A spherical balloon is being inflated so that its volume is increasing at the rate of 200 cm3/min. At what rate is the radius increasing when the radius is 15 cm.
A car is heading east toward an intersection at the rate of 40 mph. A truck is heading south, away from the same intersection at the rate of 60 mph. At what rate is the distance between the car and the truck changing when the car is 8 miles from the intersection and the truck is 15 from the intersection?
V = (4/3) pi r^3
dV/dt = 4 pi r^2 dr/dt
200 = 4 pi (225) dr/dt
To find the rate at which the radius is increasing, we need to use the relationship between the volume and radius of a sphere. The volume of a sphere is given by the formula:
V = (4/3) * π * r³
Taking the derivative of both sides with respect to time (t), we can express the rate of change of volume with respect to time:
dV/dt = 4π * r² * dr/dt
We are given that dV/dt (rate of change of volume) is 200 cm^3/min. We are asked to determine dr/dt (rate of change of radius) when r = 15 cm.
We can rearrange the equation to solve for dr/dt:
dr/dt = (1/(4π * r²)) * dV/dt
Substituting the known values, we get:
dr/dt = (1/(4π * 15²)) * 200
Let's calculate the result:
dr/dt = (1/(4 * 3.1416 * 225)) * 200
= 0.0001768 cm/min
Therefore, when the radius is 15 cm, the rate at which the radius is increasing is approximately 0.0001768 cm/min.