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### 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.

1. 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. 1
2. 0.5
3. 0.251
4. 0.2501
5. 0.25001
1. 0
2. 0.1
3. 0.249
4. 0.2499
5. 0.24999
2. 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. 1
2. 0.5
3. 0.251
4. 0.2501
5. 0.25001
1. 0
2. 0.1
3. 0.249
4. 0.2499
5. 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 $$- 6$$.