Help me with my related rates Calculus problem please!?

Question

Imagine a railroad crossing gate. For purposes of this problem we are going to treat the arm of the gate as a line that pivots on the vertical pole via a round gear whose center is exactly 4 feet off the ground. The distance between the tip of the arm and the center of the gear is exactly 28 ft. When the arm is being lowered the angle of elevation of the arm decreases at a constant rate of 6 degrees/sec. Find the rate at which the tip of the arm approaches the ground (vertically) at the instant the angle of elevation of the arm is 30 degrees.

Answer 1

Define: E(t) = Angle of elevation (as a function of time, t)
Define: h(t) = height of tip of the arm (as a function of time, t)

Given:

dE/dt = - 6 degrees / second = - pi / 30 degrees / second
h(t) = 4 + 28 sin(E(t)) feet

Related rate derivative:

dh/dt = 28 cos(E(t)) dE/dt = 28 cos(E(t)) (- pi / 30) = - (14 pi / 15) cos(E(t))

We want the value of this when E(t) = 30 degrees or pi / 6 radians, when cos(E(t)) = cos(pi/6) = (1/2) sqrt(3).

dh/dt = - (14 pi / 15) (1/2) sqrt(3) = - 14 pi sqrt(3) / 30
or approx. - 2.539 feet/second

Dan

p.s. As kirchwey states you don't need to do this as a related rate. If the bar is vertical until it starts to descend at time t = t0, then E(t) = pi/2 - (pi/30)(t - t0), and h(t) = 4 + 28 sin(E(t)) becomes

h(t) = 4 + 28 sin( pi/2 - (pi/30)(t - t0) )

So you can compute dh/dt directly.

Answer 2

I assume the gear with attached arm pivots on a horizontal axle attached to the vertical pole, right? The vertical speed is omega (angular rate in rad/s) * R * cos(theta) =
6*pi/180 * 28 * cos(30deg) ft/s.
I don't see a related rate problem or a need for calculus here, and the height of the center of rotation seems irrelevant. All you do is find the vertical component of the tangential speed of the tip. If there's more information that makes it trickier, please add it.