A 3m ladder has a bent rung 2m from its base. How fast is the distance the bent rung is from the wall changing when the base of the ladder is 1.7m from the wall and the top of the ladder is sliding down the wall 1m/s?

1 Answer
Steve M
Mar 17, 2018

the Bent Rung is decreasing at a speed of #0.485 \ ms^(-1)# (3sf).

Explanation:

Using the figure:

Steve M

Let us define the following variables:

# { (t,"time ", s), (x=BD,"distance of bent rung from wall at time "t, m), (y=OC,"distance of base of ladder from wall at time "t, m), (h=OA,"height of top of ladder from the floor at time "t, m) :} #

We seek the value of #dx/dt# when #y=1.7# and #(dh)/dt=1#

By similar triangles we have:

# AB:AC = BD:OC => 1/3 = x/y #
# :. y = 3x #

Differentiating (implicitly) wrt #t# we have:

# dy/dt = 3dx/dt #

By Pythagoras, we have:

# AC^2 = OA^2 + OC^2 => 3^2 = h^2 + y^2 #
# :. h^2 + y^2 = 9 #

Differentiating (implicitly) wrt #t# we get:

# 2h (dh)/dt + 2ydy/dt = 0 #

# :. sqrt(9-y^2) (dh)/dt + 3ydx/dt = 0 #

Substituting the values #x=1.7# and #(dh)/dt=1#, we get:

# sqrt(9-1.7^2)* 1 + 3*1.7*dx/dt = 0 #

# :. sqrt(9-2.89) + 5.1 dx/dt = 0 #

# :. sqrt(6.11) + 5.1 dx/dt = 0 #

# :. dx/dt = - sqrt(6.11)/5.1 = - 0.484674 ... #

Hence, the Bent Rung is decreasing at a speed of #0.485 \ ms^(-1)# (3sf).

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