Calculus Problem, object on rope

Question

An object with weight W is dragged along a horizontal plane by a force acting along a rope attached to the object. If the rope makes an angle t with the plane, then the magnitude of the force is F = (µW)/(µ sin (t) + cos(t)). where µ is a constant called the coefficient of friction. Let W = 50 lb and µ = 0.6.

(a) Find the rate of change of F with respect to 

F'(t) = _____

(b) When is this rate of change equal to zero?

t= ____

Answer 1

a)

F(x) = 30/(.6sin(t)+cos(t))

let u = .6sin(t)+cos(t) du/dt = .6cos(t)-sin(t)

F(u) = 30/u

F'(u) -30/u² du/dt

F'(t) = (-30/(.6sin(t)+cos(t))²) (.6cos(t)-sin(t)) = 30(sin(t)-.6cos(t))/(.6sin(t)+cos(t))²

b)

Only need to set the numerator to zero and solve for t.

30(sin(t)-.6cos(t)) = 0

sin(t) = .6cos(t)

tan(t) = .6

t = tan^-1(.6) ≈ 30.96°