question a 15 - 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 height of the top of the ladder on the wall. By the Pythagorean - theorem, $x^{2}+y^{2}=15^{2}=225$.

Step2: Differentiate with respect to time $t$

Differentiating both sides of the equation $x^{2}+y^{2}=225$ 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 the value of $y$

When $x = 12$, we can find $y$ using the Pythagorean - theorem: $y=\sqrt{225 - x^{2}}=\sqrt{225 - 144}=\sqrt{81}=9$.

Step4: Substitute the known values

We know that $\frac{dx}{dt}=2$ (the rate at which the bottom of the ladder is moving away from the wall), $x = 12$, and $y = 9$. Substitute these values into the equation $x\frac{dx}{dt}+y\frac{dy}{dt}=0$. We get $12\times2+9\times\frac{dy}{dt}=0$.

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

First, simplify the left - hand side of the equation: $24 + 9\times\frac{dy}{dt}=0$. Then, isolate $\frac{dy}{dt}$: $9\times\frac{dy}{dt}=-24$, so $\frac{dy}{dt}=-\frac{24}{9}=-\frac{8}{3}$.

Answer:

$-\frac{8}{3}$