posted by Olivis on .
A 20 foot ladder is sliding down a vertical wall at a constant rate of 2 feet per second.
a) How fast is the ladder moving away from the wall when the base of the ladder is 12 feet away from the wall?
b) Find the rate of change at which the angle between the ladder and the ground is changing when the base of the ladder is 16 feet from the wall.
Did you make your sketch?
foot of ladder is x ft from wall
top of ladder is y ft above ground
length of ladder (hypotenuse) is 20
x^2 + y^2 = 400
2x dx/dt + 2y dy/dt = 0
when x = 12
144 + y^2= 400
y = 16
24 dx/dt + 32(-2) = 0
dx/dt = 64/24 or 8/3 ft/s
b) let the angle be Ø
cosØ = x/20
x = 20cosØ
dx/dt = -20sinØ dØ/dt
when x= 16, cosØ = 16/20 = 4/5
then sinØ = 3/5 , (recognize the 3-4-5 right-angled triangle)
using a) when x= 16, y = 12 , dy/dt = -2
32 dxdt + 24(-2) = 0
dx/dt = 48/32 = 3/2
so in dx/dt = -20sinØ dØ/dt
3/2 = -20(3/5) dØ/dt
dØ/dt = (3/2)(5/3)(-1/20) = -1/8
so at that moment, the angle is decreasing at a rate of 1/8 radians/second
Just noticed that Steve had already answered the same question above.
At least we both ended up with the same answer.