a spherical balloon is filled with water. if the water is leaking out at a rate of 2 cm^3/min, find the rate at which the radius is changing when the volume is v=1000cm^3?
v = 4/3 pi r^3
when v = 1000, r = ∛(3000/(4pi))
dv/dt = 4 pi r^2 dr/dt
Now just plug in your values for dv/dt and r to get dr/dt.
To find the rate at which the radius of the balloon is changing, we can use related rates. We are given that the water is leaking out of the balloon at a rate of 2 cm^3/min. We need to determine the rate at which the radius is changing when the volume is 1000 cm^3.
Let's denote the radius of the balloon as r(t) and the volume of the balloon as V(t), where t represents time. We are given that dV/dt = -2 cm^3/min, because the volume is decreasing.
Using the formula for the volume of a sphere, V = (4/3) * π * r^3, we can find the derivative of the volume with respect to time:
dV/dt = (4/3) * π * 3r^2 * dr/dt
Now, let's substitute the known values:
-2 = (4/3) * π * 3r^2 * dr/dt
Simplifying the equation further:
-2 = 4πr^2 * dr/dt
Now, we can solve for dr/dt, which represents the rate at which the radius is changing:
dr/dt = -2 / (4πr^2)
We want to find the rate at which the radius is changing when the volume is 1000 cm^3. Therefore, when V = 1000 cm^3, we can substitute this value into the equation and solve for dr/dt:
dr/dt = -2 / (4πr^2)
1000 = (4/3) * π * r^3
r^3 = 1000 * 3 / (4π)
r^3 = 750 / π
r = (750 / π)^(1/3)
Substituting the value of r into the equation for dr/dt:
dr/dt = -2 / (4π * (750 / π)^(1/3))^2
Simplifying the expression further should give you the rate at which the radius is changing when the volume is 1000 cm^3.