V = (4/3)Ï€ r^3
dV/dt = 4Ï€ r^2 dr/dt
given : when r = 1.5 , dV/dt = 1
find dr/dt
for b),
A = 4Ï€r^2
dA = 8Ï€r dr/dt , you have dr/dt from a)
just plug in the given stuff
(a.) how fast is the radius increasing?
(b.) how fast is the surface area increasing?
dV/dt = 4Ï€ r^2 dr/dt
given : when r = 1.5 , dV/dt = 1
find dr/dt
for b),
A = 4Ï€r^2
dA = 8Ï€r dr/dt , you have dr/dt from a)
just plug in the given stuff
1 = 4Ï€(2.25) dr/dt
dr/dt = 1/(9Ï€) <----- nice, you had that
dA = 8Ï€r dr/dt
= 8Ï€(1/(9Ï€) = 8/9
you had 4/3, looks like a "sloppy" error.
(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)Ï€r^3
Differentiating both sides of the equation with respect to time (t), we get:
dV/dt = (4/3)Ï€ * 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)Ï€ * (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 = 4Ï€r^2
Differentiating both sides of the equation with respect to time (t), we get:
dA/dt = 4Ï€ * 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.