A small stone is dropped into a still pool, creating a circular ripple that moves outward from the point of impact at a constant rate

Question

How is the area enclosed by the ripple changing after 2 s if the circumference of the ripple is changing at a rate of 2π m/s at this instant?

Answer 1

given: dC/dt=2π find dA/dt at t=2 seconds. Note that Circumference = πD =C π and Area =πD²/4 =A where D=diameter. Note further that dA/dt=πD/2*(dD/dt) and dC/dt=π(dD/dt). Combine to find dD/dt= (1/π)dC/dt and dA/dt = πD. To find D at t=2seconds note that dC/dt=2π m/s =, a constant speed. To find distance use distance = speed * time thus C= 2πt = 4π when t=2 seconds noting that requires D=4 m. Now substitute this into dA/dt to find dA/dt= 4π m2/sec

Answer 2

A = πr²; C = 2πr; dC/dt = 2πdr/dt = 2π; dr/dt = 1 m/s;

dA/dt = 2πrdr/dt = 2πrdr/dt and we need r at t =2s. They not give us radius r at t = 2s to get dA/dt at this moment