Calculus II - Integration and Statistics - Term Paper Example

Task A: This task requires the creation of a real-world science question/problem that requires the application of differentiation for it to be solved by the carrying out the following tasks: 1. A description in the context of the above real-world problem of the following terms using appropriate units. a) Independent variable b) Dependent variable c) Range d) Domain 2. An explanation of what the real-world problem above is about or is addressing 3. The problem created should involve taking the second and the first derivative of the above problem which includes the following components: a) Describe how f’(x) describes the behavior of f(x) within the context of the real-world application. b) Describe how f’’(x) describes the characteristics and changes of f(x) and f’(x) in the context of the real-world application. 4. Provide an answer that comprises of all relevant mathematical justifications for each step in the real-world solution context.

Question A jet follows a path with distance in km, which is given by: Given that the horizontal velocity is expressed as V(x) = x, find the direction and magnitude of the velocity when the jet hits the ground if time taken is in minutes. The assumption made here is such that the terrain is all level (Bourne, 2011). Solution Let us first see a graph of the motion, to clarify what is going on. It can be seen that the jet hits the ground again somewhere around x = 9.5 km. At this point, the horizontal velocity is positive (the jet is from going left to right) and the vertical velocity is negative (the jet is going down). "V(x) = x" means that as x increases, the horizontal velocity also increases with the same number (different units, of course).

So for example, at x = 2 km, the horizontal speed is 2 km/min, and at x = 7 km, the horizontal speed is 7 km/min, and so on. To calculate the magnitude of the velocity as the jet hits the ground, it is important that we know the vertical and horizontal aspects of the velocity at this instance. (1) Horizontal velocity. In order to find the exact point the rocket hits the ground, it is necessary to find a solution for the following equation we only need to solve the following: Factorizing gives: And solving for 0 gives us x = 0, x = -3v10, x = 3v10 We only need the last value, x = 3v10 ? 9.4868 km (This value is consistent with the graph above).

So the horizontal speed when the rocket hits the ground is 9.4868 km/min (since V(x) = x). (2) Vertical velocity. We now need to use implicit differentiation with respect to t (not x!) to find the vertical velocity. However, we already know dx/dt and x at impact, so we simply substitute: This gives us a negative velocity, as we expected before: So now, we need to calculate the magnitude of the velocity. This considers both the horizontal and vertical components. Magnitude = Substituting, we have: Velocity has magnitude and direction.

Now for part of the direction. Angle of motion: Substituting our vertical and horizontal components, we have: In degrees, this is equivalent to -1.107148718 ? 57.25578 = -63.3907° We can see that this answer is reasonable by zooming in on that part of the graph where the jet hits the ground (with equal-axis scaling): Therefore, in summary, the velocity of the jet when it hits the ground is 21.2 km/min in the direction 63.4