Question

At a school professional development workshop, a generous administrator provided coffee at a rate modeled by c(t) =30 for 0≤t≤5 (in cups per hour). Teachers are drinking the coffee at a rate modeled by d(t) = 25 - 1/15 t^2 + 10 cos (2t) for 0≤t≤5. Find how much the rate of change of the amount of coffee available is decreasing at t = 3. Identify units and find the numerical value to proceed.

ANSWER IS 5.188 cup/hour^2 ( I HAVE NO CLUE HOW)

Answer 1

To determine the rate at which the amount of available coffee is decreasing at t = 3, we calculate the derivative of the coffee consumption rate, d'(t), at t = 3. The result is approximately 5.188 cups per hour squared, representing the acceleration of coffee consumption.

The student is asking to find the rate of change of the amount of coffee available, which is decreasing at a particular time t = 3 hours. To find this rate, we need to calculate the derivative of the net rate of change of the coffee amount at that time.

The net rate of coffee change is given by the difference between the supply and consumption rates c(t) - d(t). Thus, we need to find the derivative of d(t) since c(t) is constant.

We have
`d(t) = 25 - (1)/(15)t^2 + 10 \cos(2t)`. Taking the derivative d'(t) gives us
`-(2)/(15)t - 20 \sin(2t)`. Substituting
`t = 3 into d'(t)`, we get
`d'(3) = -(2)/(15)(3) - 20 \sin(6)`, which calculates to approximately 5.188 cups per hour squared.