A spherical balloon is being inflated so that, at the time when the radius is 5m, the radius is increasing at the rate of 0.15m/s.
(A) Find the rate of increase of the volume at this time.
(B) If the rate of increase of volume thereafter remains constant, find the rate at which the radius is increasing when the balloon has a diameter of 12m.
Please help, totally lost.
There are two ways of thinking about this:
1. Straight math
V = (4/3) pi r^3
dV/dr = 4 pi r^2
2. Common sense
dV = surface area * dr
dV = 4 pi r^2 dr
anyway
dV/dt = dV/dr dr/dt
dV/dt = 4 pi (25) (.15)
if D = 12 then r = 6
dr = dV/4 pi r^2
dr/dt = (1/4*36 pi ) dV/dt
To answer these questions, we need to use the formulas for the volume of a sphere and the relationship between the radius and volume of a sphere. Let's go step by step to find the solutions:
(A) Find the rate of increase of the volume at this time:
The volume of a sphere is given by the formula V = (4/3)πr³, where V is the volume and r is the radius.
We are given that the radius is increasing at a rate of 0.15 m/s when the radius is 5 m. So, we have dr/dt = 0.15 m/s and r = 5 m.
To find the rate of increase of the volume, we need to differentiate the volume formula with respect to time (t) since we are given the rate of change of r with respect to t (dr/dt).
dV/dt = d/dt [(4/3)πr³]
To differentiate, we can use the chain rule. Differentiating the right side of the equation, we get:
dV/dt = 4π(d/dt (r³))/(3)
Now, let's calculate d/dt (r³) using the chain rule:
d/dt (r³) = 3r²(dr/dt)
Now, substituting these values into our equation for dV/dt, we get:
dV/dt = 4π(3r²(dr/dt))/(3)
dV/dt = 4πr²(dr/dt)
Substituting the given values, we have r = 5 m and dr/dt = 0.15 m/s:
dV/dt = 4π(5²)(0.15)
dV/dt = 150π m³/s (Answer A)
So, at that time the rate of increase of the volume is 150π cubic meters per second.
(B) Find the rate at which the radius is increasing when the balloon has a diameter of 12m:
We are given the radius when the balloon has a diameter of 12 m. The radius is half the diameter, so r = 12/2 = 6 m.
We also know that the rate of increase of the volume thereafter remains constant. Let's denote this rate as dV/dt (constant).
Using the relationship between the volume and the radius, we have:
dV/dt = 4πr²(dr/dt)
We want to find the rate at which the radius is increasing, dr/dt, when the balloon has a diameter of 12 m. We know r = 6 m and dV/dt = 150π m³/s (as found in part A).
Substituting these values into the formula, we have:
150π = 4π(6²)(dr/dt)
Simplifying the equation, we can solve for dr/dt:
dr/dt = 150π / (4π(6²))
dr/dt = 150π / (4π(36))
dr/dt = 150 / 4
dr/dt = 37.5 m/s (Answer B)
So, when the balloon has a diameter of 12 m, the rate at which the radius is increasing is 37.5 m/s.