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### Section 2-1 : Tangent Lines And Rates Of Change

7.The position of an object is given by $$s\left( t \right) = \left( {8 - t} \right){\left( {t + 6} \right)^{\frac{3}{2}}}$$ . Note that a negative position here simply means that the position is to the left of the “zero position” and is perfectly acceptable. Answer each of the following questions.

1. Compute (accurate to at least 8 decimal places) the average velocity of the object between $$t = 10$$ and the following values of $$t$$.
1. 10.5
2. 10.1
3. 10.01
4. 10.001
5. 10.0001
1. 9.5
2. 9.9
3. 9.99
4. 9.999
5. 9.9999
2. Use the information from (a) to estimate the instantaneous velocity of the object at $$t = 10$$ and determine if the object is moving to the right (i.e. the instantaneous velocity is positive), moving to the left (i.e. the instantaneous velocity is negative), or not moving (i.e. the instantaneous velocity is zero).

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a Compute (accurate to at least 8 decimal places) the average velocity of the object between $$t = 10$$ and the following values of $$t$$. Show Solution
1. 10.5
2. 10.1
3. 10.01
4. 10.001
5. 10.0001
1. 9.5
2. 9.9
3. 9.99
4. 9.999
5. 9.9999

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,

$A.V. = \frac{{s\left( t \right) - s\left( {10} \right)}}{{t - 10}} = \frac{{\left( {8 - t} \right){{\left( {t + 6} \right)}^{\frac{3}{2}}} + 128}}{{t - 10}}$

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.

$$t$$ $$A.V.$$ $$t$$ $$A.V.$$
10.5 -79.11658419 9.5 -72.92931693
10.1 -76.61966704 9.9 -75.38216890
10.01 -76.06188418 9.99 -75.93813418
10.001 -76.00618759 9.999 -75.99381259
10.0001 -76.00061875 9.9999 -75.99938125

b Use the information from (a) to estimate the instantaneous velocity of the object at $$t = 10$$ and determine if the object is moving to the right (i.e. the instantaneous velocity is positive), moving to the left (i.e. the instantaneous velocity is negative), or not moving (i.e. the instantaneous velocity is zero). Show Solution

From the table of values above we can see that the average velocity of the object is moving towards a value of -76 from both sides of $$t = 10$$ and so we can estimate that the instantaneous velocity is -76 and so the object will be moving to the left at $$t = 10$$.