a 10 - foot ladder is leaning straight up against a wall when a person begins pulling the base of the ladder…

Explanation:

Step1: Establish the Pythagorean - relation

Let $x$ be the distance between the base of the ladder and the wall, and $y$ be the distance between the top of the ladder and the ground. The length of the ladder $L = 10$ feet. By the Pythagorean theorem, $x^{2}+y^{2}=L^{2}=100$.

Step2: Differentiate with respect to time $t$

Differentiating both sides of the equation $x^{2}+y^{2}=100$ with respect to $t$ gives $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$. Then we can simplify it to $x\frac{dx}{dt}+y\frac{dy}{dt}=0$.

Step3: Find $y$ when $x = 9$

When $x = 9$, from $x^{2}+y^{2}=100$, we have $y=\sqrt{100 - x^{2}}=\sqrt{100 - 81}=\sqrt{19}$.

Step4: Substitute known values into the differentiated - equation

We know that $\frac{dx}{dt}=1$ foot per second, $x = 9$, and $y=\sqrt{19}$. Substituting these values into $x\frac{dx}{dt}+y\frac{dy}{dt}=0$, we get $9\times1+\sqrt{19}\frac{dy}{dt}=0$.

Step5: Solve for $\frac{dy}{dt}$

Rearranging the equation $9+\sqrt{19}\frac{dy}{dt}=0$ to solve for $\frac{dy}{dt}$, we have $\frac{dy}{dt}=-\frac{9}{\sqrt{19}}$. The negative sign indicates that the distance $y$ is decreasing.

Answer:

B. The distance is decreasing at a rate of $\frac{9}{\sqrt{19}}$ feet per second.