Please HELP !!!!! A ladder 25 feed long is leaning against the wall of a house.
ID: 3288867 • Letter: P
Question
Please HELP !!!!!
A ladder 25 feed long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate 2 feet per second. What is the velocity of the top of the ladder when the base is given below? 7 feet away from the wall -7/12 ft/sec 15 feet away from the wall -1.5 ft/sec 20 feet away from the wall -8/3 ft/sec Consider the triangle formed by the side of 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. -527/24 ft2/sec 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 feed from the wall. rad/secExplanation / Answer
a)
dx/dt = 2 ft/sec
dy/dt = ? ft/sec
625= x^2 + y^2
625 = 7^2 + y^2
y = 24
0 = 2x(dx/dt) + 2y(dy/dt)
0 = 2(7)(2) + 2(24)(dy/dt)
a) dy/dt = -7/12 ft/sec
b)
A = .5(x)(y)
dA/dt = .5(x(dy/dt) + y(dx/dt))
dA/dt = .5(7(-7/12) + 24(2))
dA/dt = 574/24 ft^2/sec or 21.958 ft^2/sec
c)
sin(@) = x/25
sin(@) = 7/25
@=16.26* degrees
cos(@) (d@/dt) = (dx/dt)/25
.96(d@/dt) = (2)/25
d@/dt = 1/12 degrees/sec or .08333333 degrees/sec