Section 2.1 : Tangent Lines And Rates Of Change
5. The population (in hundreds) of fish in a pond is given by \(P\left( t \right) = 2t + \sin \left( {2t - 10} \right)\) answer each of the following questions.
- Compute (accurate to at least 8 decimal places) the average rate of change of the population of fish between \(t = 5\) and the following values of \(t\). Make sure your calculator is set to radians for the computations.
- 5.5
- 5.1
- 5.01
- 5.001
- 5.0001
- 4.5
- 4.9
- 4.99
- 4.999
- 4.9999
- Use the information from (a) to estimate the instantaneous rate of change of the population of the fish at \(t = 5\).
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a Compute (accurate to at least 8 decimal places) the average rate of change of the population of fish between \(t = 5\) and the following values of \(t\). Make sure your calculator is set to radians for the computations. Show Solution- 5.5
- 5.1
- 5.01
- 5.001
- 5.0001
- 4.5
- 4.9
- 4.99
- 4.999
- 4.9999
The first thing that we need to do is set up the formula for the slope of the secant lines. As discussed in this section this is given by,
\[A.R.C. = \frac{{P\left( t \right) - P\left( 5 \right)}}{{t - 5}} = \frac{{2t + \sin \left( {2t - 10} \right) - 10}}{{t - 5}}\]Now, all we need to do is construct a table of the value of \({m_{PQ}}\) for the given values of \(x\). All of the values in the table below are accurate to 8 decimal places.
\(x\) | \(A.R.C.\) | \(x\) | \(A.R.C.\) |
---|---|---|---|
5.5 | 3.68294197 | 4.5 | 3.68294197 |
5.1 | 3.98669331 | 4.9 | 3.98669331 |
5.01 | 3.99986667 | 4.99 | 3.99986667 |
5.001 | 3.99999867 | 4.999 | 3.99999867 |
5.0001 | 3.99999999 | 4.9999 | 3.99999999 |
b Use the information from (a) to estimate the instantaneous rate of change of the population of the fish at \(t = 5\). Show Solution
From the table of values above we can see that the average rate of change of the population of fish is moving towards a value of 4 from both sides of \(t = 5\) and so we can estimate that the instantaneous rate of change of the population of the fish is 400 (remember the population is in hundreds).