The rate is #65/6 "m/min"=10 5/6 "m/min"~~10.83 "m/min"#
This is a nice example of a Related Rates problem. (Perhaps better called by the full name "Related Rates of Change")
Solution:
#h# is the distance from the floor to the end of the ladder.
#l# is the length of the ladder.
#(dl)/dt=10"m/min"#
We need to find #(dh)/(dt)#
The relationship between the variables, #h# and #l# is given by the Pythagorean Theorem:
#h^2+5^2=l^2#
We need to find #(dh)/(dt)#
To find the relationship between the rates of change, differentiate (implicitly) with respect to #t#.
#2h(dh)/(dt)+0=2l(dl)/(dt)#
So:
#h(dh)/(dt)=l(dl)/(dt)#
Substitute what we know and solve for the desired value.
We are told that #(dl)/dt=10"m/min"# and we are interested in
the instant when #l=13"m"#, but we also need #h# at that instant.
Using Pythagoras to find #h# when #l=13"m"#
#h^2+5^2=13^2#
#h^2=144#
#h=12"m"#
Now we can finish:
#h(dh)/(dt)=l(dl)/(dt)# so at the instant we were asked about:
#12"m"(dh)/(dt)=13"m"(10"m/min")#
Thus
#(dh)/(dt)=130/12 "m/min"=65/6 "m/min"=10 5/6 "m/min"~~10.83 "m/min"#
