question a 13 - foot ladder is leaning against a wall. if we pull the ladder away from the wall at a rate of…

Explanation:

Step1: Establish the relationship

Let $x$ be the distance of the bottom of the ladder from the wall and $y$ be the distance of the top of the ladder from the ground. By the Pythagorean - theorem, $x^{2}+y^{2}=13^{2}=169$.

Step2: Differentiate with respect to time $t$

Differentiating both sides of the equation $x^{2}+y^{2}=169$ 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 $x = 5$

When $x = 5$, we substitute into $x^{2}+y^{2}=169$. So $5^{2}+y^{2}=169$, which gives $y^{2}=169 - 25=144$, and $y = 12$ (since $y\gt0$).

Step4: Substitute known values

We know that $\frac{dx}{dt}=5$ ft/s, $x = 5$, and $y = 12$. Substitute these values into $x\frac{dx}{dt}+y\frac{dy}{dt}=0$. We get $5\times5+12\times\frac{dy}{dt}=0$.

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

First, simplify the left - hand side of the equation: $25+12\times\frac{dy}{dt}=0$. Then, isolate $\frac{dy}{dt}$: $12\times\frac{dy}{dt}=-25$, so $\frac{dy}{dt}=-\frac{25}{12}$ ft/s.

Answer:

$-\frac{25}{12}$ ft/s