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"
