A spherical balloon is blown up so that it's volume increases constantly at the rate of 2cm^3 per second .find the rate of the increase of the radius when the volume of the balloon is 50cm^3
v = 4/3 πr^3
dv/dt = 4πr^2 dr/dt
so now, determine r when v = 50, and then plug in your numbers to find dr/dt.
To find the rate of the increase of the radius, we can use the formula for the volume of a sphere:
V = (4/3) * π * r^3
Given that the volume is increasing at a rate of 2 cm^3 per second, we can differentiate the volume formula with respect to time (t) to find the rate of change of the volume:
dV/dt = 4πr^2 * dr/dt
Where dV/dt represents the rate of change of the volume with respect to time, and dr/dt represents the rate of change of the radius with respect to time.
We are given that the volume of the balloon is 50 cm^3, so we can substitute V = 50 cm^3 into the volume formula:
50 = (4/3) * π * r^3
To find the radius (r), we can rearrange the formula:
r^3 = (3/4π) * 50
Taking the cube root of both sides:
r = (50 * 3 / (4π))^(1/3)
Now, we can substitute the value of r into the rate formula and solve for dr/dt:
2 = 4π * ([(50 * 3) / (4π)]^(1/3))^2 * dr/dt
Simplifying:
2 = (3 * 50^2 / 4π)^(2/3) * dr/dt
dr/dt = 2 / (3 * 50^2 / 4π)^(2/3)
Calculating, we get:
dr/dt ≈ 0.007 cm/s
Hence, when the volume of the balloon is 50 cm^3, the rate of increase of the radius is approximately 0.007 cm/s.
To find the rate of the increase of the radius, we can use the formula for the volume of a sphere:
V = (4/3) * π * r^3,
where V is the volume and r is the radius.
We are given that the volume of the balloon is increasing at a rate of 2 cm^3 per second. Let's differentiate the volume equation with respect to time (t) to find the rate of change with respect to time:
dV/dt = (4/3) * π * (3r^2) * (dr/dt),
where dV/dt is the rate of change of volume with respect to time (2 cm^3/s), r is the radius at any given time, and dr/dt is the rate of change of the radius with respect to time (we need to find this).
Now, we need to find dr/dt. We are given that the volume is 50 cm^3, so we substitute V = 50 in the equation:
2 = (4/3) * π * (3r^2) * (dr/dt).
Now we can solve for dr/dt:
2 = 4πr^2 * (dr/dt),
(dr/dt) = 2 / (4πr^2).
Substituting V = 50 cm^3, we can find the radius (r):
50 = (4/3) * π * r^3,
r^3 = 50 * (3/4π),
r^3 = 37.68,
r ≈ 3.29 cm.
Finally, substituting the radius (r) back into the equation for dr/dt:
(dr/dt) = 2 / (4π * (3.29)^2),
(dr/dt) ≈ 0.047 cm/s.
Therefore, the rate of increase of the radius when the volume of the balloon is 50 cm^3 is approximately 0.047 cm/s.