Question

Inside a long empty cylinder with radius R = 25 cm is put a long solid cylinder with radius r = 10 cm such that the bases of the two cylinders are attached. Then water is poured in the remaining empty region of the first cylinder until the height of the water becomes 30 cm (and then no other water is poured in the cylinder). After that, the radius of the inner solid cylinder starts to grow larger at a constant rate of 0.2 cm/s (and meanwhile the height of the water increases). How fast is the height of the water increasing at the moment the radius of the inner cylinder has become 15 cm?


I don't get the question very well should we subtract the volume of the big cylinder from the small one?

Answer 1

In the real world, one would normally subtract a smaller volume from a larger one. So, the volume in question is

v = πh(R^2-r^2)

when h=30 and r=10,
v = π(625-100)(30) = 15750π cm^3

since the volume of water does not change, dv/dt = 0. So,

(R^2-r^2) dh/dt + h(-2r dr/dt) = 0
when r=15,

(625-225) dh/dt + h(-30*0.2) = 0
but what is h when r=15? It is

15750π/400π = 39.375

So,

400π dh/dt = 39.375(6) = 236.25
dh/dt = 0.188 cm/s

Answer 2

Thank you!