Question

additional details were added for this assignment.
a ladder 18 feet long leans up against a house.
the bottom of the ladder starts to slip away from the house at 0.13 feet per second.
how fast is the tip of the ladder along the side of the house slipping when the ladder is 6.2 feet away from the house?
blank \\(\frac{ft}{sec}\\)
consider the angle the bottom of the ladder makes with the ground.
how fast is the angle changing (in radians) when the ladder is 6.2 feet away from the house?
blank
question help: video message instructor

Explanation:

Step1: Define Variables

Let \( x \) be the distance from the bottom of the ladder to the house, \( y \) be the height of the tip of the ladder on the house, and \( L = 18 \) ft (length of the ladder). By Pythagoras, \( x^2 + y^2 = L^2 = 324 \).

Step2: Differentiate with Respect to Time

Differentiate both sides with respect to \( t \): \( 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \). Simplify: \( x\frac{dx}{dt} + y\frac{dy}{dt} = 0 \). We need to find \( \frac{dy}{dt} \) when \( x = 6.2 \) ft and \( \frac{dx}{dt} = 0.13 \) ft/sec.

Step3: Find \( y \) when \( x = 6.2 \)

Using \( x^2 + y^2 = 324 \), substitute \( x = 6.2 \): \( y = \sqrt{324 - 6.2^2} = \sqrt{324 - 38.44} = \sqrt{285.56} \approx 16.898 \) ft.

Step4: Solve for \( \frac{dy}{dt} \)

Substitute \( x = 6.2 \), \( y \approx 16.898 \), \( \frac{dx}{dt} = 0.13 \) into \( x\frac{dx}{dt} + y\frac{dy}{dt} = 0 \):
\( 6.2(0.13) + 16.898\frac{dy}{dt} = 0 \)
\( 0.806 + 16.898\frac{dy}{dt} = 0 \)
\( \frac{dy}{dt} = -\frac{0.806}{16.898} \approx -0.0477 \) ft/sec. The negative sign indicates the tip is slipping down.

For the Angle (Let \( \theta \) be the angle with the ground):

Step1: Relate \( \theta \) and \( x \)

\( \cos\theta = \frac{x}{L} = \frac{x}{18} \).

Step2: Differentiate with Respect to Time

Differentiate: \( -\sin\theta\frac{d\theta}{dt} = \frac{1}{18}\frac{dx}{dt} \). So, \( \frac{d\theta}{dt} = -\frac{\frac{dx}{dt}}{18\sin\theta} \).

Step3: Find \( \sin\theta \) when \( x = 6.2 \)

\( \sin\theta = \frac{y}{L} = \frac{16.898}{18} \approx 0.9388 \).

Step4: Solve for \( \frac{d\theta}{dt} \)

Substitute \( \frac{dx}{dt} = 0.13 \), \( \sin\theta \approx 0.9388 \):
\( \frac{d\theta}{dt} = -\frac{0.13}{18 \times 0.9388} \approx -\frac{0.13}{16.898} \approx -0.0077 \) rad/sec. The negative sign indicates the angle is decreasing.

Answer:

• Tip slipping speed: \(\approx -0.048\) ft/sec (or \(0.048\) ft/sec downward)
• Angle change rate: \(\approx -0.0077\) rad/sec (or \(0.0077\) rad/sec decreasing)

(For the first part, boxed as \(\boxed{-0.048}\) ft/sec; for the angle, \(\boxed{-0.0077}\) rad/sec)