Determine the rate at which the surface area is changing with respect to time at t=2 min, assuming that r = 3 at t = 0
Assume that the radius r of a sphere is expanding at a rate of 14 in / min. Its surface area is 4(pi)r^2
Ds/dt= 4pi2rdr/dt
i got 4pi2(28)(14)
is that right or would the radius be different at t = 2 min
To determine the rate at which the surface area is changing with respect to time at t = 2 min, you need to consider the given information about the radius and its rate of change.
Given that the radius of the sphere is expanding at a rate of 14 in/min, we can find the radius at t = 2 min by using the initial radius at t = 0 and the rate of change.
Since the rate is constant, we can calculate the new radius at t = 2 min as follows:
r(t) = r(0) + (rate of change) * t
r(2) = 3 + 14 * 2
r(2) = 3 + 28
r(2) = 31 inches
Now that we have determined the radius at t = 2 min, we can calculate the rate at which the surface area is changing using the formula:
ds/dt = 4πr^2 * dr/dt
ds/dt = 4π(31^2) * 14
ds/dt = 4π(961) * 14
ds/dt = 4π(13454)
ds/dt ≈ 53323.14 square inches per minute
Therefore, the rate at which the surface area is changing with respect to time at t = 2 min is approximately 53323.14 square inches per minute.