SOLUTION: Please help me with this related rates question The top of a ladder slides down a vertical wall at a rate of 0.125 m/s. At the moment when the bottom of the ladder is 5 m from the

We let the ladder slip down the wall, then we 
"freeze" it at an instant.



z is the length of the ladder, so it is constant.

y is a variable speed, getting shorter and shorter with 
respect to time.

dy/dt is a variable speed, getting faster and faster due 
to gravity, with respect to time. 

x is a variable getting longer with respect to time.
dx/dt is a variable, getting slower and slower. 

[Important: do not make the mistake of substituting the 
"freezing' value of the variables x, until we 
have done the calculus letting them vary. Leave the variables
as they are until we have taken derivatives.]

We use the Pythagorean theorem:



We take derivatives with respect to t (time), remembering
that since the length of the ladder z is the ONLY constant, 
the derivative of z² is 0.



Divide through by 2



Now that we have taken the derivative we can substitute the 
"freezing" values:

We are given that the top of a ladder slides down the
wall at a rate of 0.125 m/s, since y is getting shorter,
the rate y is changing is negative or -0.125 m/s.
We "freeze" the variable x at 5 m. We "freeze" dx/dt at
0.3 m/s.













So the "freezing" value of y is 12 m.

Now we can calculate the length of the ladder z
from the Pythagorean theorem:








So the ladder is 13 m long.

Edwin