Section 2.1 : Tangent Lines And Rates Of Change
4. The volume of air in a balloon is given by \(\displaystyle V\left( t \right) = \frac{6}{{4t + 1}}\) answer each of the following questions.
- Compute (accurate to at least 8 decimal places) the average rate of change of the volume of air in the balloon between \(t = 0.25\) and the following values of \(t\).
- 1
- 0.5
- 0.251
- 0.2501
- 0.25001
- 0
- 0.1
- 0.249
- 0.2499
- 0.24999
- Use the information from (a) to estimate the instantaneous rate of change of the volume of air in the balloon at \(t = 0.25\).
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a Compute (accurate to at least 8 decimal places) the average rate of change of the volume of air in the balloon between \(t = 0.25\) and the following values of \(t\). Show Solution- 1
- 0.5
- 0.251
- 0.2501
- 0.25001
- 0
- 0.1
- 0.249
- 0.2499
- 0.24999
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,
\[A.R.C. = \frac{{V\left( t \right) - V\left( {0.25} \right)}}{{t - 0.25}} = \frac{{\frac{6}{{4t + 1}} - 3}}{{t - 0.25}}\]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. In several of the initial values in the table the values terminated and so the “trailing” zeroes were not shown.
\(x\) | \(A.R.C.\) | \(x\) | \(A.R.C.\) |
---|---|---|---|
1 | -2.4 | 0 | -12 |
0.5 | -4 | 0.1 | -8.57142857 |
0.251 | -5.98802395 | 0.249 | -6.01202405 |
0.2501 | -5.99880024 | 0.2499 | -6.00120024 |
0.25001 | -5.99988000 | 0.24999 | -6.00012000 |
b Use the information from (a) to estimate the instantaneous rate of change of the volume of air in the balloon at \(t = 0.25\). Show Solution
From the table of values above we can see that the average rate of change of the volume of air is moving towards a value of -6 from both sides of \(t = 0.25\) and so we can estimate that the instantaneous rate of change of the volume of air in the balloon is \( - 6\).