Is it possible to obtain a tangent from a graph using Excel?

Firstly, you've got the things on the wrong axes. Time is usually shown on the x-axis, but the way you have it there, a tangent will be measuring the instantaneous rate of change of time with respect to change in volume. (wouldn't you wanna measure how volume changes with time and not the other way round?)

Secondly, there is no tangent at any of those points since they're not "smooth" but you can approximate it by taking the difference of (volume at time t - volume at time t -1) and dividing that by the difference in time.
 
feo said:
Firstly, you've got the things on the wrong axes. Time is usually shown on the x-axis, but the way you have it there, a tangent will be measuring the instantaneous rate of change of time with respect to change in volume. (wouldn't you wanna measure how volume changes with time and not the other way round?)

Secondly, there is no tangent at any of those points since they're not "smooth" but you can approximate it by taking the difference of (volume at time t - volume at time t -1) and dividing that by the difference in time.
Click to expand...

Agreed. Dominic, if you want to find for example the approximate rate of change of time with respect to volume (assuming your axes are correct, see feo's post ;)) at a volume of 60 cm3, you could use:
(time(80)-time(40))/(80-40) where time(80) is the time when volume is 80. This is the usual formula of a gradient, change in y-value over change in x-value :)
 
wonder.gif


:D
 
No, the graph has the correct axes...our science teacher told us that much. But thanks for the 2nd part ;). Very helpful.
 
I'm with feo, your time axis should always be the x axis. You measure a change of volume with respect to time, not vice versa.

Are you supposed to work it out from a graph? Or do you have a formula? This sounds suspiciously like a differential calculus problem.
 
What is the application? For change in time with respect to volume, that is... ?
 
You'll get better results if you've got more data points. Tangents are rather meaningless if you've only got half a dozen points.
 
HavocXphere said:
You'll get better results if you've got more data points. Tangents are rather meaningless if you've only got half a dozen points.
Click to expand...

Agreed. Since you only have certain datapoints, you will only be able to work out the mean gradient between points a and b, and points b and c, but never the tangent at point b. (IIR my maths C)

edit:

were you given the data points or were you given the chart as a question? If you were given the data points, please post them here.
 
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"at the same time value" makes me think that time is your independent variable and thus, as all the forumites have pointed out, should be on the x-axis.
What exactly do you need to do? Just draw tangets on the Excel graphs? Do you need formulae?
If possible, pay attention to what HavocXphere said bout including more points
 
Code: 
Concentration (mol.dm-3)      (Volume of H2 produced)
                             20	  40	 60	80     100
2                           6.98 12.72	16.21	21.76	33.82                    
1.5                        8.71	13.56   21.04	33.95	57.71
1                          11.2	20.71	33.95	59.5	99
0.75                      16.54	37.19	64.23	101.56	143.01 
0.5                        5.48	78.43	139.34	214.59	284.84



3
Ok the 2, 1.5 etc. refers to the concentration of HCL. That's fixed. The 0,20,40 etc. is the volume of hydrogen gas produced, and the other values are the times that it took to reach the required amount of H2 (using the different concentrations of HCL which reacted with 100g of magnesium)
 
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Plot as per this

x-axis> 0,20,40,60,80,1

Legend Y-Values
2, 0,6.98,12.72,16.21,21.76,33.82
1.5, 0,8.71,13.56,21.04,33.95,57.71
1,0, 11.2,20.71,33.95,59.5,99
0.75, 0,16.54,37.19,64.23,101.56,143.01
0.5, 0,35.48,78.43,139.34,214.59,284.84
 
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