Question

the top of a ladder slides down a vertical wall at a rate of 0.25 m/s. at the moment when the bottom of the ladder is 5 m from the wall, it slides away from the wall at a rate of 0.6 m/s. how long is the ladder?

m

Explanation:

Step1: Establish the relationship

Let $x$ be the distance of the bottom of the ladder from the wall, $y$ be the distance of the top of the ladder from the ground, and $L$ be the length of the ladder. By the Pythagorean theorem, $x^{2}+y^{2}=L^{2}$. Since the length of the ladder is constant, differentiating both sides with respect to time $t$ gives $2x\frac{dx}{dt}+2y\frac{dy}{dt} = 0$.

Step2: Substitute known values

We know that $\frac{dx}{dt}=0.6$ m/s, $\frac{dy}{dt}=- 0.25$ m/s (negative because $y$ is decreasing), and $x = 5$ m. Substituting into $x\frac{dx}{dt}+y\frac{dy}{dt}=0$, we have $5\times0.6+y\times(-0.25)=0$. Solving for $y$:
\[

$$\begin{align*} 3-0.25y&=0\\ 0.25y&=3\\ y& = 12 \end{align*}$$

\]

Step3: Calculate the length of the ladder

Using the Pythagorean theorem $L=\sqrt{x^{2}+y^{2}}$. Substitute $x = 5$ m and $y = 12$ m into it. Then $L=\sqrt{5^{2}+12^{2}}=\sqrt{25 + 144}=\sqrt{169}=13$ m.

Answer:

13