A ladder 10 feet long is leaning against a wall. If the top of the ladder is sliding down the wall at 4 feet per second, how fast is the foot of the ladder being pulled away from the wall when the foot of the ladder is 8 feet away from the wall?
A) $6 \mathrm { tt } / \mathrm { s }$

B) $1 \mathrm { ft } / \mathrm { s }$
C) $3 \mathrm { ft } / \mathrm { s }$
D) $10 \mathrm { ft } / \mathrm { s }$
E) $5 \mathrm { ft } / \mathrm { s }$
F) $8 \mathrm { ft } / \mathrm { s }$

G) $2 \mathrm { ft } / \mathrm { s }$

H) $4 \mathrm { ft } / \mathrm { s }$

Question

A ladder 10 feet long is leaning against a wall. If the top of the ladder is sliding down the wall at 4 feet per second, how fast is the foot of the ladder being pulled away from the wall when the foot of the ladder is 8 feet away from the wall?
A) $6 \mathrm { tt } / \mathrm { s }$

B) $1 \mathrm { ft } / \mathrm { s }$
C) $3 \mathrm { ft } / \mathrm { s }$ 
D) $10 \mathrm { ft } / \mathrm { s }$ 
E) $5 \mathrm { ft } / \mathrm { s }$ 
F) $8 \mathrm { ft } / \mathrm { s }$

G) $2 \mathrm { ft } / \mathrm { s }$
 
H) $4 \mathrm { ft } / \mathrm { s }$

Quizplus · Accepted Answer

This problem can be solved using the Pythagorean theorem in a related rates context. Let $x$ be the distance from the foot of the ladder to the wall, $y$ the height of the ladder on the wall, and $l = 10$ feet the length of the ladder. We have $x^2 + y^2 = l^2$. Differentiating both sides with respect to time $t$, we get $2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0$. Given $\frac{dy}{dt} = -4$ ft/s (since $y$ is decreasing) and $x = 8$ ft, we can solve for $\frac{dx}{dt}$. Substituting $y = \sqrt{l^2 - x^2} = \sqrt{100 - 64} = 6$ ft, we get $2(8) \frac{dx}{dt} - 2(6)(4) = 0$, leading to $\frac{dx}{dt} = 3$ ft/s.

A Ladder 10 Feet Long Is Leaning Against a Wall 6tt/s6 \mathrm { tt } / \mathrm { s }6tt/s

A Ladder 10 Feet Long Is Leaning Against a Wall $6 \mathrm { tt } / \mathrm { s }$