A spherical balloon is being inflated and its radius is changing at the rate of 2m/minute. When the radius is 4m, find the rate at which the:
(a).volume is increasing
V= 3/4 pie r^2
(b). surface area is increasing
A = 4 pie r^2
Your formula for volume is incorrect.
V = (4/3)πr^3
dV/dt = 4πr^2 dr/dt
so when r = 4
dV/dt = 4π(16)(2) = 72π cm^3/minute
A = 4πr^2
dA/dt = 8πr dr/dt
when r=4
dA/dt = 8π(4)(2) = 64π cm^2/min
To find the rate at which the volume is increasing when the radius is 4m, we can differentiate the volume formula with respect to time:
V = (3/4)πr^3
First, differentiate both sides of the equation with respect to time (t):
dV/dt = d/dt[(3/4)πr^3]
Using the chain rule, we can differentiate the right side with respect to r and multiply it by the rate at which the radius is changing, which is given as 2m/minute:
dV/dt = (3/4)π * 3r^2 * (dr/dt)
Substituting the values given:
dV/dt = (3/4)π * 3(4^2) * 2
dV/dt = (3/4)π * 3 * 16 * 2
dV/dt = 72π m^3/minute
Therefore, the volume is increasing at a rate of 72π m^3/minute when the radius is 4m.
To find the rate at which the surface area is increasing when the radius is 4m, we can differentiate the surface area formula with respect to time:
A = 4πr^2
Differentiating both sides with respect to time (t):
dA/dt = d/dt[4πr^2]
Using the chain rule:
dA/dt = 4π * 2r * (dr/dt)
Substituting the values given:
dA/dt = 4π * 2(4) * 2
dA/dt = 4π * 8
dA/dt = 32π m²/minute
Therefore, the surface area is increasing at a rate of 32π m²/minute when the radius is 4m.
To find the rate at which the volume and surface area are increasing, we can use the formulas for volume and surface area in terms of the radius. Let's start with the volume:
(a) Volume is given by the formula V = (4/3) π r^3.
To find the rate at which the volume is increasing, we need to differentiate this equation with respect to time (t):
dV/dt = (4/3) π (3r^2)*(dr/dt).
Given that dr/dt = 2 m/minute (rate at which the radius is changing), and when the radius is 4m (r = 4), we can substitute these values into the equation:
dV/dt = (4/3) π (3(4^2))*(2) = 32π m^3/minute.
Therefore, the volume is increasing at a rate of 32π m^3/minute.
(b) Surface area is given by the formula A = 4πr^2.
To find the rate at which the surface area is increasing, we need to differentiate this equation with respect to time (t):
dA/dt = 8πr*(dr/dt).
Once again, we substitute the values: r = 4m and dr/dt = 2 m/minute.
dA/dt = 8π(4)*(2) = 64π m^2/minute.
Therefore, the surface area is increasing at a rate of 64π m^2/minute.