Question

A 20ft poll is resting on the ground. If the poll rotates upward at one end at a rate of 1degree/sec, then at what rate is the top of the poll rising from the ground when the poll is 30∘ with the ground?

Answer 1

The rate at which the top of the pole is rising from the ground when the pole is at 30 degrees with the ground is approximately -20√3/3°/sec.

Step-by-step explanation:

To solve this problem, we can use the concept of related rates. Let x represent the height of the top of the pole from the ground. Since the pole is rotating at a rate of 1 degree/sec, we can use the tangent function to relate the rate of change of x to the angle of elevation of the pole. Let's differentiate both sides of the equation to find the rate of change:

dx/dt = -20tan(θ)dθ/dt

Now, when the pole is at 30 degrees with the ground, we can plug in the values into the equation:

dx/dt = -20tan(30°)(1°/sec) = -20(√3/3)(1°/sec) = -20√3/3°/sec

Therefore, the rate at which the top of the pole is rising from the ground when the pole is at 30 degrees with the ground is approximately -20√3/3°/sec.

Answer 2

The rate at which the top of the pole is rising from the ground when the pole is at an angle of 30 degrees with the ground is (pi/180) ft/sec.

Step-by-step explanation:

To solve this problem, we can use the concept of related rates. Let's denote the height of the top of the pole from the ground as 'h'. We are given that the pole is rotating at a rate of 1 degree/sec. We need to find the rate at which the top of the pole is rising from the ground when the pole is at an angle of 30 degrees with the ground.

First, let's convert the angle from degrees to radians: 30 degrees * (pi/180) radians/degree = (pi/6) radians.

Now, let's draw a right triangle with the pole as the hypotenuse, the height of the top of the pole as the opposite side, and the distance from the bottom of the pole to the point directly below the top of the pole as the adjacent side. By trigonometry, we know that tan(angle) = opposite/adjacent.

tan(pi/6) = h/20ft.

Simplifying, we get sqrt(3)/3 = h/20.

Now, let's differentiate both sides of the equation with respect to time t: (d/dt)(sqrt(3)/3) = (d/dt)(h/20).

Since the rate of change of the angle is given as 1 degree/sec, we have (d/dt)(sqrt(3)/3) = (1 degree/sec)*(pi/180 radians/degree) = (pi/180) radians/sec.

Now, let's solve for (d/dt)(h/20):

(d/dt)(h/20) = (d/dt)(sqrt(3)/3) * (d/dt)(1/20).

Substituting the values we have, (pi/180) = (d/dt)(1/20).

Therefore, (d/dt)(h/20) = (pi/180).

So, the rate at which the top of the pole is rising from the ground is (pi/180) ft/sec.