Calculating tangents from a table

[Tarragon]

I'm given a simple table (time(s),position) to make into a graph:
0,0
2,0.6
4.,2.4
6,5.4
8,9.6
10,15
I'm asked to draw tangent lines at 4, 6, and 8 seconds.

There is no indication of how they should be drawn, are tangent lines just eyeballed? I see equation involving "lim" online and equations using functions of slopes... these don't seem relevant as I am taking grade 11 physics.

The textbook has shown one equation that involves slope;

v2-v1/t2-t = slope

If i substitute some points given it makes even less sense to me:

6-4/5.4-2.4 = 2/3 (slope?) does this mean the slope of a tangent line for 4 seconds AND 6 seconds = 2/3... something?[Moderator note: Strikeout mistake fixed.]

 

[jbriggs444]

I'm given a simple table (time(s),position) to make into a graph:

There is a recent thread, https://www.physicsforums.com/threads/use-the-data-to-construct-a-graph-etc.959452/ which is very similar, though the numeric values are different.

 

[tnich]

The textbook has shown one equation that involves slope;

v2-v1/t2-t = slope

If i substitute some points given it makes even less sense to me:

6-4/5.4-2.4 = 2/3 (slope?) does this mean the slope of a tangent line for 4 seconds AND 6 seconds = 2/3... something?

First, it would be worth your while to get in the habit of using parentheses in writing equations to make clear what operations are supposed to be done first. As you have written it, it looks you are supposed to divide 4 by 5.4, then subtract the result from 6, and then subtract 2.4. When you get beyond doing a simple calculation, those parentheses are going to prevent a lot of arithmetic errors.

So let's start with your equation for the slope and write it as
$(v_2 - v_1) / (t_2-t_1) = s$

I would suggest that to estimate the slope of the curve at $t=4$, you make $t_1=2$ and $t_2=6$, that is, use the points on either side of your target point. If you draw a line segment connecting these two points, and eyeball the tangent line at the point in between, you will see that their slopes are similar. You will notice that the question did not ask for the slope at $t=0$ or $t=10$, most likely because there are not points on both sides to use for the estimate.

 

[gneill]

There is no indication of how they should be drawn, are tangent lines just eyeballed?

When the data is just provided as a series of data points, you have a couple of options. Hand plot the data and eyeball your tangents, or use some curve-fitting algorithm to find a function that matches your data points. Then you could use calculus to find the slope at any point on that curve.

If i substitute some points given it makes even less sense to me:

6-4/5.4-2.4 = 2/3 (slope?) does this mean the slope of a tangent line for 4 seconds AND 6 seconds = 2/3... something?

That would give you the slope of a chord joining the 4 and 6 second points on the curve. It would not be a tangent to the curve at either point, but might be a good approximation for the slope of the curve somewhere between the two points; It depends upon how smooth your curve is between the selected points. Pick the points too far apart from where you want to find the slope and you'll get nonsense. Ideally you'd pick two points very close to where you want to find the slope.

When you use the "eyeball" method to draw your tangent, the usual procedure for finding the slope is to extend your tangent line until it crosses some convenient grid lines on your graph, then read off two points that the tangent passes through:

So in the above figure, the slope at x = 4 on the red curve would be given by:

$$m = \frac{\Delta y}{\Delta x} = \frac{2.9 - 1.1}{8 - 0} = 0.23 $$

 

[jbriggs444]

That would give you the slope of a chord joining the 4 and 6 second points on the curve. It would not be a tangent to the curve at either point, but might be a good approximation for the slope of the curve somewhere between the two points

The mean value theorem (if applicable) says that it'll match the slope exactly, somewhere in the interval. You just don't know where.

 

[Tarragon]

When the data is just provided as a series of data points, you have a couple of options. Hand plot the data and eyeball your tangents, or use some curve-fitting algorithm to find a function that matches your data points. Then you could use calculus to find the slope at any point on that curve.

Thank you for the reply.

But the next question asks me to calculate the slopes of these tangents and record values in a time-velocity data table and then calculate the acceleration. But if I just eyeball these tangents then a) why even draw the tangent lines if they are not exact, and b) how could there be a specific answer if the tangent lines are largely arbitrary? Shouldn't there be an exact answer based off the original table, and if so, why would the tangent line be somewhat arbitrary, or why even make a tangent line at all?

We haven't been shown any curve-fitting algorithms in this course or calculus for that matter.

Maybe this is my first question in physics where the answer is just arbitrary? Are arbitrary answers in high-school physics common?

 

[gneill]

But the next question asks me to calculate the slopes of these tangents and record values in a time-velocity data table. But if I just eyeball these tangents then a) why even draw the tangent lines if they are not exact, and b) how could there be a specific answer if the tangent lines are largely arbitrary?

They won't exactly be arbitrary, but your closest approximation by eye. Just as your curve wasn't arbitrary as you did an eyeball fit of a curve through your data points.

Until you're using software curve fitting and taking into account error ranges for the data points, you'll be doing your best to eyeball the fits and slopes.

 

[Tarragon]

They won't exactly be arbitrary, but your closest approximation by eye. Just as your curve wasn't arbitrary as you did an eyeball fit of a curve through your data points.

Until you're using software curve fitting and taking into account error ranges for the data points, you'll be doing your best to eyeball the fits and slopes.

Thank you!

 

[sophiecentaur]

I'm asked to draw tangent lines at 4, 6, and 8 seconds.

What you were given seems to be yet another badly worded question (if, indeed those are the only words in the original question). The data provided has no "tangents' because it could represent a polygon with sharp corners . Grade 11 would have the beginnings of Differential Calculus, I guess. Your comment

I see equation involving "lim" online and equations using functions of slopes... these don't seem relevant as I am taking grade 11 physics.

confirms this as the word "lim" (limit) would be in an introduction to Calculus.
Drawing the points and putting a subjectively assessed curve of best fit is all you can do, at this stage. There are ways of taking the data points and fitting them to a polynomial curve of a chosen order. (Say a quadratic or a cubic curve)

The textbook has shown one equation that involves slope;

v2-v1/t2-t = slope

This makes sense if you ignore the word "tangent". It tells you the average slope between those graph points. The slope of a curve function is taken to be the "limit" of the slope between data points as the density of data points increases and the step size approaches zero. That's what's needed to get the 'tangent'.

Maybe, next week they will start you off on Calculus?

 

[gneill]

For what it's worth, if I fit a curve of the form $x = m t^2$ to your data, I find that

$x(t) = \frac{3}{20} t^2$

fits your data quite exactly. The slope of this curve at time t is given by (a bit of calculus -- differentiation):

$slope(t) = \frac{3}{10} t$

You might wish to check your "eyeballed" slopes against this fitted curve derived slope to see how good your eye is when estimating tangents.

 

[ZapperZ]

Thank you for the reply.

But the next question asks me to calculate the slopes of these tangents and record values in a time-velocity data table and then calculate the acceleration. But if I just eyeball these tangents then a) why even draw the tangent lines if they are not exact, and b) how could there be a specific answer if the tangent lines are largely arbitrary? Shouldn't there be an exact answer based off the original table, and if so, why would the tangent line be somewhat arbitrary, or why even make a tangent line at all?

We haven't been shown any curve-fitting algorithms in this course or calculus for that matter.

Maybe this is my first question in physics where the answer is just arbitrary? Are arbitrary answers in high-school physics common?

Jbriggs444 gave you a link to a similarly-worked problem. Why haven’t you look at it? It answers all of the questions in the first post.

Zz.