Posted by **Tanya** on Friday, November 23, 2012 at 3:33pm.

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?

- Calculus! -
**Steve**, Friday, November 23, 2012 at 6:23pm
I suspect a typo somewhere. When the ladder is lying flat at t=3, the base is no longer moving.

- Calculus! -
**Reiny**, Friday, November 23, 2012 at 6:36pm
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)

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