Question

a 25 - 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 0.7 ft/s. how fast is the top of the ladder sliding down the wall when the foot of the ladder is 7 ft from the wall? (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 $L = 25$ ft. By the Pythagorean theorem, $x^{2}+y^{2}=L^{2}=25^{2}=625$.

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

Differentiating both sides of $x^{2}+y^{2}=625$ 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 $x = 7$

When $x = 7$, we can find $y$ from $x^{2}+y^{2}=625$. So $y=\sqrt{625 - x^{2}}=\sqrt{625 - 49}=\sqrt{576}=24$.

Step4: Substitute the known values into the differentiated equation

We know that $\frac{dx}{dt}=0.7$ ft/s, $x = 7$, and $y = 24$. Substituting into $x\frac{dx}{dt}+y\frac{dy}{dt}=0$, we get $7\times0.7+24\times\frac{dy}{dt}=0$.

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

First, calculate $7\times0.7 = 4.9$. Then the equation becomes $4.9+24\times\frac{dy}{dt}=0$. Rearranging, we have $24\times\frac{dy}{dt}=- 4.9$, so $\frac{dy}{dt}=-\frac{4.9}{24}\approx - 0.20$ ft/s.

Answer:

$-0.20$ ft/s