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?

Question

2551 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-17T00:57:34

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

enter image source here

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