The base of a 25-foot ladder is being pulled away from a wall at a rate of 12 feet per second. How fast is the top of the ladder sliding down the wall at the instance when the base of the ladder is 15 feet from the wall?
A)36 ft/s
B)3 ft/s
C)48 ft/s
D)16 ft/s
E)9 ft/s

Question

Quizplus · Accepted Answer

To solve this problem, we can use the Pythagorean theorem to relate the sides of the right triangle formed by the wall, the ground, and the ladder. Let $x$ be the distance from the wall to the base of the ladder, $y$ be the height of the ladder on the wall, and $l = 25$ feet be the length of the ladder. We have $x^2 + y^2 = l^2$.Given that $x = 15$ feet when we want to find the rate at which $y$ is changing ($\frac{dy}{dt}$), and we know $\frac{dx}{dt} = 12$ ft/s, we differentiate both sides of the equation with respect to $t$ to get $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$.Plugging in the known values, $2(15)(12) + 2y\frac{dy}{dt} = 0$. To find $y$ when $x = 15$, we use the Pythagorean theorem: $15^2 + y^2 = 25^2$, which gives $y = 20$ feet.Substituting $x = 15$, $y = 20$, and $\frac{dx}{dt} = 12$ ft/s into the differentiated equation, we get $2(15)(12) + 2(20)\frac{dy}{dt} = 0$, which simplifies to $360 + 40\frac{dy}{dt} = 0$. Solving for $\frac{dy}{dt}$, we find $\frac{dy}{dt} = -9$ ft/s. The negative sign indicates that the top of the ladder is sliding down the wall, and the rate is $9$ ft/s.

The Base of a 25-Foot Ladder Is Being Pulled Away