Ask a New Question
Menu

A fly starts out 2 meters from a lightbulb, flies closer to the light, then farther away (2

meters again). At this point the fly goes toward the bulb again, but hits the bulb and then
finally flies away.
A. Find a function (it may be piecewise) that gives the distance of the bug from the light as
a function of time. [There is more than one solution here—you have lots of freedom with
what your function can be, but it must satisfy the situational requirements given.]
Demonstrate that your function DOES satisfy the situational requirements.

1 answer

To model the fly's movement with respect to time, we can use a piecewise function. Let's denote the distance of the fly from the lightbulb at time \( t \) (in seconds) as \( d(t) \).

Piecewise function: Assuming we break the movement of the fly into distinct segments, we can describe its flight as follows:

  1. From \( t = 0 \) to \( t = 1 \) second, the fly moves towards the lightbulb.
  2. From \( t = 1 \) to \( t = 2 \) seconds, the fly moves away from the lightbulb.
  3. From \( t = 2 \) to \( t = 3 \) seconds, the fly again moves towards the lightbulb and hits it.
  4. From \( t = 3 \) to \( t = 4 \) seconds, the fly finally flies away from the lightbulb.

Defining the segments of \( d(t) \):

  1. For \( 0 \leq t < 1 \): Assume the fly moves towards the bulb linearly from 2 meters to 0 meters: \[ d(t) = 2 - 2t \quad (0 \leq t < 1) \] At \( t = 0, d(0) = 2 \) and at \( t = 1, d(1) = 0\).

  2. For \( 1 \leq t < 2 \): The fly moves away from the bulb linearly from 0 meters to 2 meters: \[ d(t) = 2t - 2 \quad (1 \leq t < 2) \] At \( t = 1, d(1) = 0 \) and at \( t = 2, d(2) = 2\).

  3. For \( 2 \leq t < 3 \): The fly moves towards the bulb again, hitting it at \( t = 3\): \[ d(t) = 2 - 2(t - 2) = 6 - 2t \quad (2 \leq t < 3) \] At \( t = 2, d(2) = 2 \) and at \( t = 3, d(3) = 0\).

  4. For \( 3 \leq t < 4 \): After hitting the bulb, it flies away, from 0 to 2 meters again: \[ d(t) = 2(t - 3) \quad (3 \leq t < 4) \] At \( t = 3, d(3) = 0\) and at \( t = 4, d(4) = 2\).

Complete Piecewise Function:

Combining all segments, we get: \[ d(t) = \begin{cases} 2 - 2t & \text{if } 0 \leq t < 1 \ 2t - 2 & \text{if } 1 \leq t < 2 \ 6 - 2t & \text{if } 2 \leq t < 3 \ 2(t - 3) & \text{if } 3 \leq t < 4 \ \end{cases} \]

Verification of Requirements:

  • At \( t = 0 \): \( d(0) = 2 \) (2 meters from the bulb)
  • At \( t = 1 \): \( d(1) = 0 \) (fly reaches the bulb)
  • At \( t = 2 \): \( d(2) = 2 \) (fly 2 meters away again)
  • At \( t = 3 \): \( d(3) = 0 \) (fly hits the bulb)
  • At \( t = 4 \): \( d(4) = 2 \) (fly is 2 meters away again)

The piecewise function describes the fly's journey appropriately and satisfies all the situational requirements.