At what rate is the top of the ladder sliding down the side of the house when the base is: (a) 1 foot from the house? (b) 10 feet from the house? (c) 23 feet from the house? (d) 24 feet from the house?
We apply the concept of related rates:
The relationship between \(x\), \(y\), and \(L\) is given by the Pythagorean theorem:
\[ x^2 + y^2 = L^2 \]
Given \(L = 24\) feet, we substitute values:
\[ x^2 + y^2 = 576 \]
Different both sides with respect to time \(t\):
\[ \frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(576) \]
\[ 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \]
This simplifies to:
\[ x\frac{dx}{dt} + y\frac{dy}{dt} = 0 \]
We can rearrange the equation to solve for the rate of change of \(y\) with respect to time (\(\frac{dy}{dt}\)):
\[ y\frac{dy}{dt} = -x\frac{dx}{dt} \]
\[ \frac{dy}{dt} = -\frac{x}{y} \]
\[ y = \sqrt{576 - x^2} \]
The ladder is dropping at a rate of -0.042 ft/s.
\[ y = \sqrt{576 - 1^2} = \sqrt{575} \] \[ \frac{dy}{dt} = -\frac{1}{\sqrt{575}} \approx -0.042 \text{ ft/s} \]
The ladder is dropping at a rate of -0.458 ft/s.
\[ y = \sqrt{576 - 10^2} = \sqrt{476} \] \[ \frac{dy}{dt} = -\frac{10}{\sqrt{476}} \approx -0.458 \text{ ft/s} \]
The ladder is dropping at a rate of -3.355 ft/s.
\[ y = \sqrt{576 - 23^2} = \sqrt{49} = 7 \] \[ \frac{dy}{dt} = -\frac{23}{7} \approx -3.355 \text{ ft/s} \]
At this point, the ladder is on the ground and no longer falling.
\[ y = \sqrt{576 - 24^2} = 0 \] \[ \frac{dy}{dt} = -\frac{24}{0} \text{ (undefined)} \]