Annual sales of bottled water in the country in the period 1993-2003 could be approximated by the function below, where t is time in years since 1990. Were sales of bottled water accelerating or decelerating in 2000? How fast?
R(t) = 10 t**2 + 100 t + 2400 text( million gallons ) \(3<=t<=13\)
Correct: Your answer is correct. by Incorrect: Your answer is incorrect. million gals/
Since R increases with t for any t>0 the answer to the first question requires no computation
R = 10 t^2 + 100 t + 2400 million/year
dR/dt = 20 t + 100
in 2000 t = 10
dR/dt = 20(10) + 100 = 300 million/year^2
To determine whether sales of bottled water were accelerating or decelerating in 2000, we need to find the second derivative of the sales function and evaluate it at t = 10. The second derivative will tell us the rate at which the rate of sales is changing.
First, let's find the first derivative of R(t) with respect to t:
R'(t) = d/dt (10t^2 + 100t + 2400)
= 20t + 100
Now, let's find the second derivative by taking the derivative of R'(t) with respect to t:
R''(t) = d/dt (20t + 100)
= 20
The second derivative is a constant, 20. This means that the rate at which the rate of sales is changing is constant and does not depend on time.
To determine whether sales were accelerating or decelerating in 2000, we evaluate the second derivative at t = 10:
R''(10) = 20
Since the second derivative is positive (20), sales of bottled water were accelerating in 2000.
To find out how fast the sales were accelerating, the rate of acceleration is given by the value of the second derivative. In this case, the rate of acceleration is 20 million gallons per year.