ap calc help is needed!!!?

Question

a 26ft ladder leans against a building so that its foot moves away from the building at the rate of 3ft/sec. when the foot of the ladder is 10ft from the building the top is moving down at the rate r ft/sec where r is
a. 46/3
b. 3/4
c. 5/4
d. 5/2
e. 4/5

Answer 1

h^2+x^2 = 26^2, where h is the height and x is the horizontal leg.

Differentiate both sides with respect to time,
2hh'+2xx' = 0

Solve for h',
h' = -xx'/h = -10(3)/√(26^2-10^2) = -5/4 ft/sec

|h'| = 5/4 ft/sec, the decreasing rate of the height of the ladder.

Answer 2

x^2 + y^2 = 676, y = (676 - x^2)^(1/2)
2xdx/dt + 2ydy/dt = 0
dy/dt = - (x/y)dx/dt
dy/dt = - (x/(676 - x^2)^(1/2))3ft/sec
dy/dt = - (10/(676 - 100)^(1/2))3ft/sec
dy/dt = - (10/(576)^(1/2))3ft/sec
dy/dt = - (10/24)3ft/sec
dy/dt = - (10/8)ft/sec
dy/dt = - 1.25 ft/sec
(up being +)