A Calculator, or Paul, MAY be used to help answer this question.

Be certain to limit yourself to the four permissible calculator functions.

Paul runs to catch the bus along a line in such a way that at time t, 1 ≤ t ≤ 8, his position is given by:

(a) Write a formula for the velocity of Paul at time t.

(b) At what instant does Paul reach his maximum speed.

(c) When is Paul moving to the left?

(d) Find the total distance traveled by Paul from t = 1 to t = 8.

Solution (ROUGH DRAFT)

PART A

This is an accumulation function. Differentiating this accumulation function would give us the argument of the accumulation function by The Fundamental Theorem of Calculus (since derivatives and antiderivatives are like inverses of each other). Algebraically speaking, F’(x) = f(x), where F(x) is the accumulation function.

v(t) = 1 – tcost – lntsint

PART B

Since we’re not allowed to use our calculators to determine the maximum velocity in the interval 1 ≤ t ≤ 8, we’ll have to use a differentiate approach. We’ll use The First Derivative Test!

By using The First Derivative Test, we can determine the parent function’s roots, then use a line analysis to determine where the parent function is increasing or decreasing. Remember, wherever the first derivative is negative or positive, the parent function is decreasing or increasing, respectively, which means that the roots of the first derivative are local extrema.

We’re allowed to use our calculator to determine roots. So, using the calculator, v’ = 0 at 1.2340473 and 3.7311818. Using a line analysis (that I’ll insert later after the EDIT button is fixed), we find that 1.2340473 is a local minimum and 3.7311818 is a local maximum.

But, don’t forget that this is a closed interval and, by The Extreme Value Theorem, the endpoints of a function are considered local extrema. However, v(1) = 0.45969769 and v(8) = 0.10668763.

Therefore, at t = 3.7311818, Paul is at his maximum velocity.

PART C

When the velocity function is negative, Paul is moving to the left. Graph v, determine the roots, and then deduce on what interval the velocity is negative. I’ll elaborate further when I have an image loaded up.

Paul is moving to the left at 5.2035697 < t < 7.987021.

PART D

Integrate the absolute value of the velocity function to get distance. We’re finding the absolute value of the function because, if we don’t find the absolute value, we’ll end up finding displacement. Using the fnInt function in our calculator, the absolute value of the velocity function at 1 ≤ t ≤ 8 is 21.46124296.