Gradients of Graphs with and without Rates of Change

[Jimmy87]

Homework Statement

Determining which graphs technically use gradients and which do not. This is not a homework question but will help me with it so I put it here. My instructor said you have to be careful whether or not to call what you are calculating from the graph as the "gradient" or not. For example he said that most students think the gradient of a voltage vs current graph gives you the resistance but it does not - the ratio of voltage to current gives you the resistance. If you wanted to calculate the resistance on a curved pat of the graph you would literally do V/I but if it was a distance-time graph you would calculate the gradient of the tangent at that point. My question is - for curved graphs when would you draw a tangent and technically calculate 'gradient' and when would you just divide the 'y' by the 'x' value. It did some searching on google and it seems to say only rate changing graphs involve calculating the gradient - i.e. time on the x-axis. Are there any other examples where you would have to draw a tangent on a curved graph or is it only time?

Another question I had was about the force-extension graph for a spring. If you collect multiple load forces and extensions for a spring and plot a graph you get a straight line graph through the origin (force vs extension). We were told to calculate the gradient to find the spring constant using the whole line as it is more accurate than only using one set of y and x values. Is the gradient of a force-extension the spring constant? As I thought only graphs with time on the x-axis are gradients? So the spring constant is literally the ratio of force and extension isn't it? If you go past the elastic limit you wouldn't draw a tangent to find the spring constant you would do the ratio of the force to the extension i.e. one divided by the other. So why would you do change in y divided by change in x for a force-extension graph?

Homework Equations

The Attempt at a Solution

 

[Charles Link]

Just a simple comment or two: In electronics, transistor characteristic curves are often used, and in a typical case, you might work with a graph of collector current $I_C$ vs. base current $I_B$. These curves are in general very non-linear, and most often, the transistor has bias DC voltages and/or DC bias currents applied to it without any small ac signal so that any added ac signal operates at some chosen operating point or region somewhere in the middle of the graph. In these cases, the slope of the curve at the operating point determines how the transistor will respond to a small applied ac signal. Oftentimes, the ac signal is coupled in via a capacitor which passes the ac signal without affecting the DC values.

 

[Jimmy87]

Ok, I don't think I was clear in my OP so I will give an example to show better what I am confused with. What I am confused with is when you have a CURVED graph I don't always know whether to take the gradient of the slope/tangent or just divide the y-value by the x-value. This first example is a distance-time graph:

If I make up some numbers - let's say for that point shown on the graph the y value is 10 and the x value is 2. The instantaneous speed would not be 5 m/s (10/2). You would have to draw a tangent at the point shown and calculate the gradient of that.

But if we look at a different graph you have to do something different:

To make it easier let's also say that for the point shown on this graph the y value is also 10 and x value is also 5. In this case the resistance is 5 and calculating the gradient of the slope/tangent at that point would give you the wrong answer. I did some research on google and websites seem to say that this is because resistance is not the rate of change of current with respect to voltage - it is just the ratio of the voltage to the current. I do understand both these graphs but when I see a new graph I never know which approach to take so please could someone give some guidance on how to know whether you take the gradient of the tangent or just simply divide the y value by the x value for points that sit on the curved part of the graph?

To show an example, this graph is new to me:

This is a force-extension graph for a rubber band. I know that F = kx where k is the force constant. In this graph it obviously is not a constant but 'k' tells you the stiffness so if I wanted to calculate the stiffness 'k' on the curved bits how do you know whether the gradient of the tangent would give you this or whether you simply divide the y value by the x value at that point?

Many thanks for any help!

 

[Charles Link]

In the case of the rubber band, let's do a somewhat simpler case that might help illustrate the concept. Suppose when you stretch it for the first three inches $k=.1$ units of force/inch, but then suddenly became much stiffer so that for the next 3 inches, from 3 inches to 6 inches, $k=.5$ units of force/inch . What would the graph of Force vs. Extension look like? $\\$ Also, if the rubber band was already stretched by an added weight to the 3 inch point, and you tried stretching it farther, would you have any clue that the rubber band actually had a region where it wasn't a stiff rubber band? Would an experiment that you did in this case of Added Force vs. Additional distance give you the result that $k=.5$ for this rubber band? $\\$ And suppose you then repeated the experiment, this time by removing the originally extra added weight, and then slowly added weights of your own. You should be able to generate the complete curve of Force vs. Extension, with the second part of it, from 3 to 6 inches, looking just like what you got from the graph of Added Force vs. Additional Distance of your first experiment, except that the origin will be different, both horizontally and vertically. $\\$ (i.e. what was (0,0) in the first experiment will now be located at the point (3 inches, .3 units of force))

 

[Jimmy87]

