Posted by Olivis on Sunday, October 21, 2012 at 11:46pm.
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.

Math Calculus  Reiny, Monday, October 22, 2012 at 8:49am
Did you make your sketch?
label:
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 345 rightangled 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

Arggg  Math Calculus  Reiny, Monday, October 22, 2012 at 8:53am
Just noticed that Steve had already answered the same question above.
At least we both ended up with the same answer.
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