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

 A: 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. :-)
A: First we use Pythagorean Theorem
$$5^2=x^2+y^2 \implies y = \sqrt{25-x^2}.$$
Next we use that we can write $x=0.4t=\frac{2}{5}t$:
$$y=\sqrt{25-\left(\frac{2}{5}t\right)^2} = \sqrt{25-\frac{4}{25}t^2}.$$
Now we calculate the derivative (using the chain rule)
$$ \frac{dy}{dt}= \frac{\frac{d}{dt}\left(25-\frac{4}{25}t^2\right)}{2\sqrt{25-\frac{4}{25}t^2}} = \frac{-\frac{8}{25}t}{2\sqrt{25-\frac{4}{25}t^2}}.$$
This can be simplified further, but it isn't necessary. Now for the last step: evaluating the derivative at $t=7.5$ (which comes from solving $0.4t=3$).
$$\frac{dy}{dt}|_{t=7.5} =\frac{-\frac{8}{25}\cdot 7.5}{2\sqrt{25-\frac{4}{25}\cdot7.5^2}}=-0.3. $$
So the conclusion is that the speed, when the bottom is $3\text{m}$ from the wall, is $-0.3 \text{ m}/\text{s}$.
A: 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.
