A ladder is leaning against a wall, and the floor and slipping. If the bottom of the ladder is slipping at #30 cms^(-1)# then how fast is the top of the ladder sliding down the wall when the ladder is at #45^o#?

1 Answer
Sep 29, 2017

It is sliding down at a rate (speed) of #30 \ ms^-1#.

Explanation:

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Let us define the following variables:

# { (x, "Distance of bottom of ladder from the wall " (cms^-1)), (y, "Distance of top of ladder from the floor "(cm)), (t, "Time "(s)), (theta, "Angle between Ladder and floor "angle ABO " (radians)") :} #

We are told that #dx/dt=30 \ cms^(-1)# (constant)
We aim to find #dy/dt# when #theta=45^o=pi/4#

The ladder is a fixed length, #l\ (m)# say, so by Pythagoras;

# x^2+y^2=l^2 #

Differentiating Implicitly wrt #t# we get:

# 2xdx/dt + 2ydy/dt=0 => xdx/dt + ydy/dt=0 #

# dx/dt = 30 => 30x + ydy/dt=0 #

# :. dy/dt = -30x/y # ..... [A]

Using trigonometry, we have:

# tan \ angle ABO = y/x => tantheta = y/x #
# theta = pi/4 => y/x=1 => x/y =1 #

With #x/y=1# in Eq[A] then:

# => dy/dt = -30 #

The minus sign denotes that the ladder is sliding down, i.e., the height #y# is decreasing. It is sliding down at a rate (speed) of #30 \ ms^-1#.