Bench Can Breakdance?
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Bench Can Breakdance?

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

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

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Bench Can Breakdance? Presentation Transcript

  • 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!!!