The base of a 35-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 21 feet from the wall ?
A) $9 \frac { \mathrm { ft } } { \mathrm { s } }$
B) $48 \frac { \mathrm { ft } } { \mathrm { s } }$
C) $36 \frac { \mathrm { ft } } { \mathrm { s } }$
D) $16 \frac { \mathrm { ft } } { \mathrm { s } }$
E) $3 \frac { \mathrm { ft } } { \mathrm { s } }$

Question

The base of a 35-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 21 feet from the wall  ?
A) $9 \frac { \mathrm { ft } } { \mathrm { s } }$
B) $48 \frac { \mathrm { ft } } { \mathrm { s } }$
C) $36 \frac { \mathrm { ft } } { \mathrm { s } }$
D) $16 \frac { \mathrm { ft } } { \mathrm { s } }$
E) $3 \frac { \mathrm { ft } } { \mathrm { s } }$

Quizplus · Accepted Answer

To solve this problem, we can use the Pythagorean theorem in a related rates context. Let $x$ be the distance from the wall to the base of the ladder, and $y$ be the height of the ladder on the wall. Given that the ladder's length is 35 feet, we have $x^2 + y^2 = 35^2$. Differentiating both sides with respect to time $t$, we get $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$. When $x = 21$ feet, $y = \sqrt{35^2 - 21^2} = \sqrt{1225 - 441} = \sqrt{784} = 28$ feet. Given $\frac{dx}{dt} = 12$ ft/s, we substitute the values into the differentiated equation to solve for $\frac{dy}{dt}$: $2(21)(12) + 2(28) \frac{dy}{dt} = 0$, which simplifies to $\frac{dy}{dt} = -9$ ft/s. The negative sign indicates the direction of motion (down the wall), so the speed is $9$ ft/s.

The Base of a 35-Foot Ladder Is Being Pulled Away 9fts9 \frac { \mathrm { ft } } { \mathrm { s } }9sft​

The Base of a 35-Foot Ladder Is Being Pulled Away $9 \frac { \mathrm { ft } } { \mathrm { s } }$