Practice Problems: Extremes and Inflections Solutions:

1. (moderate) Graph the motion equation given below and determine any extreme and inflection points (over the time range of -1 to +1 seconds).
x = 9t³ - 4t + 7
(with x in meters and t in seconds)


Extreme points:
v = dx/dt = 0
dx/dt = d(9t³ - 4t + 7)/dt = 0
dx/dt = 27t² - 4 = 0
t² = 4/27
t = ±0.38s
Inflection point:
a = d²x/dt² = dv/dt = 0
dv/dt = d(27t² - 4)/dt = 0
54t = 0
t = 0

2. (moderate) Find the displacement of a particle from 4 to 8 seconds if the equation below accurately describes its motion.
v = 4t³+ 4t² + 7t +1
(v is in m/s and t is in s)
Δx = ∫vdt
Δx = ∫(4t³+ 4t² + 7t +1)dt    (with limits from t = 4 to t = 8)
Δx = t⁴ + 1.33t³ + 3.5t² + t   (with limits from t = 4 to t = 8)
Δx = [8⁴ + 1.33(8)³ + 3.5(8)² + 8] - [4⁴ + 1.33(4)³ + 3.5(4)² + 4]
Δx = 5009 - 401 = 4608 m

3. (moderate) Use the graph below to determine the final position of a particle given that the initial position was at x_o = 20 m.



4. (moderate) A particle is moving along a straight line according to: x = t⁴/6 - 7t³/6 + 3t²/2 + 5
Where t is in seconds x is in meters.
a. Find the velocity of the particle at t = 1.8 seconds.
v = dx/dt = 2t³/3 - 7t²/2 + 3t
v = 2(1.8)³/3 - 7(1.8)²/2 + 3(1.8)
v = -2.0 m/s
b. Where is the particle at t = 4.1 seconds?
x = (4.1)⁴/6 - 7(4.1)³/6 + 3(4.1)²/2 + 5
x = -3.1 m
c. Create a v - t graph for the motion for the particle in the range from t = 0 to t = 3 seconds.
Extreme points: First derivative is 0.
dv/dt = a = 0 = 2t²- 7t + 3
t = {0.5s, 3.0 s}
At t = 0.5 s, v = 0.71 m/s
At t = 3.0 s, v = -4.5m/s
Inflection point: Second derivative is 0.
da/dt = J = 4t - 7
t = 1.75 s
At 1.75 s, v = -1.9 m/s



5. (moderate) v = 8t³ + 7t²/2 - 9t + 3 (with t in seconds and v in m/s)
Determine the position of the particle at t = 2.4 s if the initial position was x = 2.0 m.
v = dx/dt
∫dx = ∫(8t³ + 7t²/2 - 9t + 3)dt
x = 8t⁴/4 + 7t³/6 - 9t²/2 + 3t + C
Evaluate initial condition: C = 2.0 m
At t = 2.4 s
x = 8(2.4)⁴/4 + 7(2.4)³/6 - 9(2.4)²/2 + 3(2.4) + 2.0
x = 65.8 m

6. (moderate) Extremes and inflection points have meaning in many relationships besides those found in motion studies. For example, if two variables, A and B are related by the equation A = 5B² + 3B + 7, one can easily find the max or min for the magnitude of A by setting the first derivative dA/dB = 0. Try to use this concept to solve the following problem: You have 24 meters of fence with which to construct a rectangular rabbit enclosure. You want the enclosure to have the maximum amount of area.
a. Using calculus, determine the length of the sides of this enclosure.
b. Determine an equation that can be used to find the lengths of the sides having any length of fencing that you are given.
a. The area equation for a rectangle is Area = Length x Width
A = LW
Since the perimeter of the enclosure must be 24 meters long:
24 = 2L + 2W
L = (24 - 2W)/2
Thus: A = [(24 - 2W)/2]W
A = 12W - W²
Now find W when A is maximized:
dA/dW = 12 - 2W = 0
12 = 2W
W = 6 meters
This means that the enclosure will have maximum area if it is square, with a length of 6 meters on all sides (making the enclosure have an area of 36 m²).
b. The general equation uses P as the perimeter (the length of fencing that is given).
P = 2L + 2W
L = (P/2) - W
A = [(P/2) - W]W
A = (PW/2) - W²
To maximize the area, dA/dW = 0
dA/dW = (P/2) - 2W = 0
W = P/4
L = (P - 2(P/4))/2 = (P - P/2)/2 = P/4
Therefore, a square enclosure will always maximize the area with A_max = P²/16.