Modeling Rates of Change: Quiz
Problem 4
Consider the following table of values of an unknown distance function D(t), its first and second derived values.
| t | D(t) | first derived | second derived
|
| –4 | 9.2 | –2.5 | 0.4
|
| –3.5 | 7.95 | –2.3 | 0.4
|
| –3 | 6.8 | –2.1 | 0.4
|
| –2.5 | 5.75 | –1.9 | 0.4
|
| –2 | 4.8 | –1.7 | 0.4
|
| –1.5 | 3.95 | –1.5 | 0.4
|
| –1 | 3.2 | –1.3 | 0.4
|
| –0.5 | 2.55 | –1.1 | 0.4
|
| 0 | 2 | –0.9 | 0.4
|
| 0.5 | 1.55 | –0.7 | 0.4
|
| 1 | 1.2 | –0.5 | 0.4
|
| 1.5 | 0.95 | –0.3 | 0.4
|
| 2 | 0.8 | –0.1 | 0.4
|
| 2.5 | 0.75 | 0.1 | 0.4
|
| 3 | 0.8 | 0.3 | 0.4
|
| 3.5 | 0.95 | 0.5
|
| 4 | 1.2
|
Which of the following conclusions are supported by the table?
- D(t) is quadratic.
- The acceleration function is constant.
- Because t is increasing, D(t) describes a falling object.
- The velocity function is constant.
- The instantaneous velocity at any point is zero.
- D(t) will increase without bound as t gets larger.
Answers, problem 4
- D(t) is quadratic. Yes, because a quadratic's first derivative will be linear and its second derivatives will be constant.
- The acceleration function is constant. Yes, because the second derivative, which represents acceleration, is constant for all values of t.
- Because t is increasing, D(t) describes a falling object. No, D(t) is not given to be the distance an object falls or moves vertically. D(t) might describe the position of a point on a line.
- The velocity function is constant. No, because the first derived values are not constant.
- The instantaneous velocity at any point is zero. No, we cannot calculate instantaneous velocities from this table.
- D(t) will increase without bound as t gets larger. Yes, because that is the trend we can deduce from the data in the table.