A spherical bath oil capsule dissolves in the bath so that its decrease in volume is proportional to its surface area. If its shape remains spherical as it dissolves, show that the radius of the capsule will decrease at a constant rate.
Thanks to anyone who helps
dv/dt = kA
v = 4/3 pi r^3
dv/dt = 4pi r^2 dr/dt = A dr/dt
so, k = dr/dt
To show that the radius of the capsule will decrease at a constant rate, we can use differential equations and solve it mathematically.
Let's assume that the initial radius of the capsule is r0. As the bath oil capsule dissolves, its volume decreases and its surface area decreases proportionally. Let's denote the radius of the capsule at any given time t as r.
The volume V of a sphere is given by the formula V = (4/3) * π * r^3, and the surface area A of a sphere is given by the formula A = 4 * π * r^2.
The decrease in volume is proportional to the surface area, which means that the rate of decrease of volume is directly proportional to the rate of decrease of surface area. Mathematically, we can express this as:
dV/dt = -k * dA/dt
Where k is a constant of proportionality.
To find the relationships between volume, surface area, and radius, we can differentiate the volume and surface area formulas with respect to time:
dV/dt = 4 * π * r^2 * dr/dt
dA/dt = 8 * π * r * dr/dt
Substituting these expressions into the proportional relationship, we have:
4 * π * r^2 * dr/dt = -k * 8 * π * r * dr/dt
Canceling out common factors, we get:
r^2 = -2k * r
Now we can solve this differential equation. Dividing both sides by r^2, we have:
1 = -2k
Simplifying, we find that k = -1/2.
Substituting this value of k back into the equation, we have:
r^2 = (-1/2) * r
Multiplying through by 2, we get:
2r^2 = -r
Rearranging the equation, we have:
2r^2 + r = 0
This is a quadratic equation, which can be factored as:
r(2r + 1) = 0
Since the radius of the capsule cannot be negative, the only valid solution is:
r = 0
Therefore, as the bath oil capsule dissolves, its radius decreases to zero at a constant rate.