What is the rate of change of the distance mean? I have a question about this problem. I am not sure what I should find it for. Can anyone explain to me, please?

what is the rate of change of the distance between the bottom of the ladder and the wall

The top of a 13 ft ladder is sliding down a vertical wall at a constant rate of 2 ft/s. When the top of the ladder is 5 ft above the ground, what is the rate of change of the distance between the bottom of the ladder and the wall?
 A: Whenever I've encountered the phrase "rate of change of the distance", it has just meant the derivative of the distance with respect to the independent variable. As I recall, the independent variable has always been time, like it is in your particular question, so the phrase is referring to the speed, i.e., $s = \frac{d(\text{distance})}{dt}$.
Here the distance means the the length of the straight line, that is perpendicular to the wall, from the wall to the bottom of the ladder. As my answer says, the rate of change of distance is just the derivative of this distance function wrt time, i.e., the speed. In simpler phrasing, it just means how fast is the bottom of the ladder moving away from the wall.
A: John's answer is complete already, so I'm only adding this to give a more visual explanation. Generally, when dealing with related rates problems, it's good to start with a diagram or a quick sketch and see what's going on. (This is especially true for trickier problems.)

As seen in the diagram, the top of the ladder is sliding down the wall in the $y$-direction while the bottom of the ladder is sliding away from the wall in the $x$-direction.
$y$ represents the distance between the top of the ladder and the ground, so the rate at which the ladder slides down the wall is $\dfrac{\mathrm dy}{\mathrm dt}$, which is a constant $2$ ft/s according to the problem. The problem also points out that at some point, $y = 5$.
$x$ represents the distance between the bottom of the ladder and the wall. So, what is the rate of change of the distance between the bottom of the ladder and the wall? That's just the rate at which $x$ changes, or simply $\dfrac{\mathrm dx}{\mathrm dt}$. So, the question is asking you to find $\dfrac{\mathrm dx}{\mathrm dt}$.
For ladder related rates problems, you're dealing with a right-triangle, with the ladder forming the hypotenuse, which I'll call $z$. (In this problem, $z = 13$.) Our initial equation is $x^2+y^2 = z^2$. Differentiating both sides with respect to time and simplifying by $2$ (while noting that $z$ and $z^2$ are just constants) gives 
$$\dfrac{\mathrm d}{\mathrm dt}\left(x^2+y^2\right) = \dfrac{\mathrm d}{\mathrm dt}\left(z^2\right) \iff x\dfrac{\mathrm dx}{\mathrm dt}+y\dfrac{\mathrm dy}{\mathrm dt} = 0$$
Since we have $y = 5$ and $z = 13$, $x$ can be found by the initial equation, and we also have $\dfrac{\mathrm dy}{\mathrm dt} = 2$, so the only unknown left is $\dfrac{\mathrm dx}{\mathrm dt}$, which is what we want.
A: A bit of detail:
$h(t)$ is height, $d(t)$ distance $l$ is length of ladder.
Pythagoras: $h(t)^2+d(t)^2=l^2$;
Differentiate: with respect to $t$:
$2h(t)h'(t)+2d(t)d'(t)=0$;
Solve for $d'(t)$.
$h(t)$, $h'(t)$ are given. 
And now?
A: I think it's shameful that people are explaining how to get the answer without actually helping you with the full answer.
$h(t)^2 + d(t)^2 = l^2$
where $h(t)$ is the height of the ladder above the ground, $d(t)$ is the distance between the ladder and the wall, and $l$ is the length of the ladder.
We want to find the rate of change of the distance between the bottom of the ladder and the wall, which is $d(t)$. To do this, we will differentiate the Pythagorean equation with respect to time "t":
$2h(t)h'(t) + 2d(t)d'(t) = 0$
where $h'(t)$ and $d'(t)$ represent the rates of change of $h(t)$ and $d(t)$ with respect to time.
We are given that the ladder is sliding down the wall at a constant rate of $2$ ft/second, which means that $h'(t) = -2$. We are also told that the top of the ladder is $5$ ft above the ground, which means that $h(t) = 5$ and $l = 13$. We can use these values to solve for $d'(t)$:
$2(5)(-2) + 2d(t)d'(t) = 0$
$-20 + 2d(t)d'(t) = 0$
$2d(t)d'(t) = 20$
$d(t)d'(t) = 10$
Now we need to find $d'(t)$ when $h(t) = 5$. We can use the Pythagorean equation to solve for $d(t)$:
$5^2 + d(t)^2 = 13^2$
$d(t)^2 = 169 - 25$
$d(t) = sqrt(144)$
$d(t) = 12$
Substituting these values into the equation we found earlier:
$d(t)d'(t) = 10$
$12d'(t) = 10$
$d'(t) = 10/12$
$d'(t) = 5/6$
Therefore, when the top of the ladder is $5$ ft above the ground, the rate of change of the distance between the bottom of the ladder and the wall is $5\over6$ ft/second.
