Calculus I - Tangent Lines and Rates of Change

Show Mobile Notice Show All Notes Hide All Notes

Mobile Notice

You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 2.1 : Tangent Lines And Rates Of Change

Back to Problem List

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.

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
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 All Solutions Hide All Solutions

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

10.5
10.1
10.01
10.001
10.0001

9.5
9.9
9.99
9.999
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 \(t = 10\).