Calculus I - Tangent Lines and Rates of Change

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Section 2.1 : Tangent Lines And Rates Of Change

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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.
1. 5.5
2. 5.1
3. 5.01
4. 5.001
5. 5.0001
1. 4.5
2. 4.9
3. 4.99
4. 4.999
5. 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).