Question

a 10 - foot ladder is leaning straight up against a wall when a person begins pulling the base of the ladder away from the wall at the rate of 1 foot per second. which of the following is true about the distance between the top of the ladder and the ground when the base of the ladder is 9 feet from the wall?
a the distance is increasing at a rate of $\frac{9}{sqrt{19}}$ feet per second.
b the distance is decreasing at a rate of $\frac{9}{sqrt{19}}$ feet per second.
c the distance is increasing at a rate of $\frac{sqrt{19}}{9}$ feet per second.
d the distance is decreasing at a rate of $\frac{sqrt{19}}{9}$ feet per second.

Explanation:

Step1: Establish the Pythagorean - theorem relationship

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 both sides with respect to time $t$

Differentiating $x^{2}+y^{2}=100$ with respect to $t$ gives $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$. Then simplify to $x\frac{dx}{dt}+y\frac{dy}{dt}=0$.

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

When $x = 9$, we substitute into $x^{2}+y^{2}=100$. So $9^{2}+y^{2}=100$, which gives $y=\sqrt{100 - 81}=\sqrt{19}$.

Step4: Substitute known values to find $\frac{dy}{dt}$

We know that $\frac{dx}{dt}=1$ foot - per - second, $x = 9$, and $y=\sqrt{19}$. Substitute into $x\frac{dx}{dt}+y\frac{dy}{dt}=0$. We get $9\times1+\sqrt{19}\frac{dy}{dt}=0$. Solving for $\frac{dy}{dt}$, we have $\frac{dy}{dt}=-\frac{9}{\sqrt{19}}$. The negative sign indicates that $y$ is decreasing.

Answer:

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