In the case of the rubber band, let's do a somewhat simpler case that might help illustrate the concept. Suppose when you stretch it for the first three inches $k=.1$ units of force/inch, but then suddenly became much stiffer so that for the next 3 inches, from 3 inches to 6 inches, $k=.5$ units of force/inch . What would the graph of Force vs. Extension look like? $\\$ Also, if the rubber band was already stretched by an added weight to the 3 inch point, and you tried stretching it farther, would you have any clue that the rubber band actually had a region where it wasn't a stiff rubber band? Would an experiment that you did in this case of Added Force vs. Additional distance give you the result that $k=.5$ for this rubber band? $\\$ And suppose you then repeated the experiment, this time by removing the originally extra added weight, and then slowly added weights of your own. You should be able to generate the complete curve of Force vs. Extension, with the second part of it, from 3 to 6 inches, looking just like what you got from the graph of Added Force vs. Additional Distance of your first experiment, except that the origin will be different, both horizontally and vertically. $\\$ (i.e. what was (0,0) in the first experiment will now be located at the point (3 inches, .3 units of force))

Thanks for all the info. I am still unsure how you tell whether to take the gradient of a tangent or the ratio of the co-ordinates?

 

[Charles Link]

Thanks for all the info. I am still unsure how you tell whether to take the gradient of a tangent or the ratio of the co-ordinates?

In the example I gave you, the gradient will simply be the slope of the graph. The slope is constant in that example for quite a wide range=e.g. from x=0 to x=3 where it is m=.1, and then from x=3 to x=6, where it is m=.5. $\\$ In the case of the rubber band experiment, do you want to measure the whole distance displaced and know the total weight, or do you want to know what will happen if you start with some weight already attached, and then see how far it stretches from that point with additional weight? $\\$ The slope will give you what happens with smaller weights. If you want to attach a larger weight in a region where the slope isn't constant, you may need to take the complete graph, and find where adding that much additional weight takes you on the graph. The difference in the x coordinate is how much it stretches. For constant slope $k$ , you can just use $\Delta Weight=k \Delta x$ , but this only works if the slope is constant.

 

[Jimmy87]

In the example I gave you, the gradient will simply be the slope of the graph. The slope is constant in that example for quite a wide range=e.g. from x=0 to x=3 where it is m=.1, and then from x=3 to x=6, where it is m=.5. $\\$ In the case of the rubber band experiment, do you want to measure the whole distance displaced and know the total weight, or do you want to know what will happen if you start with some weight already attached, and then see how far it stretches from that point with additional weight?

Starts with no weight attached and gradually add one weight at a time and measure the extension. I just want to know how you would calculate the stiffness at a point on the curved parts of the graph?

 

[Charles Link]

Starts with no weight attached and gradually add one weight at a time and measure the extension. I just want to know how you would calculate the stiffness at a point on the curved parts of the graph?

The slope of the graph tells you the shape the graph has (=stiffness) at that point if you add small weights. See also the last sentence or two of my previous post, where the case of non-constant slope and larger weights is considered.

 

[Jimmy87]

The slope of the graph tells you the shape the graph has (=stiffness) at that point if you add small weights. See also the last sentence or two of my previous post, where the case of non-constant slope and larger weights is considered.

Why doesn't the slope of the graph on a voltage vs current graph not tell you the resistance?

 

[Charles Link]

Why doesn't the slope of the graph on a voltage vs current graph not tell you the resistance?

In the case of a small (e.g. ac signal) added to the D.C. value, the slope would indeed give the small signal resistance. For D.C. signals though , resistance is defined as $R=V_{ D.C. \,total }/I_{D.C. \, total }$.

 

[Jimmy87]

Why doesn't the slope of the graph on a voltage vs current graph not tell you the resistance?

In the case of a small (e.g. ac signal) added to the D.C. value, the slope would indeed give the small signal resistance. For D.C. signals though , resistance is defined as $R=V_{ D.C. \,total }/I_{D.C. \, total }$.

Does this sound right - the slope will always be the correct method if it is the derivative of y and x? Since resistance is not the derivative of voltage and current the slope doesn't give you the resistance? is that correct?

 

[Charles Link]

Does this sound right - the slope will always be the correct method if it is the derivative of y and x? Since resistance is not the derivative of voltage and current the slope doesn't give you the resistance? is that correct?

Yes, that is correct. :) Except of course in the case where the resistance is constant everywhere. Then the slope is the same as $R=V/I$.

 

[Jimmy87]

Yes, that is correct. :) Except of course in the case where the resistance is constant everywhere. Then the slope is the same as $R=V/I$.

