A uniform layer of ice covers a spherical water-storage tank. As the ice melts, the volume V of ice decreases at a rate that varies directly as the surface area A. Show that the outside radius decreases at a constant rate.

Question

A uniform layer of ice covers a spherical water-storage tank. As the ice melts, the volume V of ice decreases at a rate that varies directly as the surface area A. Show that the outside radius decreases at a constant rate.

Answer 1

Consider:

r = the radius of the tank
R = the radius of the combined sphere
k, k' = constants

Volume of the layer of ice = Volume of the large sphere - Volume of the small sphere
= (4π/3)*(R^3 - r^3)

Now, it's given that the rate of decrease of this volume is proportional to the surface area, which is 4π(R^2)

=> d[(4π/3)*(R^3 - r^3)]/dt = k*4π(R^2)
=> (4π/3)*[d(R^3)/dt - d(r^3)/dt] = k*4π(R^2)

r is a constant term, and the derivative of its cube is hence zero.

=> d(R^3)/dt = k' * (R^2) (Where k' is another constant including pi and the other numbers)
=> 3(R^2)*(dR/dt) = k' * (R^2) (Using the product rule for differenciation)
=> dR/dt = k' / 3

The R.H.S is constant.

Answer 2

or alternatively

Given
d Vice/dt = - c * 4pi r^2
but if the thickness h is << r then Vice = 4 pi r^2 h
then
dh/dt = (1/4pi r^2) dVice/dt
so
dh/dt = (1/4pi r^2)(-c * 4 pi r^2)
dh/dt = -c

Answer 3

To show that the outside radius decreases at a constant rate as the volume of ice decreases, we need to express the volume and surface area of the ice layer in terms of the outside radius.

Let's denote the outside radius of the spherical water-storage tank as R and the radius of the ice layer as r.

The volume V of the ice layer can be expressed as the difference between the volume of the spherical tank with radius R and the volume of the hollow space inside the tank with radius r. The volume of a sphere is given by the formula V = (4/3)πr^3.

So, the volume of the ice layer can be written as:

V = (4/3)π(R^3 - r^3) (1)

The surface area A of the ice layer is equal to the curved surface area of the spherical tank:

A = 4πR^2 (2)

Given that the rate of change of volume V is directly proportional to the surface area A, we can write:

dV/dt = k * A

where dV/dt is the rate of change of volume, k is the constant of proportionality, and A is the surface area.

Differentiating equation (1) with respect to time (t), we get:

dV/dt = 4π(R^2 - r^2) * (dR/dt - dr/dt) (3)

Substituting equation (2) into equation (3), we have:

4π(R^2 - r^2) * (dR/dt - dr/dt) = k * 4πR^2

Simplifying the equation further, we get:

(R^2 - r^2) * (dR/dt - dr/dt) = kR^2

Expanding the equation, we have:

R^2 * dR/dt - r^2 * dR/dt - R^2 * dr/dt + r^2 * dr/dt = kR^2

Combining similar terms, we get:

R^2 * dR/dt - r^2 * dR/dt = R^2 * dr/dt - r^2 * dr/dt

Factoring out dR/dt and dr/dt, we obtain:

(dR/dt)(R^2 - r^2) = (dr/dt)(R^2 - r^2)

Since (R^2 - r^2) is not equal to zero (since there is an ice layer), we can divide both sides by (R^2 - r^2):

dR/dt = dr/dt

Therefore, we can conclude that the outside radius decreases at a constant rate as the volume of ice decreases.

Answer 4

To show that the outside radius decreases at a constant rate as the volume of ice decreases, we need to establish a relationship between the volume of ice, the surface area of the ice, and the radius.

Let's denote the volume of ice as V, the surface area as A, and the outside radius as r. We are given that the rate at which the volume of ice decreases is directly proportional to the surface area, which can be written as:

V' = kA,

where V' is the rate of change of the volume of ice and k is a constant of proportionality.

For a sphere, the volume is given by the formula:

V = (4/3)πr^3.

We can differentiate this equation with respect to time (t) to find the rate of change of volume:

V' = dV/dt = 4πr^2(dr/dt).

Since the volume of ice is decreasing, V' is negative. Therefore, we can rewrite the equation as:

-4πr^2(dr/dt) = kA.

Now, let's relate the surface area to the radius. The surface area of a sphere is given by:

A = 4πr^2.

We can solve this equation for r^2:

r^2 = A / (4π).

Substituting this expression into the equation for the rate of change of volume, we get:

-4π(A / (4π))(dr/dt) = kA.

Simplifying the equation, we find:

-1/4 (dr/dt) = k.

This equation shows that the rate of change of the outside radius, dr/dt, is constant (denoted by k).

Therefore, we have shown that as the volume of ice decreases at a rate proportional to the surface area, the outside radius of the spherical water-storage tank decreases at a constant rate.

A uniform layer of ice covers a spherical​ water-storage tank. As the ice​ melts, the volume V of ice decreases at a rate that varies directly as the surface area A. Show that the outside radius decreases at a constant rate.

Consider:

or alternatively

To show that the outside radius decreases at a constant rate as the volume of ice decreases, we need to express the volume and surface area of the ice layer in terms of the outside radius.

To show that the outside radius decreases at a constant rate as the volume of ice decreases, we need to establish a relationship between the volume of ice, the surface area of the ice, and the radius.

A uniform layer of ice covers a spherical water-storage tank. As the ice melts, the volume V of ice decreases at a rate that varies directly as the surface area A. Show that the outside radius decreases at a constant rate.