Question

a ladder 25 feet long is leaning against the wall of a house. the base of the ladder is pulled away from the wall at a rate of 2 feet per second.
(a) what is the velocity (in ft/sec) of the top of the ladder when the base is given below?
7 feet away from the wall
ft/sec
20 feet away from the wall
ft/sec
24 feet away from the wall
ft/sec
(b) consider the triangle formed by the side of the house, ladder, and the ground. find the rate (in ( ft^{2}/sec )) at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall.
( ft^{2}/sec )
(c) find the rate (in rad/sec) at which the angle between the ladder and the wall of the house is changing when the base of the ladder is 7 feet from the wall
rad/sec

Explanation:

Step1: Establish the relationship

Let \(x\) be the distance of the base of the ladder from the wall and \(y\) be the height of the top of the ladder from the ground. By the Pythagorean theorem, \(x^{2}+y^{2}=25^{2}\). Differentiating both sides with respect to time \(t\), we get \(2x\frac{dx}{dt}+2y\frac{dy}{dt} = 0\), so \(\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}\). Given \(\frac{dx}{dt}=2\) ft/sec.

For \(x = 7\)

First, find \(y\) when \(x = 7\). \(y=\sqrt{25^{2}-7^{2}}=\sqrt{625 - 49}=\sqrt{576}=24\).
Then \(\frac{dy}{dt}=-\frac{7}{24}\times2=-\frac{7}{12}\) ft/sec.

For \(x = 20\)

Find \(y\) when \(x = 20\). \(y=\sqrt{25^{2}-20^{2}}=\sqrt{625 - 400}=\sqrt{225}=15\).
Then \(\frac{dy}{dt}=-\frac{20}{15}\times2=-\frac{8}{3}\) ft/sec.

For \(x = 24\)

Find \(y\) when \(x = 24\). \(y=\sqrt{25^{2}-24^{2}}=\sqrt{625 - 576}=\sqrt{49}=7\).
Then \(\frac{dy}{dt}=-\frac{24}{7}\times2=-\frac{48}{7}\) ft/sec.

Step2: Area of the triangle

The area of the triangle \(A=\frac{1}{2}xy\). Differentiating with respect to \(t\), \(\frac{dA}{dt}=\frac{1}{2}(x\frac{dy}{dt}+y\frac{dx}{dt})\).
When \(x = 7\), \(y = 24\), \(\frac{dy}{dt}=-\frac{7}{12}\), \(\frac{dx}{dt}=2\)
\(\frac{dA}{dt}=\frac{1}{2}(7\times(-\frac{7}{12})+24\times2)=\frac{1}{2}(-\frac{49}{12}+48)=\frac{1}{2}(\frac{- 49 + 576}{12})=\frac{527}{24}\) ft²/sec.

Step3: Angle \(\theta\) (between the ladder and the wall)

Let \(\theta\) be the angle between the ladder and the wall. \(\sin\theta=\frac{x}{25}\). Differentiating with respect to \(t\), \(\cos\theta\frac{d\theta}{dt}=\frac{1}{25}\frac{dx}{dt}\).
When \(x = 7\), \(y = 24\), \(\cos\theta=\frac{y}{25}=\frac{24}{25}\)
\(\frac{24}{25}\frac{d\theta}{dt}=\frac{1}{25}\times2\), so \(\frac{d\theta}{dt}=\frac{1}{12}\) rad/sec.

Answer:

(a)

• \(7\) feet away: \(-\frac{7}{12}\) ft/sec
• \(20\) feet away: \(-\frac{8}{3}\) ft/sec
• \(24\) feet away: \(-\frac{48}{7}\) ft/sec

(b) \(\frac{527}{24}\) ft²/sec
(c) \(\frac{1}{12}\) rad/sec