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?

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!

Answer 3

No, we don't need to subtract the volumes of the cylinders. Let me explain the problem to you step by step.

We have a long empty cylinder with a radius of R = 25 cm. Inside this cylinder, we place a solid cylinder with a radius of r = 10 cm, such that the bases of the two cylinders are attached. This means that the height of the inner cylinder is the same as the height of the outer cylinder.

Next, water is poured into the remaining empty region of the outer cylinder until the height of the water becomes 30 cm. At this point, the radius of the inner cylinder starts increasing at a constant rate of 0.2 cm/s. We need to find the rate at which the height of the water is increasing when the radius of the inner cylinder reaches 15 cm.

To solve this problem, we can apply the principles of similar triangles. Let's call the height of the water h cm at a given time t seconds, and let's call the radius of the inner cylinder x cm at that time.

We know that the radius of the outer cylinder is R = 25 cm, and the initial radius of the inner cylinder is r = 10 cm. Since the height of the outer and inner cylinders is the same, we can write the following proportion:

(R + h) / R = (x + h) / x

Simplifying this equation, we get:

x(R + h) = R(x + h)

Expanding and rearranging, we get:

xR + xh = Rx + Rh

Now, substituting the initial values of R = 25 cm and r = 10 cm, we can simplify further:

10(25 + h) = 25(10 + h)

250 + 10h = 250 + 25h

Simplifying, we get:

15h = 0

This implies that h = 0. Therefore, the height of the water remains constant at 30 cm while the radius of the inner cylinder increases.

Therefore, at the moment the radius of the inner cylinder reaches 15 cm, the height of the water is not changing. The rate at which the height of the water is increasing is 0 cm/s.