Question
A 20ft long ladder is leaning against a wall. The bottom of the ladder is sliding away from the wall a a rate of 2.5 ft/sec. a. How fast is the top of the ladder sliding down the wall when x=12ft? b. How fast is the angle θ changing when x=12ft 2 c. How fast is the area of the triangle changing when x=12ft?
Answers
GPT-4o mini
To solve the problem, we’ll use related rates and the Pythagorean theorem.
Let:
- \( L \) = length of the ladder = 20 ft
- \( x \) = distance from the wall to the bottom of the ladder
- \( y \) = height of the top of the ladder on the wall
- \( \frac{dx}{dt} \) = rate at which the bottom slides away from the wall = 2.5 ft/sec
- \( \frac{dy}{dt} \) = rate at which the top slides down the wall
- \( \theta \) = angle between the ladder and the ground
- Area \( A \) of the triangle formed by the wall, ground, and the ladder
### a. How fast is the top of the ladder sliding down the wall when \( x = 12 \) ft?
**Using the Pythagorean theorem:**
\[
x^2 + y^2 = L^2
\]
Substituting \( L = 20 \) ft:
\[
x^2 + y^2 = 20^2
\]
\[
x^2 + y^2 = 400
\]
Now differentiate implicitly with respect to time \( t \):
\[
2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0
\]
\[
x\frac{dx}{dt} + y\frac{dy}{dt} = 0
\]
Solving for \( \frac{dy}{dt} \):
\[
y\frac{dy}{dt} = -x\frac{dx}{dt}
\]
\[
\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}
\]
**Now we need to find \( y \) when \( x = 12 \) ft using the Pythagorean theorem:**
\[
12^2 + y^2 = 400
\]
\[
144 + y^2 = 400
\]
\[
y^2 = 256 \implies y = 16 \text{ ft}
\]
Now substituting the values into the rate equation:
\[
\frac{dy}{dt} = -\frac{12}{16}(2.5) = -\frac{12}{16} \cdot 2.5 = -\frac{30}{16} = -1.875 \text{ ft/sec}
\]
Thus, the top of the ladder is sliding down the wall at \( 1.875 \) ft/sec.
### b. How fast is the angle \( \theta \) changing when \( x = 12 \) ft?
Using the relationship involving the tangent of the angle:
\[
\tan(\theta) = \frac{y}{x}
\]
Differentiating both sides with respect to \( t \):
\[
\sec^2(\theta) \frac{d\theta}{dt} = \frac{x\frac{dy}{dt} - y\frac{dx}{dt}}{x^2}
\]
At \( x = 12 \) ft and \( y = 16 \) ft:
\[
\tan(\theta) = \frac{16}{12} \implies \theta = \tan^{-1}\left(\frac{4}{3}\right)
\]
To find \( \sec^2(\theta) \):
\[
\sec^2(\theta) = 1 + \tan^2(\theta) = 1 + \left(\frac{16}{12}\right)^2 = 1 + \frac{256}{144} = \frac{400}{144} = \frac{25}{9}
\]
Now substitute:
\[
\frac{dy}{dt} = -1.875 \text{ ft/sec}, \, \frac{dx}{dt} = 2.5 \text{ ft/sec}
\]
\[
\frac{d\theta}{dt} = \frac{12(-1.875) - 16(2.5)}{12^2} \cdot \frac{9}{25}
\]
Calculating:
\[
= \frac{-22.5 - 40}{144} \cdot \frac{9}{25} = \frac{-62.5}{144} \cdot \frac{9}{25} = \frac{-562.5}{3600} = -0.15625 \text{ radians/sec}
\]
### c. How fast is the area of the triangle changing when \( x = 12 \) ft?
The area \( A \) of the triangle formed by the ladder, wall, and ground is given by:
\[
A = \frac{1}{2}xy
\]
Differentiating with respect to \( t \):
\[
\frac{dA}{dt} = \frac{1}{2}\left(x\frac{dy}{dt} + y\frac{dx}{dt}\right)
\]
Substituting \( x = 12 \), \( y = 16 \), \( \frac{dy}{dt} = -1.875 \), and \( \frac{dx}{dt} = 2.5 \):
\[
\frac{dA}{dt} = \frac{1}{2}\left(12(-1.875) + 16(2.5)\right)
\]
Calculating:
\[
= \frac{1}{2}\left(-22.5 + 40\right) = \frac{1}{2}\left(17.5\right) = 8.75
\]
Thus, the area of the triangle is increasing at a rate of \( 8.75 \) ft²/sec when \( x = 12 \) ft.
