Calculus help

Question

1. A Ferris wheel with a radius of 10m is rotating at a rate of one revolution every 2 minutes. How fast is a rider rising when the rider is 18m above ground level?

2. Sand falls from a conveyor belt onto a conical pile at a rate of 10m^3 per minute. If the height of the pile is 3/8 of the base radius, how fast is the radius changing when the pile is 4 meters high? You can answer in the rational form in terms of π; if you choose to write as a decimal, round the answer to the thousandths column

Answer 1

1. Ferris Wheel

 

The height of a rider on a Ferris wheel is

y(t) = h + r + r sinθ(t) = h + r[1 + sinθ(t)]

where h is the clearance above the ground of the Ferris wheel.

 

The problem does not include the height of the wheel at its lowest point, so h = 0.

 

For this problem

 

y(t) = 10 [1 + sinθ(t)]

dy/dt = 10 cosθ dθ/dt = 10 cosθ π/min

 

When the rider is 18m above ground level

y = 10 [1 + sinθ] = 18

⇒ sinθ = 0.8

⇒ cosθ = 0.6 {sin²θ + cos²θ = 1, and the rider is rising}

 

Therefore, at 18m above ground level, the rider is rising at the rate

dy/dt =10 (0.6) π = 6π m/min or π/10 m/sec.

 

2. Radius changing rate.

 

The volume of a cone is

V = π/3 R²h = π/3 R²(3/8)R = π/8 R³

 

The change in volume is dV/dt = 3π/8 R² dR/dt

⇒ dR/dt = 8/(3πR²) dV/dt

 

When h = 4 m, R = 8/3 h = 8/3 (4) = 32/3 m.

 

So the radius of the conical pile, when h = 4 m, is increasing at the rate

dR/dt = 8/[3π(32/3)²] dV/dt

         = 24/(1024π)10

         = 15/(64π) m/min or 0.075 m/min