Question

Your neighbor is growing a slightly different watermelon. It also has a rind whose thickness is one tenth of the radius of that watermelon.

However, the rind of your neighbor's water melons grows at a constant rate of 20 cubic centimeters a week. The radius of your neighbor's
watermelon after 5 weeks is_____ centimeters and at that time it is growing at _____ centimeters per week.

Answer 1

Let's denote the radius of your neighbor's watermelon as r(t), where t is the time in weeks. We know that the thickness of the rind is one-tenth of the radius, so the inner radius of the watermelon (i.e., the radius of the edible part) is 9/10 of the total radius.

The volume of the watermelon can be expressed as the difference between the volumes of the outer sphere and the inner sphere:

V = (4/3)πr(t)^3 - (4/3)π(9/10r(t))^3

= (4/3)πr(t)^3 - (27/1000)πr(t)^3

= (1073/750)πr(t)^3

The rate at which the radius is growing is given by the derivative of r(t) with respect to t:

r'(t) = 20/((4/3)π(1.1r(t))^2)

= 750/221πr(t)^2

We can use this expression to find the rate at which the radius is growing after 5 weeks:

r'(5) = (750/221π)r(5)^2

We don't have enough information to solve for r(5) exactly, but we can use the fact that the rind grows at a constant rate of 20 cubic centimeters a week. The volume of the rind is given by the difference between the volumes of the outer sphere and the inner sphere with radii r(t) and 9/10r(t), respectively:

V_rind = (4/3)πr(t)^3 - (4/3)π(9/10r(t))^3

= (1073/750)πr(t)^3 - (243/1000)πr(t)^3

= (673/750)πr(t)^3

The rate at which the rind is growing is given by the derivative of V_rind with respect to t:

dV_rind/dt = (673/250)πr(t)^2 dr/dt

We know that dV_rind/dt = 20, so we can solve for dr/dt:

dr/dt = 20 / ((673/250)πr(t)^2)

We still don't know the value of r(5), but we can use the fact that the thickness of the rind is one-tenth of the radius to relate the volume of the rind to the radius:

V_rind = (1/10)³ (4/3)πr(t)^3

= (4/3)πr(t)^3/1000

We know that V_rind = 20(5) = 100, so we can solve for r(5):

(4/3)πr(5)^3/1000 = 100

r(5) = 1000(750/221π)^(1/3)

This gives an approximate value of r(5) ≈ 13.825 centimeters. We can use this value to find the rate at which the radius is growing:

r'(5) = (750/221π)r(5)^2

≈ 4.942 centimeters per week

Therefore, after 5 weeks, your neighbor's watermelon has a radius of approximately 13.825 centimeters and is growing at a rate of approximately 4.942 centimeters per week