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) Consider the triangle formed by the side of the house, ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 24 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 24 feet from the wall.
x = along ground to wall
y = height up wall
x^2 + y^2 = 625
2 x dx/dt + 2 y dy/dt = 0
dx/dt = 2 ft/s
so
4 x + 2 y dy/dt = 0
area = A = (1/2) x y
so
dA/dt = (1/2) (x dy/dt + y dx/dt)
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now if x = 24
y = sqrt (625 - 576) =sqrt (49) = 7
but we know when dx/dt = 2
4 x + 2 y dy/dt = 0
96 + 2*7 dy/dt = 0
dy/dt = - 6.86 ft/s
so
we have everything for
dA/dt = (1/2) (x dy/dt + y dx/dt)
= (1/2)[ 24(-6.86) + 7(2) ]
now I think you can do the sin of the angle = x/25 thing
da/dt = (625-2x^2)/(2√(625-x^2)) dx/dt
= (625-2*24^2)/(2*7) * 2
well I suppose both angles are changing at he same rate, one + and the other -
To answer these questions, we'll need to apply some basic concepts from geometry and calculus.
a) To find the rate at which the area of the triangle is changing, we need to differentiate the equation for the area of a triangle with respect to time using the chain rule.
Let's denote the height of the triangle as h(t) and the base of the triangle as b(t), where t represents time.
Given that the base of the ladder is pulled away from the wall at a rate of 2 feet per second, we can express b(t) as 24 - 2t.
The area of a triangle is given by the formula A = (1/2) * base * height. So, the area of the triangle is A(t) = (1/2) * b(t) * h(t).
Now, differentiate the equation A(t) with respect to time (t) using the chain rule:
dA/dt = (1/2) * (db/dt * h(t) + b(t) * dh/dt)
We need to find dA/dt when the base of the ladder is 24 feet from the wall, which means b(t) = 24 and db/dt = -2 (since it is moving away from the wall).
Let's also assume that the ladder is fixed to the ground, which means dh/dt = 0 (since the height of the triangle is not changing).
Substituting these values in the equation:
dA/dt = (1/2) * (-2 * h(t) + 24 * 0)
= -h(t)
Therefore, the rate at which the area of the triangle is changing when the base of the ladder is 24 feet from the wall is equal to the negative of the height of the triangle.
b) To find the rate at which the angle between the ladder and the wall of the house is changing, we need to differentiate the equation for the tangent of the angle with respect to time.
Let's denote the angle between the ladder and the wall as θ(t).
Using trigonometry, we know that tan(θ(t)) = h(t)/b(t).
Differentiating both sides of the equation with respect to time, we get:
sec^2(θ(t)) * dθ/dt = (dh/dt * b(t) - h(t) * db/dt) / (b(t))^2
Again, assuming that the ladder is fixed to the ground, we have dh/dt = 0.
Substituting the other values we know (b(t) = 24, db/dt = -2, and dh/dt = 0):
sec^2(θ(t)) * dθ/dt = (0 * 24 - h(t) * (-2)) / (24)^2
= 2h(t) / 576
= h(t) / 288
Therefore, the rate at which the angle between the ladder and the wall of the house is changing is equal to the height of the triangle divided by 288.
agree about -75
you know that x^2+y^2 = 625
so, when x=24, y=7
(a) a = 1/2 xy = 1/2 x√(625-x^2)
da/dt = (625-2x^2)/(2√(625-x^2)) dx/dt
= 49/(2*7) * 2
(b) tanθ = y/x = √(625-x^2)/x
sec^2θ dθ/dt = -625/(x^2 √(625-x^2)) dx/dt
(25/24)^2 dθ/dt = -625/(24^2 * 7) * 2