Problem Statement
A ladder $5\text{ m}$ long is leaning against a vertical wall. The bottom of the ladder is pulled along the ground away from the wall at the rate of $2\text{ m/s}$. How fast is its height on the wall decreasing when the foot of the ladder is $4\text{ m}$ away from the wall?
Verified Solution & Marking Scheme
Model with Pythagorean Theorem
Let the distance of foot of ladder from wall be $x$, height on wall be $y$. Since ladder length is $5\text{ m}$:
$x^2 + y^2 = 5^2 = 25$
When $x = 4\text{ m}$: $4^2 + y^2 = 25 \implies y^2 = 9 \implies y = 3\text{ m}$.
Differentiate with Respect to Time t
$\frac{d}{dt}(x^2 + y^2) = \frac{d}{dt}(25) \implies 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \implies x\frac{dx}{dt} + y\frac{dy}{dt} = 0$
Substitute Values and Solve for dy/dt
Given $\frac{dx}{dt} = 2\text{ m/s}$, $x = 4$, $y = 3$:
$4(2) + 3\frac{dy}{dt} = 0 \implies 8 + 3\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{8}{3}\text{ m/s}$
The negative sign indicates height is decreasing.