The rate is 65/6 "m/min"=10 5/6 "m/min"~~10.83 "m/min"656m/min=1056m/min≈10.83m/min
This is a nice example of a Related Rates problem. (Perhaps better called by the full name "Related Rates of Change")
Solution:
hh is the distance from the floor to the end of the ladder.
ll is the length of the ladder.
(dl)/dt=10"m/min"dldt=10m/min
We need to find (dh)/(dt)dhdt
The relationship between the variables, hh and ll is given by the Pythagorean Theorem:
h^2+5^2=l^2h2+52=l2
We need to find (dh)/(dt)dhdt
To find the relationship between the rates of change, differentiate (implicitly) with respect to tt.
2h(dh)/(dt)+0=2l(dl)/(dt)2hdhdt+0=2ldldt
So:
h(dh)/(dt)=l(dl)/(dt)hdhdt=ldldt
Substitute what we know and solve for the desired value.
We are told that (dl)/dt=10"m/min"dldt=10m/min and we are interested in
the instant when l=13"m"l=13m, but we also need hh at that instant.
Using Pythagoras to find hh when l=13"m"l=13m
h^2+5^2=13^2h2+52=132
h^2=144h2=144
h=12"m"h=12m
Now we can finish:
h(dh)/(dt)=l(dl)/(dt)hdhdt=ldldt so at the instant we were asked about:
12"m"(dh)/(dt)=13"m"(10"m/min")12mdhdt=13m(10m/min)
Thus
(dh)/(dt)=130/12 "m/min"=65/6 "m/min"=10 5/6 "m/min"~~10.83 "m/min"dhdt=13012m/min=656m/min=1056m/min≈10.83m/min