A 30-ft ladder leaning against a wall begins to slide. How fast is the top of the ladder sliding down the wall at the instant of time when the bottom of the ladder is 18 ft from the wall and sliding away from the wall at the rate of 3 ft/sec?
Hint: Refer to the accompanying figure. By the Pythagorean theorem, . Find when x=18 and

a = 30; b = 3
A) 4ft/sec
B) ft/sec
C) ft/sec
D) 3 ft/ssec

Question

A 30-ft ladder leaning against a wall begins to slide. How fast is the top of the ladder sliding down the wall at the instant of time when the bottom of the ladder is 18 ft from the wall and sliding away from the wall at the rate of 3 ft/sec? ​
Hint: Refer to the accompanying figure. By the Pythagorean theorem,  . Find  when x=18 and

a = 30; b = 3
A) 4ft/sec 
B) ft/sec 
C) ft/sec 
D) 3 ft/ssec

Quizplus · Accepted Answer

We can use the Pythagorean theorem to find that the height of the ladder (the distance from the top of the ladder to the ground) is 24 ft when the bottom of the ladder is 18 ft from the wall.

Then, we can use the chain rule to find the rate of change of the height with respect to time (dh/dt) when the bottom of the ladder is moving away from the wall at 3 ft/sec:

a^2 + b^2 = c^2
30^2 + 24^2 = c^2
c = 36 ft

c^2 = a^2 + b^2
(36)^2 = (18 + dx)^2 + 24^2

Differentiate both sides with respect to time:

2(36)(dc/dt) = 2(18 + dx)(dx/dt) + 2(24)(dh/dt)

Simplify and plug in known values:

2(36)(-3) = 2(18 + 18)(-3) + 2(24)(dh/dt)
dh/dt = (216/48) ft/sec
dh/dt = 11.25 ft/sec (approx)

Therefore, the top of the ladder is sliding down the wall at a rate of approximately 11.25 ft/sec.

A 30-Ft Ladder Leaning Against a Wall Begins to Slide