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.
I assume you do calculus.
V= 4/3 PI r^3
dV/dt= 4 PI r^2 dr/dt
you are given dV/dt= 50cm^3/s; find dr/dt
To find the rate at which the radius of the balloon is changing, we need to use the formula for the volume of a sphere and apply related rates.
The volume of a sphere is given by the formula:
V = (4/3)πr³
Where V represents the volume and r represents the radius of the sphere.
Given that the balloon is being blown at a rate of 50 cm³/s, we can express dV/dt, the rate at which the volume is changing, as 50 cm³/s.
We are asked to find the rate at which the radius, dr/dt, is changing when the radius is 10 cm.
To find this rate, we need to differentiate the volume formula with respect to t (time) by applying the chain rule:
dV/dt = (dV/dr) * (dr/dt)
We already know that dV/dt = 50 cm³/s and we need to find dr/dt.
Differentiating the volume formula with respect to r, we get:
dV/dr = 4πr²
Now we can solve for dr/dt:
50 = (4π(10)²) * dr/dt
dr/dt = 50 / (4π(10)²)
Now, we can calculate the rate at which the radius is changing:
dr/dt = 50 / (4π(10)²) ≈ 0.40 cm/s
Therefore, the rate at which the radius of the balloon is changing when the radius is 10 cm is approximately 0.40 cm/s.