Asked by Alex
A ladder 30 ft. long is leaning up against a building. If the top of the ladder is being pulled up the wall of the building at a rate of 1.5 feet per minute, find the rate at which the base of the ladder is moving toward the building when it is 18 feet from the wall.
Answers
Answered by
Steve
we have
x^2+y^2 = 30
so,
2x dx/dt + 2y dy/dt = 0
when x=18, y=24
2*18 dx/dt + 2*24 dy/dt = 0
. . .
x^2+y^2 = 30
so,
2x dx/dt + 2y dy/dt = 0
when x=18, y=24
2*18 dx/dt + 2*24 dy/dt = 0
. . .
Answered by
Alex
So what's next to find the rate at which the ladder is moving toward the building when it is 18 ft from the wall?
Answered by
Alex
36 dx/dt + 48 dy/dt = 0
48 dy/dt = -36 dx/dt
dy/dt = -.75 or -3/4
48 dy/dt = -36 dx/dt
dy/dt = -.75 or -3/4
Answered by
Alex
Is this right?
1.5 ft = dr/dt
2x dx/dt + 2y dy/dt = 0
2(18)(2) + 2(24)= 0
72 + 48 dy/dt = 0
48 dy/dt = -72
dy/dt = -72/48 = -3/4 feet per minute
1.5 ft = dr/dt
2x dx/dt + 2y dy/dt = 0
2(18)(2) + 2(24)= 0
72 + 48 dy/dt = 0
48 dy/dt = -72
dy/dt = -72/48 = -3/4 feet per minute
Answered by
Alex
-3/2 feet per minute I meant
Answered by
Steve
You need to read the question more carefully. They gave you dy/dt, and you need to find dx/dt.
2x dx/dt + 2y dy/dt = 0
when x=18, y=24
The ladder is being pulled up the wall at 1.5 ft/min. That means dy/dt = 3/2
Plugging all that into the equation gives
2(18) dx/dt + 2(24)(3/2) = 0
36 dx/dt = -72
dx/dt = -2
As expected, the distance of the base of decreases as the top of the ladder gets pulled up.
2x dx/dt + 2y dy/dt = 0
when x=18, y=24
The ladder is being pulled up the wall at 1.5 ft/min. That means dy/dt = 3/2
Plugging all that into the equation gives
2(18) dx/dt + 2(24)(3/2) = 0
36 dx/dt = -72
dx/dt = -2
As expected, the distance of the base of decreases as the top of the ladder gets pulled up.
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