Related Rates Ladder problem

A 10 foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the building at 3 ft/sec, how fast is the top of the ladder moving down when the foot of the ladder is 3 feet from the wall?

Length of Ladder, L = 10 ft.
Distance from Building, x = 3 ft.
dx/dt = 3 ft./sec.
Height on Wall = y

Find dy/dt:

y² = L² - x²
y² = (10)² - (3)²
y² = 100 - 9
y² = 91
y = √91
y = 9.54

y² = 100 - x²

Since the length of the ladder does not change, it is a constant and the derivative of a constant is zero, so

Differentiating Implicitly Over Time,

2y(dy/dt) = 0 - 2x(dx/dt)
y(dy/dt) = - x(dx/dt)
9.54(dy/dt) = - 3(3)
9.54(dy/dt) = - 9
dy/dt = - 9 / 9.54
dy/dt = - 0.944

The ladder is moving down the wall at a rate of 0.944 ft./sec. or 11.32 in./sec.
 

the shape of the ladder leaning against the wall is right triangle
let A= distance between the bottom of ladder and wall .
let B= the ladder length
let C = the distance from the top of the ladder right down to the ground
we have
B^2 = A^2 + C^2

C^2 = B^2- A^2

take derivative ===> 2C dc = 2Bdb - 2Ada ====> Cdc= Bdb- Ada

we are looking for dc :

dc = ( Bdb-Ada)/C . but we now that the length of the ladder is a constant , so Bdb=0

dc= -Ada/C
we know A= 3 . Da = 3ft/s what is C?
C = saqrt of B^2- A^2 = sqrt of ( 100 - 9 )= sqrt of 91 =====> C =9

dc = -3x3/9 = -1 ft/s

the - sign is correct because the distance is getting smaller