A ladder 20 ft long leans against a vertical wall. If the top end slides downwards at the rate of 2ft per second, what will be the rate at which the slope of the ladder changes?

The correct option is -25/54.

To understand why this is correct, we can model the ladder leaning against the vertical wall using a right-angled triangle. Let the horizontal ground be represented by the x-axis and the vertical wall by the y-axis.

Let $x$ be the distance of the foot of the ladder from the wall at any time $t$, and $y$ be the height of the top of the ladder from the ground at any time $t$. Since the length of the ladder is constant at $20\text{ft}$, we can apply the Pythagorean theorem:

$x^2+y^2={20}^2=400$

We are given that the top of the ladder slides downwards at a rate of $2\text{ft/s}$. Since $y$ is decreasing, the rate of change of $y$ with respect to time $t$ is:

$\frac{dy}{dt}=-2\text{ft/s}$

To find the rate of change of the slope of the ladder, let $m$ denote the slope of the ladder. The slope of the line segment connecting the points $(x,0)$ and $(0,y)$ is given by:

$m=\frac{y}{x}$

Differentiating both sides of the relation $x^2+y^2=400$ with respect to $t$, we get:

$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$

Solving for $\frac{dx}{dt}$ gives:

$\frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}$

Now, we differentiate the slope equation $m=\frac{y}{x}$ with respect to $t$ using the quotient rule:

$\frac{dm}{dt}=\frac{x\frac{dy}{dt}-y\frac{dx}{dt}}{x^2}$

Substituting $\frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}$ into the equation for $\frac{dm}{dt}$, we have:

$\frac{dm}{dt}=\frac{x\frac{dy}{dt}-y\left(-\frac{y}{x}\frac{dy}{dt}\right)}{x^2}=\frac{\left(x+\frac{y^2}{x}\right)\frac{dy}{dt}}{x^2}=\frac{\left(x^2+y^2\right)\frac{dy}{dt}}{x^3}$

Substituting the values $x^2+y^2=400$ and $\frac{dy}{dt}=-2$ into this expression, we get:

$\frac{dm}{dt}=\frac{400(-2)}{x^3}=-\frac{800}{x^3}$

At the specific instant when the foot of the ladder is $12\text{ft}$ from the wall (meaning $x=12\text{ft}$, and consequently $y=\sqrt{400-144}=16\text{ft}$), we compute the rate of change of the slope as follows:

$\frac{dm}{dt}=-\frac{800}{{12}^3}=-\frac{800}{1728}$

Simplifying the fraction by dividing the numerator and denominator by their greatest common divisor, $32$, we obtain:

$\frac{dm}{dt}=-\frac{25}{54}\text{per second}$

Thus, the rate at which the slope of the ladder changes is $-25/54$ units per second.