Question
A 15-ft ladder leans against a wall, forming a right triangle. If the ladder's base is 9 ft away from the wall, how fast is the top of the ladder sliding down the wall when the base is moving away from the wall at a rate of 0.25 ft/sec?
Answers
GPT-4o mini
To solve this problem, we can use related rates in the context of a right triangle formed by the wall, the ground, and the ladder. Let's set up our variables:
- Let \( x \) be the distance from the base of the ladder to the wall (in feet). We are given that \( \frac{dx}{dt} = 0.25 \) ft/sec (the base of the ladder is moving away from the wall).
- Let \( y \) be the height of the top of the ladder on the wall (in feet).
- The length of the ladder is constant at 15 ft. Therefore, we have the Pythagorean theorem:
\[
x^2 + y^2 = 15^2 = 225.
\]
We need to find how fast \( y \) is changing with respect to time (\( \frac{dy}{dt} \)) when \( x = 9 \) ft.
1. **Differentiate the Pythagorean theorem with respect to time \( t \)**:
\[
2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0.
\]
Dividing the entire equation by 2, we can simplify it to:
\[
x \frac{dx}{dt} + y \frac{dy}{dt} = 0.
\]
2. **Isolate \( \frac{dy}{dt} \)**:
\[
y \frac{dy}{dt} = -x \frac{dx}{dt} \quad \Rightarrow \quad \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}.
\]
3. **Find \( y \) when \( x = 9 \)**:
Using the Pythagorean theorem:
\[
x^2 + y^2 = 225 \quad \Rightarrow \quad 9^2 + y^2 = 225 \quad \Rightarrow \quad 81 + y^2 = 225.
\]
Solving for \( y^2 \):
\[
y^2 = 225 - 81 = 144 \quad \Rightarrow \quad y = 12 \text{ ft}.
\]
4. **Substituting known values into the equation for \( \frac{dy}{dt} \)**:
We have \( x = 9 \), \( y = 12 \), and \( \frac{dx}{dt} = 0.25 \) ft/sec. Now, substitute these values:
\[
\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} = -\frac{9}{12} \cdot 0.25 = -\frac{3}{4} \cdot 0.25 = -\frac{3 \cdot 0.25}{4} = -\frac{0.75}{4} = -0.1875 \text{ ft/sec}.
\]
Thus, the top of the ladder is sliding down the wall at a rate of approximately \(\boxed{-0.1875}\) ft/sec when the base is moving away from the wall at a rate of 0.25 ft/sec. The negative sign indicates that the height \( y \) is decreasing, which is expected as the ladder slides down.
- Let \( x \) be the distance from the base of the ladder to the wall (in feet). We are given that \( \frac{dx}{dt} = 0.25 \) ft/sec (the base of the ladder is moving away from the wall).
- Let \( y \) be the height of the top of the ladder on the wall (in feet).
- The length of the ladder is constant at 15 ft. Therefore, we have the Pythagorean theorem:
\[
x^2 + y^2 = 15^2 = 225.
\]
We need to find how fast \( y \) is changing with respect to time (\( \frac{dy}{dt} \)) when \( x = 9 \) ft.
1. **Differentiate the Pythagorean theorem with respect to time \( t \)**:
\[
2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0.
\]
Dividing the entire equation by 2, we can simplify it to:
\[
x \frac{dx}{dt} + y \frac{dy}{dt} = 0.
\]
2. **Isolate \( \frac{dy}{dt} \)**:
\[
y \frac{dy}{dt} = -x \frac{dx}{dt} \quad \Rightarrow \quad \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}.
\]
3. **Find \( y \) when \( x = 9 \)**:
Using the Pythagorean theorem:
\[
x^2 + y^2 = 225 \quad \Rightarrow \quad 9^2 + y^2 = 225 \quad \Rightarrow \quad 81 + y^2 = 225.
\]
Solving for \( y^2 \):
\[
y^2 = 225 - 81 = 144 \quad \Rightarrow \quad y = 12 \text{ ft}.
\]
4. **Substituting known values into the equation for \( \frac{dy}{dt} \)**:
We have \( x = 9 \), \( y = 12 \), and \( \frac{dx}{dt} = 0.25 \) ft/sec. Now, substitute these values:
\[
\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} = -\frac{9}{12} \cdot 0.25 = -\frac{3}{4} \cdot 0.25 = -\frac{3 \cdot 0.25}{4} = -\frac{0.75}{4} = -0.1875 \text{ ft/sec}.
\]
Thus, the top of the ladder is sliding down the wall at a rate of approximately \(\boxed{-0.1875}\) ft/sec when the base is moving away from the wall at a rate of 0.25 ft/sec. The negative sign indicates that the height \( y \) is decreasing, which is expected as the ladder slides down.