Question

T (HOURS) 2 5 9 11 12

L(t) (cars per hours) 15 40 24 68 18

The rate at which cars enter a parking lot is modeled by E(t)=30+5(t−2)(t−5)e⁻⁰.²ᵗ. The rate at which cars leave the parking lot is modeled by the differentiable function L. Selected values of L(t) are given in the table above. Both E(t) and L(t) are measured in cars per hour, and time t is measured in hours after 5 A.M. (t=0). Both functions are defined for 0≤t≤12.

(a) What is the rate of change of E(t) at time t=7 ? Indicate units of measure.

Answer 1

The rate of change of E(t) at time t=7 is found by differentiating E(t) with respect to t, applying product and chain rules, then evaluating the derivative at t=7. The units for the rate of change are cars/hour^2.

Step-by-step explanation:

The rate of change of E(t) at time t=7 can be found by calculating the derivative E'(t) and then substituting t=7 into that derivative. The given model for E(t) is E(t) = 30 + 5(t-2)(t-5)e^{-0.2t}. First, we need to apply the product rule and the chain rule to differentiate E(t). After computing the derivative, we plug in t=7 to get the rate of change at that specific time. As we are asked for the rate of change, which is the derivative, our units of measure would be cars per hour squared or cars/hour2.

Here's a step-by-step outline of what the differentiation might look like:

1. Identify the parts of the function where the product rule and chain rule must be applied.
2. Apply the product rule to the term 5(t-2)(t-5)e^{-0.2t}.
3. Within the product rule application, apply the chain rule to the exponential function e^{-0.2t}.
4. Simplify the resulting derivative.
5. Plug in t=7 into the simplified derivative to find the rate of change at that time.

Note that the student will need to complete the actual differentiation themselves as part of their learning process.