Tangent Lines/Instaneous Velocity

In summary: I will fix my calculations and try again. In summary, the conversation revolved around creating a position-time graph and using it to draw tangents and calculate instantaneous velocity. The person had difficulty drawing the tangents correctly and calculating the velocities accurately. It was suggested to choose a tangent at t=1.25 instead of t=1.50 due to the lack of data points after t=1.5. There was also a mistake in the calculations, where the slope was being calculated incorrectly.
  • #1
pebbles3
9
0

Homework Statement


I need to use the following data table to:
1. make a position-time graph
2. draw tangents (it isn't specified how many, but a previous practice question only drew 3 out of a possible 5)
3. create a time-velocity table.

0 --- 0
0.25 --- 0.29
0.50 --- 1.15
0.75 --- 2.59
1.00 --- 4.60
1.25 --- 7.19
1.50 --- 10.35

Homework Equations


vinst = d/t

The Attempt at a Solution


The position-time graph is attached. I think that's right, but what I'm really having trouble with is drawing the tangent lines and calculating the instantaneous velocity. I feel like I didn't draw the tangents right but I'm not sure how to do it differently.
Here's my calculations:
0.50: 7 / 1.75-0.25 = 3.7
1.00: 10 / 1.65-0.50 = 5.5
1.50: 10 / 1.50-0.75 = 6.2
 

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  • #2
pebbles3 said:

Homework Statement



The position-time graph is attached. I think that's right, but what I'm really having trouble with is drawing the tangent lines and calculating the instantaneous velocity. I feel like I didn't draw the tangents right but I'm not sure how to do it differently.
Here's my calculations:
0.50: 7 / 1.75-0.25 = 3.7
1.00: 10 / 1.65-0.50 = 5.5
1.50: 10 / 1.50-0.75 = 6.2
A tangent line at a certain point on the graph is a straight line that touches the graph only at that point.

What does the slope of the tangent line on a distance-time graph represent? Does that help you determine points for a velocity-time graph?

AM
 
  • #3
pebbles3 said:

Homework Statement


I need to use the following data table to:
1. make a position-time graph
2. draw tangents (it isn't specified how many, but a previous practice question only drew 3 out of a possible 5)
3. create a time-velocity table.

0 --- 0
0.25 --- 0.29
0.50 --- 1.15
0.75 --- 2.59
1.00 --- 4.60
1.25 --- 7.19
1.50 --- 10.35

Homework Equations


vinst = d/t

The Attempt at a Solution


The position-time graph is attached. I think that's right, but what I'm really having trouble with is drawing the tangent lines and calculating the instantaneous velocity. I feel like I didn't draw the tangents right but I'm not sure how to do it differently.
Here's my calculations:
0.50: 7 / 1.75-0.25 = 3.7
1.00: 10 / 1.65-0.50 = 5.5
1.50: 10 / 1.50-0.75 = 6.2

It seems to me you did an excellent job, and have the right idea. Nice work.
 
  • #4
Your tangent lines (supposedly tangent at .5, 1.0 and 1.5) are not touching at those points. This affects the calculation of velocity at those points. If you plot the velocity vs. time you will see this. I would choose a tangent at t=1.25 since you do not have any points after t=1.5.

AM
 
  • #5
Andrew Mason said:
Your tangent lines (supposedly tangent at .5, 1.0 and 1.5) are not touching at those points. This affects the calculation of velocity at those points. If you plot the velocity vs. time you will see this. I would choose a tangent at t=1.25 since you do not have any points after t=1.5.

AM

Thanks for the help.
Do you mean do a tangent at 1.25 instead of at 1.50 or do both?
 
  • #6
Andrew Mason said:
Your tangent lines (supposedly tangent at .5, 1.0 and 1.5) are not touching at those points. This affects the calculation of velocity at those points. If you plot the velocity vs. time you will see this. I would choose a tangent at t=1.25 since you do not have any points after t=1.5.

AM

Gee. It looks to me like they are pretty tangent at these points. The tangent lines aren't perfect, but they look pretty good. Yet, when I calculated the velocities using central finite difference approximations, I got a straight line as a function of time, with all the velocities calculated from the drawn tangents lying below the (more accurate) numerically calculated velocities.

Chet
 
  • #7
pebbles3 said:
Thanks for the help.
Do you mean do a tangent at 1.25 instead of at 1.50 or do both?
The problem with drawing a tangent at 1.5 is that you don't know what the line looks like after. I would use 1.25 but you can use as many as you like: you can use any points in between 0 and 1.5.

AM
 
  • #8
pebbles3:

Actually, the real problem is with your calculations. The slope of the tangent you have drawn at t=1.0 is 10/(1.65-.50) = 9.1 not 5.5. The slope of the tangent as you have drawn it at 1.5 is 10.35/(1.5-.75) = 13.8 (the point is (10.35, 1.5)). Check the slope calculation for t=.5. That should give you an accurate v-t graph.

You seem to be dividing by the first term in the denominator and then subtracting the second! Slope is rise/run

AM
 
Last edited:
  • #9
Andrew Mason said:
pebbles3:

Actually, the real problem is with your calculations. The slope of the tangent you have drawn at t=1.0 is 10/(1.65-.50) = 9.1 not 5.5. The slope of the tangent as you have drawn it at 1.5 is 10.35/(1.5-.75) = 13.8 (the point is (10.35, 1.5)). Check the slope calculation for t=.5. That should give you an accurate v-t graph.

You seem to be dividing by the first term in the denominator and then subtracting the second! Slope is rise/run

AM

Ah I see, thank you!
 

1. What is a tangent line?

A tangent line is a straight line that touches a curve at one point without crossing it. It represents the slope of the curve at that point.

2. How is a tangent line related to instantaneous velocity?

In calculus, the slope of a tangent line at a point on a curve is equal to the instantaneous velocity at that point. This is because the tangent line represents the rate of change of the curve at that specific point.

3. Can a tangent line have multiple points of intersection with a curve?

No, a tangent line can only intersect a curve at one point. This is because the tangent line represents the slope of the curve at that point, and a curve can only have one slope at any given point.

4. How can tangent lines be used to find the maximum and minimum points of a curve?

The maximum and minimum points of a curve occur when the tangent line is horizontal, or when the slope of the curve is equal to 0. By finding the points where the tangent line is horizontal, we can determine the maximum and minimum points of the curve.

5. Can tangent lines be used to find the direction of motion of a curve?

Yes, the direction of motion of a curve can be determined by the direction of the tangent line. If the tangent line is sloping upwards, the curve is moving in a positive direction, while a downward-sloping tangent line indicates a negative direction of motion.

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