The speed of the top of a sliding ladder A $5$m ladder is leaning against a wall. If the bottom of the ladder is pulled along the ground away from the wall at a constant rate of $0.4$m/s, how fast will the top of the ladder be moving down the wall when its bottom is $3$m away from the wall.


Is my solution wrong? How come I can't get the correct answer?

x=horizontal axis , y=vertical axis
Area=XY/2, when X=3,  A=6m^2, dx/dt=0.4m/s 
Y=2A/X 
dy/dx=-2A/x^2.....A=6,x=3
     =-12/9

dy/dt=dy/dx * dx/dt
     =-12/9 * 0.4
     =-0.533333
You can't use an area relation because we're not told what $\dfrac{dA}{dt}$ is.  If you draw out the situation, we see that we can actually use a different relation to solve this problem.

The equation that you want to use comes directly from the Pythagorean theorem; in particular, if $x$ is the distance the base of the ladder is from the wall and $y$ is the the height of the ladder along the wall, then for the 5 m long ladder, we have that $x^2+y^2 = 25$.  

Differentiating this will then give you the appropriate equations involving $\dfrac{dx}{dt}$ (i.e. the rate the ladder is sliding away from the wall) and $\dfrac{dy}{dt}$ (i.e. the rate the ladder is sliding down the wall).  You can then plug in all the known values at the appropriate instant in time (when $x=3$, $y=\ldots$ [which can be found using the Pythagorean theorem], and $dx/dt = 0.4\text{ m/s}$) and then solve for $\dfrac{dy}{dt}$.

Hopefully this is enough information to help you get the correct solution. :-)
This problem can be solved 'numerically' without resorting to other methods. First we have:  

$$
w^2 + g^2 = L
$$
$$
(w + f)^2 + (g + p)^2 = L  
$$

where $w$ is the wall, $g$ is the ground, $L$ is the ladder squared, $f$ is the fall rate and $p$ is the pull rate. Note that the units of f and p are metres - time is implicit. Also note that we do not presume that the fall rate is negative - it should come out that way. Equating, canceling common terms and solving the quadratic in $f$ yields:

$$
f = \sqrt{w^2 - 2gp - p^2} - w
$$

Now we can calculate a series of increasingly accurate approximations of the fall rate if the pull rate is $0.4 m/s$. Here is a Python function that does this:

def fallrate(wall, ground, pullrate):  
  for i in [2**n for n in range(15)]:  
    print(i * ((wall**2 - 2 * ground * (pullrate/i) - (pullrate/i)**2)**.5 - wall))  
  input("\nEnter to exit. ")  

fallrate(4, 3, 0.4)  


Note that if we halve the distance we use to calculate the pull rate we will get a rate for half the distance. We must therefore double the result for the purpose of comparison - hence the i * etc at the start of the approximation formula. The fall rate seems to converge on $-0.3 m/s$. This approach can be seen as a type of calculus - if we divide the rates on both sides of the approximation formula by $i$ we are in effect turning them into infinitesimals; and it agrees with the result given by Eff worked out using conventional methods.