A ladder 10 ft long rests against a wall. If the bottom of the ladder slides away at a rate of 0.9 ft/s, how fast is the angle changing when the bottom of the ladder is 6 ft from the wall?

Answer 1

Tough to explain without a photo, but I will try.

This is a 3-4-5 triangle (6-8-10 reduces to 3-4-5). For simplicity, I am going to use 3-4-5. Solve for theta now - you'll need it later.

a^2+b^2=c^2

Do implicit with respect to t.

2a da/dt + 2b db/dt = 2c dc/dt, simplify

a da/dt + b db/dt = c dc/dt. dc/dt cannot be changing since the ladder cannot be getting taller/shorter

a da/dt + b db/dt = 0; db/dt = .9 ft/s, solve for da/dt.

Once you have da/dt, you can solve for dtheta/dt.

tan(theta) = opp/hyp = a/b= ab^-1

implicit derivative of both sides with respect to t

sec^2(theta) dtheta/dt = da/dt*b^-1 - a*b^-2 db/dt

Your only known is dtheta/dt since you know theta, a, b, da/dt, and db/dt. Solve.

Feel free to jump on a call if you have questions - this is a bit tough to explain via message board!