18 foot ladder is leaning against a house while the base is pulled away at a constant rate of 1.5 ft/s. At what rate is the top of the ladder sliding down the side of the house when the base is 5 feet from the house?

This is a related rates problem where the Pythagorean theorem plays a key role. To solve this, we let:

- \( x \) be the distance from the base of the ladder to the house,
- \( y \) be the height of the top of the ladder from the ground,
- \( L = 18 \) feet be the length of the ladder (constant),
- \( \frac{dx}{dt} = 1.5 \, \text{ft/s} \) is the rate at which the base is moving away from the house.

We are tasked with finding \( \frac{dy}{dt} \), the rate at which the top of the ladder is sliding down the house, when the base of the ladder is 5 feet from the house (i.e., \( x = 5 \)).

### Step 1: Relating \( x \), \( y \), and \( L \)
By the Pythagorean theorem, we know:
\[
x^2 + y^2 = L^2
\]
Since \( L = 18 \), this becomes:
\[
x^2 + y^2 = 18^2 = 324
\]

### Step 2: Differentiate with respect to time \( t \)
We implicitly differentiate both sides of the equation \( x^2 + y^2 = 324 \) with respect to \( t \):
\[
2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0
\]
Simplifying:
\[
x \frac{dx}{dt} + y \frac{dy}{dt} = 0
\]

### Step 3: Solving for \( \frac{dy}{dt} \)
Rearranging the equation to solve for \( \frac{dy}{dt} \):
\[
y \frac{dy}{dt} = -x \frac{dx}{dt}
\]
\[
\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}
\]

### Step 4: Finding \( y \) when \( x = 5 \)
From the Pythagorean theorem:
\[
5^2 + y^2 = 324
\]
\[
25 + y^2 = 324
\]
\[
y^2 = 299
\]
\[
y = \sqrt{299} \approx 17.3 \, \text{feet}
\]

### Step 5: Substituting values
Now, substitute \( x = 5 \), \( y = \sqrt{299} \approx 17.3 \), and \( \frac{dx}{dt} = 1.5 \) ft/s into the equation for \( \frac{dy}{dt} \):
\[
\frac{dy}{dt} = -\frac{5}{\sqrt{299}} \cdot 1.5
\]
\[
\frac{dy}{dt} \approx -\frac{5 \cdot 1.5}{17.3} \approx -\frac{7.5}{17.3} \approx -0.43 \, \text{ft/s}
\]

Thus, the top of the ladder is sliding down the side of the house at approximately \( 0.43 \, \text{ft/s} \) when the base is 5 feet away.

Would you like more details or have any questions?

Here are 5 related questions to expand on this:

1. What if the base was moving at 2 ft/s instead of 1.5 ft/s?
2. How does the rate of change of the top's descent vary as the base moves farther from the house?
3. What happens when the base of the ladder reaches 18 feet from the house?
4. Can we explore how the rate changes if the ladder is shorter or longer?
5. How does the rate of descent change when the base of the ladder is very close to the house?

**Tip:** When working with related rates, always start by identifying the variables and finding an equation that relates them before differentiating.

Solve a related rates problem where an 18-foot ladder is leaning against a house, and the base is pulled away at 1.5 ft/s. Learn how to calculate the rate at which the top of the ladder slides down the wall using differentiation and the Pythagorean theorem. Suitable for Grades 11-12.

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