Bench Can Breakdance?

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THE CALCULUS CRUSADERS Analyzing A Derivative: Break Dancing Question breakdancing squirrel by Flickr user jhoc and breakdancing mime by Flickr

description

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

Transcript of Bench Can Breakdance?

Page 1: Bench Can Breakdance?

THE CALCULUS CRUSADERS

Analyzing A Derivative: Break Dancing Question

breakdancing squirrel by Flickr user jhocand breakdancing mime by Flickr user katiew

Page 2: Bench Can Breakdance?

.:.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.

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.:. 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, .

Page 4: Bench Can Breakdance?

.:. 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.

Page 5: Bench Can Breakdance?

.:. 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

Page 6: Bench Can Breakdance?

.:. 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.

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.:. 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.

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.:. 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.

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.:. 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.

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.:. 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.

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.:. 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:

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

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.:. 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

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.:. SOLUTIONS .:.Cont’d…

(e) Using The Mean Value Theorem of Integrals…

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AWESOME!!!!

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