Calculus... Related Rates and Differentiation?

Question

A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/sec, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 ft from the wall?

Answer 1

Let
x = horizontal distance from wall
y = vertical height from floor

dx/dt = 1 ft/sec

Find dy/dt when x = 6 ft.

We have:

x² + y² = 100

Differentiate implicitly.

2x(dx/dt) + 2y(dy/dt) = 0
2y(dy/dt) = -2x(dx/dt)

dy/dt = [-2x(dx/dt)] / (2y) = [-x(dx/dt)] / [√(100 - x²)]

dy/dt = -6*1 / √(100 - 36) = -6/√64 = -6/8 = -3/4 ft/sec

Answer 2

x^2+y^2=100

derivate
2x dx/dt +dy/dt =0
then u get have to do sqrt of 10^2 +sqrt of 5^2 and u get 5root(5)
now all u do is plug in