Bench is moving quite fast. Can the Calculus Crusaders catch up with him?

1. Analyzing A Derivative: Break Dancing Question
breakdancing squirrel by Flickr user jhoc
and breakdancing mime by Flickr user katiew

2. .:.The SITUATION.:.
The tremendous trio and their pets decide to take
a break from their adventure. To pass time, Bench
decides to break-dance! Thinking that they should
be constantly on their feet and thinking
mathematically, Zeph records Bench’s movements
in a velocity-time graph shown below, where
velocity is measured in metres per second.

3. .:. The GRAPH.:.
 The graph of the velocity function, x’(t), shown is a piece-
wise function that consists of three line segments:
 At π ≤ t ≤ 0, x’(t) is a semi-circle;
 At 0 ≤ t ≤ 2π, x’(t) is a period of a sine wave;
 At 2π ≤ t ≤ 7, x’(t) is a segment of the square root
function, .

4. .:. THE QUESTION .:.
(a) Determine when Bench is farthest from the origin.
(b) Find where the position function, x(t), has an inflection
point.
(c) Determine where x(t) is concave up with a negative slope.
(d) Find the average rate of change of x’(t)
(e) Find the average value of x’(t) using seven trapezoidal
Riemann sums.

5. .:. THE SOLUTIONS .:.
(a) The question is asking where Bench’s furthest
position would be.
We can start by finding the critical points of the
derivative. This indicates where the parent function
has local extrema [minimum or maximum].
To find critical points: x’(t) = 0
This is true at t = 0, π, 2π

6. .:. THE SOLUTIONS .:.
cont’d…
(a) Looking at the graph, we see x’(t) is
negative at –π < t < 0, positive 0 < t <
π, negative π < t < 2π, and positive 2π
< t < 7. We can then visualize the
direction of the graph of the parent
function using a line analysis like shown
below.

7. .:. THE SOLUTIONS .:.
Cont’d…
(a) **NOTE: Where x’(t) is positive, x(t) is increasing. Where
x’(t) is negative, x(t) is decreasing.
x(t) has a relative maximum at t = π and relative
minimum at t = 0, 2π.
By examining the graph once more, we can see that the
area of the graph at –π < t < 0 is larger than the area
found in the domain of 0 < t < π, π < t < 2π, and at 2π <
t < 7.
Because the region has a larger area than others it is
shown that Bench is farthest from the origin at t = -π
seconds.

8. .:. THE SOLUTIONS .:.
(b) A point of inflection is a point on the graph where
the function changes concavity. To determine
where x(t) changes concavity, we firstly need to
know where x’(t) is increasing or decreasing.
Where x’(t) is increasing, x(t) is concave up.
Where x’(t) is decreasing, x(t) is concave down.

9. .:. THE SOLUTIONS .:.
Cont’d…
(b) By looking at the graph, x’(t) decreases at –π <
t < -π/2, increasing at -π/2 < t < π/2, decreasing
at π/2 < t < 3π/2, and increases at 3π/2 < t < 7.
Therefore, x(t) is concave down at –π < t < -
π/2, concave up at -π/2 < t < π/2, concave
down at π/2 < t < 3π/2, and concave up at 3π/2
< t < 7.
x(t) has a point of inflection at t = -
π/2, π/2, π, 2π seconds.

10. .:. THE SOLUTIONS .:.
(c) Referring to the previous two questions
we know that...
 According to (b), x(t) is concave up at -π/2 <
t < π/2 and at 3π/2 < t < 7.
 According to (a), it’s negative at -π < t < 0
and at π < t < 2π.
Therefore, x(t) is concave up with a
negative slope where x’(t) is increasing
and negative: -π/2 < t < 0 and at 3π/2 <
t < 2π, where t is measured in seconds.

11. .:. THE SOLUTIONS .:.
(d) Here, we are trying to find the average rate of
change of x’(t). To do this, we can use The
Mean Value Theorem of Derivatives.
The Mean Value Theorem of Derivatives:
If the function f is continuous on a closed
interval [a, b] and differentiable on the open
interval (a, b), then at least one number, c,
exists in the open interval (a, b) where:

12. .:. THE SOLUTIONS .:.
(d) Using The Mean Value Theorem…

13. .:. THE SOLUTIONS .:.
(e) To determine the average value of x’(t),
we use The Mean Value Theorem of
Integrals.
The Mean Value Theorem of Integrals:
If f is a continuous function on an
interval [a, b], then the average value of
f on [a, b] is

14. .:. SOLUTIONS .:.
Cont’d…
(e) Using The Mean Value Theorem of
Integrals…

15. AWESOME!
!!!
Bench has had his
breakdancing fun
while Zeph has
sharpened his math
skills and the whole
gang is rested!!!