Average Velocity & Instantaneous Velocity...2

[nycmathguy]

Calculus 1
Section 1.4

For part (a), just do (i).
I don't understand part (b).

 

[MathLover1]

a.
average velocity is [f(x2) - f(x1)]/(x2 - x1), or, is this case, [s(t2) - s(t1)]/(t2 - t1) so for:


t = [2, 4], v(avg )= [s(4) - s(2)]/(4 - 2) = (79.2 - 20.6)/2 = 29.3ft/s

t = [3, 4], v(avg) = [s(4) - s(3)]/(4 - 3) = (79.2- 46.5)/1 = 32.7ft/s

t = [4, 5], v(avg) = [s(5) - s(4)]/(5- 4) = (124.8 - 79.2)/1 = 45.6 ft/s

t = [4, 6], v(avg) = [s(6) - s(4)]/(6- 4) = (176.7 - 79.2)/2 = 48.75 ft/s


b.

To curve fit the data, plug it into a list in your TI-84 calculator and try the various types of curves to find the one with the best r^2 value. A quartic function gives the r^2 closest to 1 (r^2 = 0.99993). It is:

s(t) = 0.00625t^4 - 0.1300925926t^3 + 1.475694444t^2 + 0.1171957672t - 0.0115079365

Using this function and using the limit equation, for the following values of "h" I get values closer and closer to the instantaneous velocity:

For h = 0.1, v = 6.19755 ft/s

For h = 0.01, v = 6.14028 ft/s

For h = 0.001, v = 6.1345 ft/s

For h = 0.0001, v = 6.1339 ft/s

For h = 0.00001, v = 6.1339 ft/s

For h = 0.000001, v = 6.1339ft/s

So the approximation of the instantaneous velocity, at t = 3 seconds, using this curve-fit method is 6.1339 ft/s

 

[nycmathguy]

a.
average velocity is [f(x2) - f(x1)]/(x2 - x1), or, is this case, [s(t2) - s(t1)]/(t2 - t1) so for:


t = [2, 4], v(avg )= [s(4) - s(2)]/(4 - 2) = (79.2 - 20.6)/2 = 29.3ft/s

t = [3, 4], v(avg) = [s(4) - s(3)]/(4 - 3) = (79.2- 46.5)/1 = 32.7ft/s

t = [4, 5], v(avg) = [s(5) - s(4)]/(5- 4) = (124.8 - 79.2)/1 = 45.6 ft/s

t = [4, 6], v(avg) = [s(6) - s(4)]/(6- 4) = (176.7 - 79.2)/2 = 48.75 ft/s


b.

To curve fit the data, plug it into a list in your TI-84 calculator and try the various types of curves to find the one with the best r^2 value. A quartic function gives the r^2 closest to 1 (r^2 = 0.99993). It is:

s(t) = 0.00625t^4 - 0.1300925926t^3 + 1.475694444t^2 + 0.1171957672t - 0.0115079365

Using this function and using the limit equation, for the following values of "h" I get values closer and closer to the instantaneous velocity:

For h = 0.1, v = 6.19755 ft/s

For h = 0.01, v = 6.14028 ft/s

For h = 0.001, v = 6.1345 ft/s

For h = 0.0001, v = 6.1339 ft/s

For h = 0.00001, v = 6.1339 ft/s

For h = 0.000001, v = 6.1339ft/s

So the approximation of the instantaneous velocity, at t = 3 seconds, using this curve-fit method is 6.1339 ft/s

So sorry but part (b) remains unclear to me. I don't have a calculator to start with.

 

[MathLover1]

finding a function is little complicated
I suggest you to avoid this type of problems now, leave it for later

 

[nycmathguy]

finding a function is little complicated
I suggest you to avoid this type of problems now, leave it for later

I will do part (a) only. Finding a function from written information has been a struggle for me all my academic life.

You said:

"s(t) = 0.00625t^4 - 0.1300925926t^3 + 1.475694444t^2 + 0.1171957672t - 0.0115079365"

The calculator gave you this function after plugging the data table. Yes?


You also said:

"Using this function and using the limit equation, for the following values of "h" I get values closer and closer to the instantaneous velocity:"

1. What limit equation?

2. What is a limit equation?

3. I am to plug different values for "h" into the limit equation. Yes?

 

[MathLover1]

yes, the calculator gave you this function

here is explanation for 1,1,3