6. a 25 - ft ladder is leaning against a wall. if we push the ladder toward the wall at a rate of 1 ft/sec…

Explanation:

Step1: Establish the Pythagorean - relation

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, $x^{2}+y^{2}=25^{2}=625$.

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 initial values of $x$ and $y$

Initially, $x = 20$ ft. Substituting $x = 20$ into $x^{2}+y^{2}=625$, we get $20^{2}+y^{2}=625$, so $y=\sqrt{625 - 400}=\sqrt{225}=15$ ft.

Step4: Determine the values of $x$ and $\frac{dx}{dt}$ after 5 seconds

The bottom of the ladder is pushed toward the wall at a rate of $\frac{dx}{dt}=- 1$ ft/sec (negative because $x$ is decreasing). After $t = 5$ seconds, $x=20-5\times1 = 15$ ft.

Step5: Find the value of $y$ when $x = 15$

Substitute $x = 15$ into $x^{2}+y^{2}=625$, we have $15^{2}+y^{2}=625$, so $y=\sqrt{625 - 225}=\sqrt{400}=20$ ft.

Step6: Solve for $\frac{dy}{dt}$

Substitute $x = 15$, $y = 20$, and $\frac{dx}{dt}=-1$ into $x\frac{dx}{dt}+y\frac{dy}{dt}=0$. We get $15\times(-1)+20\times\frac{dy}{dt}=0$. Then $20\times\frac{dy}{dt}=15$, and $\frac{dy}{dt}=\frac{15}{20}=\frac{3}{4}$ ft/sec.

Answer:

$\frac{3}{4}$ ft/sec