a 5 m ladder leans against a house on a flat ground. the base of the ladder slides away from the house at 1…

Explanation:

Step1: Set up the right - triangle relationship

Let the height of the top of the ladder on the house be $y$, the distance of the base of the ladder from the house be $x$, and the length of the ladder $L = 5$m. By the Pythagorean theorem, $x^{2}+y^{2}=L^{2}=25$.

Step2: Differentiate with respect to time $t$

Differentiating both sides of $x^{2}+y^{2}=25$ with respect to $t$ gives $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$. We know that $\frac{dx}{dt}=1$m/s.

Step3: Find $y$ when $x = 4$

When $x = 4$, from $x^{2}+y^{2}=25$, we have $y=\sqrt{25 - x^{2}}=\sqrt{25 - 16}=3$m.

Step4: Substitute values into the differentiated equation

Substitute $x = 4$, $y = 3$, and $\frac{dx}{dt}=1$ into $x\frac{dx}{dt}+y\frac{dy}{dt}=0$. So, $4\times1+3\times\frac{dy}{dt}=0$.

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

We can solve the equation $4 + 3\frac{dy}{dt}=0$ for $\frac{dy}{dt}$. First, subtract 4 from both sides: $3\frac{dy}{dt}=-4$. Then, $\frac{dy}{dt}=-\frac{4}{3}$m/s.

Step6: Find the angle $\theta$ between the ladder and the ground

Let $\theta$ be the angle between the ladder and the ground. Then $\tan\theta=\frac{y}{x}$.

Step7: Differentiate $\tan\theta$ with respect to $t$

We know that $\sec^{2}\theta\frac{d\theta}{dt}=\frac{x\frac{dy}{dt}-y\frac{dx}{dt}}{x^{2}}$. Also, $\sec\theta=\frac{L}{x}=\frac{5}{4}$ when $x = 4$.

Step8: Substitute values to find $\frac{d\theta}{dt}$

Substitute $x = 4$, $y = 3$, $\frac{dx}{dt}=1$, and $\frac{dy}{dt}=-\frac{4}{3}$ into $\sec^{2}\theta\frac{d\theta}{dt}=\frac{x\frac{dy}{dt}-y\frac{dx}{dt}}{x^{2}}$. $\left(\frac{5}{4}\right)^{2}\frac{d\theta}{dt}=\frac{4\times\left(-\frac{4}{3}\right)-3\times1}{4^{2}}$. First, simplify the right - hand side: $\frac{4\times\left(-\frac{4}{3}\right)-3\times1}{4^{2}}=\frac{-\frac{16}{3}-3}{16}=\frac{-\frac{16 + 9}{3}}{16}=\frac{-\frac{25}{3}}{16}=-\frac{25}{48}$. Then, since $\left(\frac{5}{4}\right)^{2}=\frac{25}{16}$, we have $\frac{25}{16}\frac{d\theta}{dt}=-\frac{25}{48}$. Solving for $\frac{d\theta}{dt}$, we get $\frac{d\theta}{dt}=-\frac{1}{3}$ rad/s.

Answer:

$-\frac{1}{3}$ rad/s