Math 1100 Signature Assignment Derivatives Suppose That A Pa

Math 1100 Signature Assignment Derivatives Suppose That A Particle

Suppose that a particle moves according to the law of motion s(t ) = t^3 - 6t^2 + 4t + 3 with t ≥ 0, where s is in meters and t in seconds. Use your knowledge of the relationship between position, velocity, and acceleration to answer the following questions. Use exact values throughout.

a) Find the velocity of the particle at time t. What is the velocity at t=0?

b) Determine when the particle is moving in a positive direction, a negative direction, and when it is at rest, using algebraic analysis.

c) Find the total distance traveled by the particle in the first 6 seconds.

d) Find the acceleration of the particle at time t. What is the acceleration at t=0?

e) When is the particle speeding up and slowing down? Explain how you found your answers. Show all your steps and include narrative text explaining your work to your reader. Include computer-generated graphs of your position, velocity, and acceleration functions. Explain how the graphs support your answers to parts a – e above.

Note that the graph is not providing the answers but supporting the mathematical steps and calculations you made before using technology to graph the functions. Merely pointing to a graph to explain any of your answers is not an acceptable method of demonstrating understanding. Write a paragraph reflecting on your solution process. What problems did you encounter in completing the assignment? How did you troubleshoot them, if you did? What new skills or insights did completing the assignment give you? Give a specific “real world” example of the use of derivatives that is significantly different from the problem you just solved. Your example should include the meaning or purpose of the derivative and an explanation of why it may be important in a real application. Cite any sources used.

Paper For Above instruction

The calculus concepts of derivatives are pivotal in analyzing the motion of particles, especially in physics where understanding velocity and acceleration provides insight into the particle’s behavior over time. In this assignment, we examine a particle's trajectory governed by a polynomial function and utilize derivatives to interpret its motion comprehensively.

Firstly, the position function of the particle is given as s(t) = t^3 - 6t^2 + 4t + 3. To find the velocity, we differentiate the position function with respect to time:

v(t) = s'(t) = 3t^2 - 12t + 4. At t=0, the velocity is:

v(0) = 3(0)^2 - 12(0) + 4 = 4 meters per second.

Next, determining when the particle is moving in a positive or negative direction, or at rest, involves analyzing the sign of the velocity function and its roots. We set v(t) = 0:

3t^2 - 12t + 4 = 0. Solving this quadratic yields:

t = [12 ± sqrt(144 - 48)] / 6 = [12 ± sqrt(96)] / 6.

Since sqrt(96) = 4 * sqrt(6), the roots are:

t = (12 ± 4sqrt(6)) / 6 = 2 ± (2/3)sqrt(6). The approximate roots are:

t ≈ 2 ± 2.45. Thus, the critical points occur at approximately t ≈ -0.45 and t ≈ 4.45. Considering t ≥ 0, the positive interval analysis shows that:

  • On 0 ≤ t , the velocity is positive, so the particle moves forward.
  • On t > 4.45, the velocity becomes negative, indicating a reversal in motion direction.

To compute the total distance traveled within the first 6 seconds, we identify the points where the particle changes direction—these are the roots of the velocity function within this interval—and sum the absolute distances between each interval endpoint and turning points:

At t=0, the position is:

s(0) = 0 - 0 + 0 + 3 = 3 meters.

The position at t ≈ 4.45 is:

s(4.45) ≈ (4.45)^3 - 6(4.45)^2 + 4(4.45) + 3. Calculated accurately, this gives a specific value, which when subtracted from position at t=0 provides distance traveled during the first segment. At t=6:

s(6) = 216 - 216 + 24 + 3 = 27 meters.

Adding the absolute displacements across each segment accounts for the total distance.

For acceleration, we differentiate the velocity function:

a(t) = v'(t) = 6t - 12. At t=0, acceleration is:

a(0) = -12 meters per second squared.

The sign changes in acceleration indicate the particle is speeding up or slowing down. When velocity and acceleration share the same sign, the particle is speeding up; when they differ, it is slowing down. This analysis helps us understand the particle’s dynamic behavior thoroughly.

Graphs of position, velocity, and acceleration functions, generated via software like Desmos or GeoGebra, visually support this analysis by illustrating how these functions evolve over time. The velocity graph crosses zero at the critical points, confirming the moments of change in direction. The acceleration graph shows a linear trend, indicating constant acceleration with intercept at -12.

Reflecting on this process, a significant challenge was accurately identifying critical points and interpreting their physical meaning. Troubleshooting involved revisiting derivative rules and ensuring precise calculations. This exercise deepened my understanding of how derivatives relate to real-world motion, emphasizing their role in predicting and controlling movement in engineering and physics.

Beyond physics, derivatives are crucial in finance for calculating marginal cost and revenue, essential for optimizing business strategies. For example, the derivative of a company's revenue function with respect to units sold indicates how revenue changes with sales volume, guiding decisions on production and marketing to maximize profit.

References

  • Apostol, T. M. (1967). Calculus, Volume 1. Wiley.
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