A 16 ft. ladder leans against a wall, with the top being 12 ft. from the wall when time t=0. The top is sliding down at a constant rate of 4 ft./second. Given that the top of the ladder touches the ground at 3 seconds, what is the velocity of the bottom of the ladder when t=3?
2 answers
I suspect a typo somewhere. When the ladder is lying flat at t=3, the base is no longer moving.
Make a sketch:
x^2 + y^2 = 256
when t=0 , y=12, x = √112
when t = 1, y = 8 , x = √192
when t = 2 , y = 4 , x = √240
when t = 3, y = 0, x = √256 = 16
given dy/dt = -4 ft/s
2x dx/dt + 2y dy/dt = 0
dx/dt = -y (dy/dt) / x
= 0/16 = 0
check:
let t=1
y = 8, y = √192
dx/dt = -8(-4)/√192 = appr 2.3
let t = 2
y = 4, x = √240
dx/dt = -4(-4)/√240 = appr 1.03
slowing up ....
let t = 2.99 sec
y = 12 - 2.99(4) = .04 , x = √255.9984
dx/dt = -.04(-4)/√255.9984 = appr .01 (close to zero)
x^2 + y^2 = 256
when t=0 , y=12, x = √112
when t = 1, y = 8 , x = √192
when t = 2 , y = 4 , x = √240
when t = 3, y = 0, x = √256 = 16
given dy/dt = -4 ft/s
2x dx/dt + 2y dy/dt = 0
dx/dt = -y (dy/dt) / x
= 0/16 = 0
check:
let t=1
y = 8, y = √192
dx/dt = -8(-4)/√192 = appr 2.3
let t = 2
y = 4, x = √240
dx/dt = -4(-4)/√240 = appr 1.03
slowing up ....
let t = 2.99 sec
y = 12 - 2.99(4) = .04 , x = √255.9984
dx/dt = -.04(-4)/√255.9984 = appr .01 (close to zero)
3 answers