Thank you. So I have drawn a point on the elastic band graph:

If I just did the y-value divided by the x value at this point I would be dividing the total force by the total extension which would give an average stiffness from 0N to how ever many Newtons the y value is at that point. If I took the gradient of the tangent at that point I would be calculating the stiffness at that exact point? Is that correct?

 

[Charles Link]

Yes. Very good. In addition, starting from any point, the y-coordinate tells you how much weight is already attached. If you want to add additional weight, increase the $y$ by that additional weight, and find the change in x that occurs. For small weights $w$ this will be $w=k \Delta x$ where $k$ is the slope at that point. For larger weights, the line connecting the two points on the graph will be the average slope between those two points, but it won't be perfectly tangent to the graph.

 

[Jimmy87]

Yes. Very good. In addition, starting from any point, the y-coordinate tells you how much weight is already attached. If you want to add additional weight, increase the $y$ by that additional weight, and find the change in x that occurs. For small weights $w$ this will be ## $w=k \Delta x$ $where$ k ## is the slope at that point. For larger weights, the line connecting the two points on the graph will be the average slope between those two points, but it won't be perfectly tagent to the graph.

Great - thanks! Do most mainstream equations in physics require the gradient of the slope to be calculate e.g. force-extension, distance-time, acceleration-time etc all do. The only mainstream one I know that doesn't is voltage vs current. Are there any other ones where the variable you want requires the total y divided by the total x?

 

[Charles Link]

Great - thanks! Do most mainstream equations in physics require the gradient of the slope to be calculate e.g. force-extension, distance-time, acceleration-time etc all do. The only mainstream one I know that doesn't is voltage vs current. Are there any other ones where the variable you want requires the total y divided by the total x?

Magnetization is another place where such a calculation is done. Often the magnetic susceptibility is computed as $\chi_m=\frac{dM}{dH }$, where $M$ is the magnetization, and $H$ is the applied magnetic field. It also can be very useful to compute $\chi_{total}=M/H$. The complete graph of $M$ vs. $H$ supplies all of this information. In many cases, the part that gets measured experimentally is $\chi_m$. $\\$ Even in a distance vs. time graph, do you want to know how fast you are going, or the average speed of the complete trip? In one case, you use a gradient type operation $v=\frac{ds}{dt}$, and in the other you simply compute $v_{average}=\frac{s}{t}$.

 

[Jimmy87]

Magnetization is another place where such a calculation is done. Often the magnetic susceptibility is computed as $\chi_m=\frac{dM}{dH }$, where $M$ is the magnetization, and $H$ is the applied magnetic field. It also can be very useful to compute $\chi_{total}=M/H$. The complete graph of $M$ vs. $H$ supplies all of this information. In many cases, the part that gets measured experimentally is $\chi_m$. $\\$ Even in a distance vs. time graph, do you want to know how fast you are going, or the average speed of the complete trip? In one case, you use a gradient type operation $v=\frac{ds}{dt}$, and in the other you simply compute $v_{average}=\frac{s}{t}$.

Great. Thanks a lot for your time to help me out - it is very much appreciated!

 

[CWatters]

Great - thanks! Do most mainstream equations in physics require the gradient of the slope to be calculate e.g. force-extension, distance-time, acceleration-time etc all do. The only mainstream one I know that doesn't is voltage vs current. Are there any other ones where the variable you want requires the total y divided by the total x?

Consider what happens when you heat water. The specific heat capacities for ice,water and steam are different. So if you plot a graph of temperature Vs energy you find the slope changes in the different phases (edit: not to mention some steps due to the latent heat of melting and vapourisation).

Suppose you wanted to know the total amount of energy in the water, you would effectively be interested in the slope from absolute zero to the temperature of interest.

However if you are interested in the energy required to heat water from say 20C to 40C then you are only interested in the slope over that narower range.

 

[CWatters]

Consider a car climbing an irregular shape hill. Let's say it takes some time to climb right to the top.

The average power required would depend on the height of the hill (gain in PE) divided by the time.

The peak power required would depend on how fast it goes up the steepest part of the slope.

In short there are many cases where you might want to calculate either the slope relative to the origin or the slope at a particular point.

 

[haruspex]

Just a comment on how intuition might be playing into this.
Gradients are commonly useful where the independent variable would typical vary continuously and unidirectionally. Thus, anything plotted against time, or height plotted against distance along a line (not necessarily straight). In these contexts local gradient has obvious meaning and utility. Average gradient, Δy/Δx, is also meaningful/useful, but no more so than local gradient.
In the voltage/current context (more logically I/V, since V would be the independent variable) local gradient still has a meaning but is less obviously useful in general. Thus we speak of resistance (conductance for I/V) rather than specifying local resistance. Were it the case that standard terminology is "local resistance" and "average resistance" I think your confusion would not arise.