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 0.3 ft/s. how fast is the top of the ladder sliding down the wall when the foot of the ladder is 24 ft from the wall? when the foot of the ladder is 24 ft from the wall, the top of the ladder is sliding down the wall at a rate of (round to two decimal places as needed.)

Explanation:

Step1: Establish the relationship

Let $x$ be the distance of the foot of the ladder from the wall and $y$ be the distance of the top of the ladder from the ground. By the Pythagorean theorem, $x^{2}+y^{2}=25^{2}=625$.

Step2: Differentiate with respect to time $t$

Differentiating both sides of the equation $x^{2}+y^{2}=625$ with respect to $t$, we get $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$. Then simplify it to $x\frac{dx}{dt}+y\frac{dy}{dt}=0$.

Step3: Find the value of $y$

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

Step4: Substitute known values

We know that $\frac{dx}{dt}=0.3$ ft/s, $x = 24$, and $y = 7$. Substitute these values into $x\frac{dx}{dt}+y\frac{dy}{dt}=0$. We have $24\times0.3+7\times\frac{dy}{dt}=0$.

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

First, calculate $24\times0.3 = 7.2$. Then the equation becomes $7.2+7\times\frac{dy}{dt}=0$. Rearranging to solve for $\frac{dy}{dt}$, we get $\frac{dy}{dt}=-\frac{7.2}{7}\approx - 1.03$ ft/s. The negative sign indicates that $y$ is decreasing.

Answer:

$1.03$ ft/s