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 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

- 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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