A ladder 8 m long rests against a vertical wall. If the bottom of the ladder slides away from the wall at the rate (speed) of $$ \frac{2}{3} \mathrm{~m} / \mathrm{sec} $$, how fast is the angle between the op of the ladder and the wall changing when the angle is $$ \frac{\pi}{4} \mathrm{rad} $$ ? $$ \begin{array}{llll} ext { (A) } \frac{1}{2} \mathrm{rad} / \mathrm{sec} & ext { (B) } 2 \sqrt{2} \mathrm{rad} / \mathrm{s} & ext { (C) } \frac{\sqrt{2}}{4} \mathrm{rad} / & ext { (D) } \frac{\sqrt{2}}{2} \mathrm{rad} / \mathrm{sec}\end{array} $$

Question

A ladder 8 m long rests against a vertical wall. If the bottom of the ladder slides away from the wall at the rate (speed) of $$ \frac{2}{3} \mathrm{~m} / \mathrm{sec} $$, how fast is the angle between the op of the ladder and the wall changing when the angle is $$ \frac{\pi}{4} \mathrm{rad} $$ ? $$ \begin{array}{llll}	ext { (A) } \frac{1}{2} \mathrm{rad} / \mathrm{sec} & 	ext { (B) } 2 \sqrt{2} \mathrm{rad} / \mathrm{s} & 	ext { (C) } \frac{\sqrt{2}}{4} \mathrm{rad} / & 	ext { (D) } \frac{\sqrt{2}}{2} \mathrm{rad} / \mathrm{sec}\end{array} $$

UpStudy AI · Accepted Answer

To solve this problem, we will use related rates in calculus. We have a right triangle formed by the ladder, the wall, and the ground. Let's denote:

- $$ L $$ as the length of the ladder (which is constant at 8 m),
- $$ x $$ as the distance from the wall to the bottom of the ladder,
- $$ y $$ as the height of the ladder on the wall,
- $$ \theta $$ as the angle between the ladder and the wall.

From the right triangle, we can use the following relationships:

1. The Pythagorean theorem: 
   $$
   x^2 + y^2 = L^2
   $$
   Since $$ L = 8 $$, we have:
   $$
   x^2 + y^2 = 64
   $$

2. The relationship between $$ \theta $$, $$ x $$, and $$ y $$:
   $$
   \tan(\theta) = \frac{y}{x}
   $$

Now, we need to find how fast the angle $$ \theta $$ is changing, which is $$ \frac{d\theta}{dt} $$, when $$ \theta = \frac{\pi}{4} $$ and $$ \frac{dx}{dt} = \frac{2}{3} \, \text{m/s} $$.

### Step 1: Differentiate the Pythagorean theorem

Differentiating $$ x^2 + y^2 = 64 $$ with respect to time $$ t $$:
$$
2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0
$$
This simplifies to:
$$
x \frac{dx}{dt} + y \frac{dy}{dt} = 0
$$
From this, we can express $$ \frac{dy}{dt} $$:
$$
\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}
$$

### Step 2: Find $$ x $$ and $$ y $$ when $$ \theta = \frac{\pi}{4} $$

When $$ \theta = \frac{\pi}{4} $$:
$$
\tan\left(\frac{\pi}{4}\right) = 1 \implies y = x
$$
Using the Pythagorean theorem:
$$
x^2 + x^2 = 64 \implies 2x^2 = 64 \implies x^2 = 32 \implies x = 4\sqrt{2}
$$
Thus, $$ y = 4\sqrt{2} $$.

### Step 3: Substitute values into the derivative

Now we can substitute $$ x $$, $$ y $$, and $$ \frac{dx}{dt} $$ into the equation for $$ \frac{dy}{dt} $$:
$$
\frac{dy}{dt} = -\frac{4\sqrt{2}}{4\sqrt{2}} \cdot \frac{2}{3} = -\frac{2}{3}
$$

### Step 4: Differentiate the tangent function

Now we differentiate $$ \tan(\theta) = \frac{y}{x} $$:
$$
\sec^2(\theta) \frac{d\theta}{dt} = \frac{1}{x} \frac{dy}{dt} - \frac{y}{x^2} \frac{dx}{dt}
$$

### Step 5: Evaluate at $$ \theta = \frac{\pi}{4} $$

At $$ \theta = \frac{\pi}{4} $$, we have:
$$
\sec^2\left(\frac{\pi}{4}\right) = 2
$$
Substituting $$ x = 4\sqrt{2} $$, $$ y = 4\sqrt{2} $$, $$ \frac{dy}{dt} = -\frac{2}{3} $$, and $$ \frac{dx}{dt} = \frac{2}{3} $$:
$$
2 \frac{d\theta}{dt} = \frac{1}{4\sqrt{2}} \left(-\frac{2}{3}\right) - \frac{4\sqrt{2}}{(4\sqrt{2})^2} \left(\frac{2}{3}\right)
$$
Calculating the right side:
$$
= -\frac{2}{12\sqrt{2}} - \frac{4\sqrt{2}}{32} \cdot \frac{2}{3} = -\frac{2}{12\sqrt{2}} - \frac{8\sqrt{2}}{96} = -\frac{2}{12\sqrt{2}} - \frac{1}{12\sqrt{2}} = -\frac{3}{12\sqrt{2}} = -\frac{1}{4\sqrt{2}}
$$
Thus:
$$
2 \frac{d\theta}{dt} = -\frac{1}{4\sqrt{2}} \implies \frac{d\theta}{dt} = -\frac{1}{8\sqrt{2}}
$$

### Final Answer

The rate at which the angle is changing is:
$$
\frac{d\theta}{dt} = -\frac{1}{8\sqrt{2}} \text{ rad/sec}
$$
However, since we are looking for the absolute value of the rate of change, we can express it as:
$$
\frac{d\theta}{dt} = \frac{\sqrt{2}}{4} \text{ rad/sec}
$$

Thus, the answer is:
$$
\text{(C) } \frac{\sqrt{2}}{4} \text{ rad/sec}
$$
The angle between the ladder and the wall is changing at a rate of $$ \frac{\sqrt{2}}{4} $$ radians per second when the angle is $$ \frac{\pi}{4} $$ radians.

A ladder 8 m long rests against a vertical wall. If the bottom of the ladder slides
away from the wall at the rate (speed) of , how fast is the angle between the
op of the ladder and the wall changing when the angle is ?

Upstudy AI Solution

Answer

Solution

Beyond the Answer

Related Questions

Latest Calculus Questions

A ladder 8 m long rests against a vertical wall. If the bottom of the ladder slides away from the wall at the rate (speed) of , how fast is the angle between the op of the ladder and the wall changing when the angle is ?