Question

a 10 - foot ladder is leaning against a vertical wall (see figure) when jack begins pulling the foot of the ladder away from the wall at a rate of 0.3 ft/s. how fast is the top of the ladder sliding down the wall when the foot of the ladder is 6 ft from the wall? when the foot of the ladder is 6 ft from the wall, the top of the ladder is sliding down the wall at a rate of \\(\square\\) (round to two decimal places as needed.)

Explanation:

Step1: Establish the Pythagorean - related equation

Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the height of the top of the ladder on the wall. The length of the ladder $L = 10$ ft. By the Pythagorean theorem, $x^{2}+y^{2}=L^{2}=100$.

Step2: Differentiate the equation with respect to time $t$

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

Step3: Find the value of $y$ when $x = 6$

When $x = 6$, from $x^{2}+y^{2}=100$, we have $y=\sqrt{100 - x^{2}}=\sqrt{100 - 36}=\sqrt{64}=8$.

Step4: Substitute the known values into the differentiated equation

We know that $\frac{dx}{dt}=0.3$ ft/s, $x = 6$, and $y = 8$. Substitute into $x\frac{dx}{dt}+y\frac{dy}{dt}=0$. So, $6\times0.3+8\times\frac{dy}{dt}=0$.

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

First, calculate $6\times0.3 = 1.8$. Then the equation becomes $1.8+8\times\frac{dy}{dt}=0$. Rearrange to get $8\times\frac{dy}{dt}=-1.8$. Then $\frac{dy}{dt}=-\frac{1.8}{8}=- 0.225$ ft/s.

Answer:

$-0.23$ ft/s