A 24 ft ladder is leaning against a house while the base is pulled away at a constant rate of 1 ft/s. At what rate is the top of the ladder sliding down the side of the house when the base is:
We are given, \(\frac{dy}{dt} = 1 ft/s\) and length of the ladder is 24ft. The relation between x and y is the pythagorean theorem since the ladder forms a right angle with house and the ground. \[ x^2 + y^2 = 24^2 \\ \frac{d}{dt} x^2 + \frac{d}{dt} y^2 = \frac{d}{dt} 24^2 \\ 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \\ \Rightarrow \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} \]
When x = 1, \[ y = \sqrt{24^2 - 1^2} = \sqrt{575} \]
Plugging in the equation,
\[ \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} = -\frac{1ft}{\sqrt{575}ft} 1 ft/s = -\frac{1}{\sqrt{575}}ft/s \]
So, when x = 1, the top of the ladder is sliding at a rate of \(-\frac{1}{\sqrt{575}}ft/s\).
When x = 10, \[ y = \sqrt{24^2 - 10^2} = \sqrt{476} \]
Plugging in the equation,
\[ \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} = -\frac{10ft}{\sqrt{476}ft} 1 ft/s = -\frac{10}{\sqrt{476}}ft/s \]
So, when x = 10, the top of the ladder is sliding at a rate of \(-\frac{10}{\sqrt{476}}ft/s\).
When x = 23, \[ y = \sqrt{24^2 - 23^2} = \sqrt{47} \]
Plugging in the equation,
\[ \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} = -\frac{23ft}{\sqrt{47}ft} 1 ft/s = -\frac{23}{\sqrt{47}}ft/s \]
So, when x = 23, the top of the ladder is sliding at a rate of \(-\frac{23}{\sqrt{47}}ft/s\).
When x = 24, \[ y = \sqrt{24^2 - 24^2} = 0 \]
Plugging in the equation,
\[ \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt} = -\frac{24ft}{0ft} 1 ft/s = \text{undefined} \]
So, when x = 24, the top of the ladder is no longer sliding since it is now flat on the ground thus there is no rate of change in y.