### Summary of Answers:
- **a.** \(\frac{dy}{dt} = -1.875\) ft/sec
- **b.** \(\frac{d\theta}{dt} \approx -0.15625\) radians/sec
- **c.** \(\frac{dA}{dt} = 8.75\) ft²/sec
Let:
- \( L \) = length of the ladder = 20 ft
- \( x \) = distance from the wall to the bottom of the ladder
- \( y \) = height of the top of the ladder on the wall
- \( \frac{dx}{dt} \) = rate at which the bottom slides away from the wall = 2.5 ft/sec
- \( \frac{dy}{dt} \) = rate at which the top slides down the wall
- \( \theta \) = angle between the ladder and the ground
- Area \( A \) of the triangle formed by the wall, ground, and the ladder
### a. How fast is the top of the ladder sliding down the wall when \( x = 12 \) ft?
**Using the Pythagorean theorem:**
\[
x^2 + y^2 = L^2
\]
Substituting \( L = 20 \) ft:
\[
x^2 + y^2 = 20^2
\]
\[
x^2 + y^2 = 400
\]
Now differentiate implicitly with respect to time \( t \):
\[
2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0
\]
\[
x\frac{dx}{dt} + y\frac{dy}{dt} = 0
\]
Solving for \( \frac{dy}{dt} \):
\[
y\frac{dy}{dt} = -x\frac{dx}{dt}
\]
\[
\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}
\]
**Now we need to find \( y \) when \( x = 12 \) ft using the Pythagorean theorem:**
\[
12^2 + y^2 = 400
\]
\[
144 + y^2 = 400
\]
\[
y^2 = 256 \implies y = 16 \text{ ft}
\]
Now substituting the values into the rate equation:
\[
\frac{dy}{dt} = -\frac{12}{16}(2.5) = -\frac{12}{16} \cdot 2.5 = -\frac{30}{16} = -1.875 \text{ ft/sec}
\]
Thus, the top of the ladder is sliding down the wall at \( 1.875 \) ft/sec.
### b. How fast is the angle \( \theta \) changing when \( x = 12 \) ft?
Using the relationship involving the tangent of the angle:
\[
\tan(\theta) = \frac{y}{x}
\]
Differentiating both sides with respect to \( t \):
\[
\sec^2(\theta) \frac{d\theta}{dt} = \frac{x\frac{dy}{dt} - y\frac{dx}{dt}}{x^2}
\]
At \( x = 12 \) ft and \( y = 16 \) ft:
\[
\tan(\theta) = \frac{16}{12} \implies \theta = \tan^{-1}\left(\frac{4}{3}\right)
\]
To find \( \sec^2(\theta) \):
\[
\sec^2(\theta) = 1 + \tan^2(\theta) = 1 + \left(\frac{16}{12}\right)^2 = 1 + \frac{256}{144} = \frac{400}{144} = \frac{25}{9}
\]
Now substitute:
\[
\frac{dy}{dt} = -1.875 \text{ ft/sec}, \, \frac{dx}{dt} = 2.5 \text{ ft/sec}
\]
\[
\frac{d\theta}{dt} = \frac{12(-1.875) - 16(2.5)}{12^2} \cdot \frac{9}{25}
\]
Calculating:
\[
= \frac{-22.5 - 40}{144} \cdot \frac{9}{25} = \frac{-62.5}{144} \cdot \frac{9}{25} = \frac{-562.5}{3600} = -0.15625 \text{ radians/sec}
\]
### c. How fast is the area of the triangle changing when \( x = 12 \) ft?
The area \( A \) of the triangle formed by the ladder, wall, and ground is given by:
\[
A = \frac{1}{2}xy
\]
Differentiating with respect to \( t \):
\[
\frac{dA}{dt} = \frac{1}{2}\left(x\frac{dy}{dt} + y\frac{dx}{dt}\right)
\]
Substituting \( x = 12 \), \( y = 16 \), \( \frac{dy}{dt} = -1.875 \), and \( \frac{dx}{dt} = 2.5 \):
\[
\frac{dA}{dt} = \frac{1}{2}\left(12(-1.875) + 16(2.5)\right)
\]
Calculating:
\[
= \frac{1}{2}\left(-22.5 + 40\right) = \frac{1}{2}\left(17.5\right) = 8.75
\]
Thus, the area of the triangle is increasing at a rate of \( 8.75 \) ft²/sec when \( x = 12 \) ft.
### Summary of Answers:
- **a.** \(\frac{dy}{dt} = -1.875\) ft/sec
- **b.** \(\frac{d\theta}{dt} \approx -0.15625\) radians/sec
- **c.** \(\frac{dA}{dt} = 8.75\) ft²/sec