• f(1), according to the graph, is approximately 0.64
• To approximate f '(0.7), I used these points on the graph of f to calculate a good secant slope which is close to the slope of the tangent line: (0.7, 0.74) and (0.8, 0.7):
• The slope of the tangent line is parallel to the slope of the secant line only one time on that interval, as indicated by the below diagram. In it, the red segment is the secant line and the black segment represents the tangent line at the approximately location that the Mean Value Theorem is satisfied

• The graph of f appears to change concavity at x = 0.3, so the second derivative there is zero.

For the sake of ease, let's identify the four elements we are to order as A, B, C, and D respectively as they are listed in the problem.The correct order will be: B, D, A, C.

Answer the following questions about a particle with position equation , where t is in seconds, and position is in feet. Provide justifications for each answer.

(a) What is the speed of the particle at time = 1.8 seconds?

(b) At what time t does the particle's velocity first match its acceleration?

(c) How many times does the particle change direction on [1,4]?

(d) Find the total distance the particle travels on [1,4].

(e) Give the domain and range of s.

Difficulty:

Solution:

Before I started this problem, I found the first and second derivatives, and typed them into my function list on my calculator as follows: Y1 = original function, Y2 = first derivative, Y3 = second derivative. The equation in Y4 is used to help solve part (b).

(a) Remember that speed is the absolute value of velocity. Since s'(1.8) = -1.032 ft/sec, the speed will be the absolute value, 1.032 ft/sec. Don't forget units.

(b) The speed and acceleration are equal where Y2 = Y3, or where Y2 - Y3 = 0, which is my Y4. Notice that the domain of s is x > 0, so the first positive root of Y4 is the correct answer: t = .407 sec.

(c) The particle changes direction each time the velocity function crosses the x-axis. Note: it must completely cross the axis, not merely touch it. This happens 4 times on [1, 4]: at t = 1.647867, 2.3014442, 3.1810515, and 3.894808 sec.

(d) The total distance traveled counts time moving forward and backward separately, For example, if I move one foot forward and two feet backward, my total distance traveled is 3 feet although I have only endured a one foot net change (or displacement change). Thus, I am to integrate the velocity function separately for each time it crosses the x-axis, as shown by the below calculation.

The answer is 4.104 feet. Note: you have to subtract the second and fourth integrals because the particle is moving backward there; adding those integrals will give you the displacement, not the total distance.

(e) The domain is x > 0 (because of the ln x component of the function) and the range is all real numbers.