Question

1. A weather balloon that contains 540 m' of helium springs a leak and empties in 60 seconds. Suppose that the volume of helium in the balloon as a function of time is given by V(t)=540 (1-t÷60)^3 (a) the average rate of change of volume during the 60 seconds it takes to empty. (b) the volume of the balloon after 40 seconds. (c) the instantaneous rate of change of the volume at 40 seconds. ​

Answer 1

(a) Average rate of change of volume during the 60 seconds it takes to empty: -9 m^3/s

(b) Volume of the balloon after 40 seconds: 270 m^3

(c) Instantaneous rate of change of volume at 40 seconds: -67.5 m^3/s

Explanation:

(a) The average rate of change of volume during the 60 seconds it takes to empty can be calculated by finding the change in volume over the change in time:

Average rate of change = (V(60) - V(0)) / (60 - 0)

where V(60) is the volume of the balloon at 60 seconds and V(0) is the volume at 0 seconds. Substituting these values into the equation V(t) = 540(1 - t/60)^3, we get:

V(60) = 0

V(0) = 540

So, the average rate of change of volume is:

Average rate of change = (0 - 540) / (60 - 0) = -9 m^3/s

(b) The volume of the balloon after 40 seconds can be found by substituting t = 40 into the equation V(t) = 540(1 - t/60)^3:

V(40) = 540(1 - 40/60)^3 = 270 m^3

Therefore, the volume of the balloon after 40 seconds is 270 m^3.

(c) The instantaneous rate of change of volume at 40 seconds can be found by taking the derivative of the volume function with respect to time:

dV/dt = -540(3/60)(1 - t/60)^2

Substituting t = 40 into this equation, we get:

dV/dt | t=40 = -540(3/60)(1 - 40/60)^2 = -67.5 m^3/s

Therefore, the instantaneous rate of change of volume at 40 seconds is -67.5 m^3/s.