Rates of Change Trig - How Fast an Angle Changes

A ladder 10 m long rests against a vertical wall wall. If the bottom of the ladder slides away from the wall at a speed of 2 m/s, how fast is the angle between the top of the ladder and the wall changing when the angle is 45 degrees?

The book tells me the answer is 2/ squareroot 5.

This is what I have so far:
x=10sin(a)
y=10cos(a)

And I need to somehow use that information with 2x(dx/dt) + 2y(dy/dt) = 0

Please use this method!!

Suppose that the distance between the bottom of the ladder and the wall is x. The angle between the top of the ladder and the wall is opposite to x and the hypotenuse, which is the length of the ladder, is 10.

Using SOHCAHTOA, if θ is the angle between the top of the ladder and the wall:
sinθ = opp/hyp = x/10.

By differentiating implicitly:
cosθ(dθ/dt) = (dx/dt)/10
==> dθ/dt = (dx/dt)/(10cosθ), by solving for dθ/dt.

Given that dx/dt = 2 and θ = 45°:
dθ/dt = 2/[10(√2/2)] = 2/(5√2) = √2/5 radians/second.

I hope this helps!

x = horizontal
y = vertical
r = 10
dx/dt = 2 m/s


x = rCos(θ)
y = rSin(θ)

dx/dt = -rSin(θ)(dθ/dt)
dy/dt = rCos(θ)(dθ/dt)

This is all you need.

Actually, all you need from this is the eq'n for dy/dt, because it contains the angle you're interested in.

=> 2 m/s = 10Cos(θ)(dθ/dt)
=> Cos(θ)(dθ/dt) = 1/5

note: Cos(45) = 1/√2

=> dθ/dt = √2/5