# Santa Claus and his trouble on a ladder

1. Dec 17, 2009

### rpullo

1. The problem statement, all variables and given/known data
Santa Claus and his wife place a 10 m long ladder against a wall (which is vertical). Santa Claus climbs to the top while Mrs. Claus steadies the bottom. However, Mrs. Claus left because she had to give birth, so the ladder began to slip. The ladder kept in contact w/ the wall and the floor (which are perpendicular to each other). At one instance, the angle of the ladder was 60o, and the top end was sliding down the wall w/ a speed of 2 m/s. Calculate the speed of the bottom.

2. Relevant equations

x2 + y2 = 102

3. The attempt at a solution

i.Length from the top of the ladder to the ground (which I will represent as "y"):

y = 10 sin60 = 8.66 m

ii.Length from the base of the ladder to the wall (which I will represent as "x"):

x = 10 cos60 = 5 m

iii.Differentiate both sides of the equation 'x2 + y2 = 102' w.r.t. t:

(2x)dx/dt + (2y)dy/dt = 0

iv.Isolate for dx/dt:

dx/dt = (-x/y)*dx/dt

v.When x = 5, y = 8.66, dy/dt = 2.
Therefore...

dx/dt = (-5/8.66)*2 = 1.15 m/s

I was just wondering if anyone could verify my steps and make sure it is done correctly. Thanks!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 17, 2009

### wmrunner24

I think your algebras wrong. When you isolated dx/dt, you put the x in the top and the y in the bottom. I'm pretty sure it should be the other way around. I got 3.4641m/s.

3. Dec 17, 2009

### denverdoc

+1. Sometimes folks forget that calculus results can be checked just like agebra results by choosing a suitably small interval, in this case of say 0.01 sec. The original answer fell substantially short.

At PF are happy to help, but developing the habit of checking your own answer is a great skill to develop! When you are working in a high tech engineering firm and are a lead engineer, who do you ask then?