a boy is blowing a spherical balloon at a rate of 50 cm3/s.at what rate is the radius of the balloon changing when the radius is 10 cm?give your answer to 2 decimal places.
dV/dt = 50 cm^3/s
V = (4/3)πr^3
dV/dt = 4πr^2 dr/dt
so when r = 10
50 = 4π(100) dr/dt
dr/dt = 1/(8π) cm/s = ... you do the button pushing,
(no calculator needed up to this point. )
To find the rate at which the radius of the balloon is changing, we can use the formula for the volume of a sphere. The formula for the volume of a sphere is V = (4/3)πr³, where V is the volume and r is the radius.
Since we know that the volume is changing at a rate of 50 cm³/s, we can take the derivative of the volume equation to find the rate at which the radius is changing.
dV/dt = (4/3)π(3r²(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 the radius with respect to time.
Substituting the given values, we have:
50 = (4/3)π(3(10²)(dr/dt))
Simplifying:
50 = (4/3)π(300)(dr/dt)
Now we can solve for dr/dt:
dr/dt = 50 / ((4/3)π(300))
Using the value of π as 3.14 and evaluating this expression, we get:
dr/dt ≈ 0.053 cm/s (rounded to 2 decimal places)
Therefore, the rate at which the radius of the balloon is changing when the radius is 10 cm is approximately 0.053 cm/s.