a toy balloon is in the shape of a sphere. it is being inflated at the rate of 20 in^3 / min. at the moment that the sphere has volume 64 cubic inches. what is the rate of change of the radius ?
64 = (4/3)pi R^3
from that find R
dV = surface area * dR (easy way :)
so
dV/dt = 4 pi R^2 dR/dt
20 in^3/min = 4 pi R^2 dR/dt
solve for dR/dt
To find the rate of change of the radius, we will use the formula for the volume of a sphere:
V = (4/3) * π * r^3
Given that the rate of change of the volume is 20 in^3/min, we can differentiate the volume formula with respect to time to find the rate of change of the radius.
Differentiating both sides of the equation with respect to t (time):
dV/dt = d/dt[(4/3) * π * r^3]
The derivative of (4/3) is zero since it is a constant. Using the power rule for differentiation, the derivative of r^3 is 3r^2 times the derivative of r with respect to t. So, the equation becomes:
dV/dt = (4/3) * π * (3r^2) * dr/dt
We are given that dV/dt = 20 in^3/min and the volume V = 64 in^3. Plugging these values into the equation:
20 = (4/3) * π * (3r^2) * dr/dt
Now we can solve for dr/dt, the rate of change of the radius:
dr/dt = 20 / [(4/3) * π * (3r^2)]
Simplifying the equation further:
dr/dt = 15 / (4πr^2)
Now we have the rate of change of the radius in terms of r.
To find the rate of change of the radius, we can use the formula for the volume of a sphere:
V = (4/3) * π * r³,
where V is the volume and r is the radius of the sphere.
We are given that the sphere has a volume of 64 cubic inches. So we can write the equation as:
64 = (4/3) * π * r³.
To find the rate of change of the radius (dr/dt), we differentiate both sides of the equation with respect to time (t), using the chain rule:
d/dt(64) = d/dt((4/3) * π * r³).
The left side becomes 0, since 64 is constant with respect to time. On the right side, we differentiate each term separately:
0 = (4/3) * π * 3r² * (dr/dt).
Simplifying the equation, we have:
0 = 4πr² * (dr/dt).
Now we can solve for (dr/dt):
dr/dt = 0 / (4πr²).
Since we are interested in the rate of change of the radius at the moment when the volume is 64 cubic inches, we can substitute the given volume into the equation.
At V = 64 cubic inches, we know r = (3V / (4π))^(1/3):
r = (3 * 64 / (4π))^(1/3)
= (192 / (4π))^(1/3)
= 3 / (π^(1/3)).
Now we can substitute this value of r into our equation for (dr/dt):
dr/dt = 0 / (4π(3 / (π^(1/3)))²)
= 0.
Hence, the rate of change of the radius at the moment when the volume is 64 cubic inches is 0.
V= (4/3)π r^3
dV/dt = 4π r^2 dr/dt
when V = 64
64 = (4/3)π r^3
48/π = r^3
r = (48/π)^(1/3) = appr 2.4814 ---> I stored the 10 decimal answer
in dV/dt = 4π r^2 dr/dt
20 = 4π(2.4814..)^2 dr/dt
dr/dt = appr .2585 inches/min