A spherical balloon is inflated at the rate of 1 cm^3 per minuter. At the instant when the radius r=1.5,
(a.) how fast is the radius increasing?
(b.) how fast is the surface area increasing?
V = (4/3)π r^3
dV/dt = 4π r^2 dr/dt
given : when r = 1.5 , dV/dt = 1
find dr/dt
for b),
A = 4πr^2
dA = 8πr dr/dt , you have dr/dt from a)
just plug in the given stuff
So dr/dt is equals to 1/9pi? or i have miscalculated again?
Is dA=4/3?
dV/dt = 4π r^2 dr/dt
1 = 4π(2.25) dr/dt
dr/dt = 1/(9π) <----- nice, you had that
dA = 8πr dr/dt
= 8π(1/(9π) = 8/9
you had 4/3, looks like a "sloppy" error.
To find the rates at which the radius and surface area are changing, we can use the formulas for the volume and surface area of a sphere.
(a.) To find how fast the radius is increasing, we need to differentiate the volume formula with respect to time.
The volume of a sphere can be given by the formula: V = (4/3)πr^3
Differentiating both sides of the equation with respect to time (t), we get:
dV/dt = (4/3)π * 3r^2 * dr/dt
Here, dV/dt represents the rate of change of volume with respect to time, and dr/dt represents the rate of change of radius with respect to time.
We are given that dV/dt = 1 cm^3/min.
When the radius (r) is 1.5 cm, we can substitute these values into the equation:
1 = (4/3)π * (1.5)^2 * dr/dt
Now we can solve for dr/dt, the rate at which the radius is increasing.
(b.) To find how fast the surface area is increasing, we need to differentiate the surface area formula with respect to time.
The surface area of a sphere can be given by the formula: A = 4πr^2
Differentiating both sides of the equation with respect to time (t), we get:
dA/dt = 4π * 2r * dr/dt
Here, dA/dt represents the rate of change of surface area with respect to time, and dr/dt represents the rate of change of radius with respect to time.
We can substitute the known values into the equation and solve for dA/dt when r = 1.5 cm.
By following these steps, we can determine both the rate at which the radius is increasing and the rate at which the surface area is increasing.