Related rates help

bl4ke360

New member
Joined
Nov 10, 2011
Messages
1
This is the problem:
A 16-ft ladder is sliding down a wall at a rate of 4 ft/sec. Find the velocity of the top of the ladder at t=s if the bottom is 5ft from the wall at t=0.


In my notes, I have a similar type of problem with the work shown:

16bmzie.jpg


But I still can't figure it out, and more specifically I don't know if 4 ft/sec is dy/dt or dx/dt, and how I'm supposed to use t=s.
Help please? This is due in a few hours so I would appreciate a fast response.
 
This is the problem:
A 16-ft ladder is sliding down a wall at a rate of 4 ft/sec. Find the velocity of the top of the ladder at t=s if the bottom is 5ft from the wall at t=0.
I don't know if 4 ft/sec is dy/dt or dx/dt, and how I'm supposed to use t=s.

"sliding down a wall at a rate of 4 ft/sec" means the top of the ladder is moving downward at that speed, so dy/dt = -4 ft/s.

I can only guess that "t=s" is supposed to be t = 1 sec. If this is so, calculate the geometry of the wall and ladder at t = 0 sec, then again at t = 1 sec. (after the top has slid down a little at -4 ft/sec). Then use the position of the ladder at t = 1s to define your x and y values.

Hope that helps.
 
Top