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
. . .
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.
6 answers
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?
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
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
-3/2 feet per minute I meant
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.