Question

You are tracking the velocity and position of a rocket-propelled object near the surface of Mars. The velocity is v(t) and the position is s(t), where t is measured in seconds, s in meters, and v in meters per second. It is known that the v(t) = ds/dt = 4.94 – 3.72t and s(0) = 5.

Explain why the condition “f is continuous over [a, b]” from the Evaluation Theorem is fulfilled by this scenario.

I followed an expert and I got asked to revise my assignment, so I found another expert and got another revise. This is what I had:

Given that the velocity function is vd494-3.72t s(0) 5 The condition F is continuous over [a,b] because we are calculating antiderivative of f means summation of f in the interval [a,b] as tiny values. So that f should exist in [a,b] for any value between a and b. The Evaluation theorem is fulfilled because the velocity function v(t) is continuous from the value of t20 Therefore f is to continuous over [a,b]

This what they wrote, "The first two sentences of the discussion ("The condition F is continuous over [a,b] because we are calculating antiderivative of f means summation of f in the interval [a,b] as tiny values. So that f should exist in [a,b] for any value between a and b") are not clear and do not explain how the first condition of the theorem is satisfied. The third sentence states that v(t) is continuous but does not give a supporting explanation about how this is known. This supporting explanation is what is needed to show that the first condition of the Evaluation Theorem is satisfied in the scenario described."

Please help with suggestions.

Answer 1

acuous Evaluation theorem : If unction and F is antiderivative o fie unction and-F__ts antiderivative- f(x) | F'(x) = then Here eeant to ind sit) and at stt) is th thus-_-by-evaluation んderiva- We knoo that e 07. UCt)" , theorem S(B): tbe aD ttveO dt Hence sG)-51)4.94t-3.72t2 4-94 (3-1)-(1.86)(g-+) = 5