Expert: Scotto - 11/13/2008

Question
A ladder 25 feet long is leaning against the wall of the house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second.

(a) Consider the triangle formed by the side of the house, the ladder, and the ground. Find the rate at which  the area of the triangle is changing when the base of the ladder is 7 feet from the wall.

(b) Find the rate 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.

Answer
When the ladder is 7 feet from the wall, the height is known to be
25² - 7² = 625 - 49 = 576, which is 24².

(a) In the original problem, we will give the the base b, the height h, and the length of the ladder l.  Now l is a constant but b and h are both functions of time, so now I will write them as b(t) and h(t).  It is known that since we are dealing with a right triangle that
b²(t) + h²(t) = l².  Taking the derivative with respect to t gives us
2b(t)(db/dt) + 2h(t)(dh/dt) = 0.

From the first paragraph, we know b(t)=7, h(t)=24, and db/dt=2.  Using these in the equation above, the only one left is dh/dt, for which we can solve.  Note that it turns out to be negative, which means that the height is decreasing.  This is good, since we know that ladder is sliding down the wall.

(b) The angle I will call Θ.  This Θ is really Θ(t).  It is known that tan[Θ(t)] is h(t)/b(t).  From here we can take the derivative, giving sec²[Θ(t)] dΘ/dt = [b(t)h'(t) - h(t)b'(t)]/b²(t).
The following are know:  b(t) is known to be 7, h(t) is known to be 24, b'(t) is known to be 2, h'(t) we just found in (a), and sec(Θ) is known to be 7/25.

All that needs to be done is to substitute these values into the equation and solve for dΘ/dt.  Note that the one value that has to be changed a bit is we know that sec(Θ)=7/25, but in the equation we have sec²(Θ).  Now when 7/25 is squared and then both sides multiplied by the invers, that would be the same as multiplying both sides by 625/49.