Assume the bottom of a 16 ft ladder is pulled out at a rate of 3 ft/s. How do you find the rate at the top of the ladder when it is 10 ft from the ground?

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Answer 1 · 2016-12-17T00:57:34

It is sliding down at a rate of #3.75# ft/sec

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Let us define the following variables:

# {(x, "Distance of bottom of ladder from the wall (ft)"), (y, "Distance of top of ladder from the floor (ft)") :} #

We are told that #dx/dt=3 " ft/s"# and we aim to find #dy/dt# when #y=10#

The ladder is a fixed length (#16 ft#) so by Pythagoras;

# x^2+y^2=16^2 #
# :. x^2+y^2=256 #

Differentiating Implicitly wrt #t# we get:

# 2xdx/dt + 2ydy/dt=0#
# :. 2x*3 + 2ydy/dt=0#
# :. 3x + ydy/dt=0#

When #y=10 => x^2+100=256 #

# :. x^2=156 #
# :. x=sqrt(156) " as " (x >0) #

And so substituting #x=sqrt(156), y=10# into our derivative gives:

# 3sqrt(156) + 10dy/dt=0#
# :. dy/dt = -3/10sqrt(156)#
# :. dy/dt = -3.74699 ...#
# :. dy/dt = -3.75 (2dp)#

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 #3.75# ft/sec