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)
=======================
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
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.
5 answers
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
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
da/dt = (625-2x^2)/(2√(625-x^2)) dx/dt
= (625-2*24^2)/(2*7) * 2
= (625-2*24^2)/(2*7) * 2
well I suppose both angles are changing at he same rate, one + and the other -
agree about -75
3 answers