the height from the top of the ladder to the ground.
• Let $\theta$ be the angle between the floor and the ladder.

We are given $\frac{dy}{dt} = -10 \, \text{m/s}$ and need to find $\frac{d\theta}{dt}$ when $x = 2$ meters.

Step 1: Relate $x$, $y$, and the ladder length

Since the ladder forms a right triangle with the wall and the floor, we can use the Pythagorean theorem: $x^2 + y^2 = 5^2$ $x^2 + y^2 = 25$

Step 2: Differentiate with respect to time

Differentiate both sides of the equation with respect to time $t$: $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$ $x \frac{dx}{dt} + y \frac{dy}{dt} = 0$ We are given $\frac{dy}{dt} = -10 \, \text{m/s}$, and we need to find $\frac{dx}{dt}$ when $x = 2$.

Step 3: Solve for $y$

Using the Pythagorean theorem at the moment when $x = 2$: $2^2 + y^2 = 25$ $4 + y^2 = 25$ $y^2 = 21 \quad \Rightarrow \quad y = \sqrt{21} \approx 4.58 \, \text{m}$

Step 4: Find $\frac{dx}{dt}$

Now, substitute the known values into the differentiated equation: $2 \times 2 \times \frac{dx}{dt} + 2 \times \sqrt{21} \times (-10) = 0$ $4 \frac{dx}{dt} - 20 \sqrt{21} = 0$ $\frac{dx}{dt} = \frac{20 \sqrt{21}}{4} \approx 10 \, \text{m/s}$

Step 5: Relate $\theta$ and $x$

We now use trigonometry to relate $\theta$ and $x$. From the triangle, we know: $\cos \theta = \frac{x}{5}$ Differentiate both sides with respect to time: $-\sin \theta \frac{d\theta}{dt} = \frac{1}{5} \frac{dx}{dt}$ Solve for $\frac{d\theta}{dt}$: $\frac{d\theta}{dt} = -\frac{1}{5} \frac{dx}{dt} \frac{1}{\sin \theta}$

Step 6: Find $\sin \theta$

We know that $\sin \theta = \frac{y}{5}$. From step 3, when $x = 2$, $y = \sqrt{21}$, so: $\sin \theta = \frac{\sqrt{21}}{5}$

Step 7: Final calculation

Now substitute the known values into the equation for $\frac{d\theta}{dt}$: $\frac{d\theta}{dt} = -\frac{1}{5} \times 10 \times \frac{1}{\frac{\sqrt{21}}{5}}$ $\frac{d\theta}{dt} = -\frac{10}{\sqrt{21}} \approx -2.5 \, \text{radians per second}$

Thus, the rate at which the angle between the ladder and the floor is decreasing is $-2.5 \, \text{radians per second}$, which matches the given answer.

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Would you like more details or have any questions?

Related Questions:

1. How would the rate change if the ladder were longer or shorter?
2. What happens to the rate if the ladder slides slower, say at $\frac{dy}{dt} = -5 \, \text{m/s}$?
3. How would you find the time when the ladder completely hits the ground?
4. What is the significance of the negative sign in the rate $\frac{d\theta}{dt}$?
5. Could you calculate the speed of the ladder's bottom end as it touches the wall?

Tip:

When solving related rates problems, always keep the variables related by a geometric or physical constraint equation like the Pythagorean theorem in this case.