Finding an acceleration of the line with tangent.

It's been 10 years since I learned this stuff, but I *think* it's through either differntiation or integration of the formula which describes the graph. I always get them mixed up.

Hopefully I have been of minor help, sorry I couldn't help further.

Say a graph measures some dependent variable with respect to some independent variable (say y with respect to x). The differential is the rate of change of that dependant variable with respect to the independent variable. This value at a point is the same as the slope of the tangent line at that point.

This slope; however, is not an acceleration. It is a velocity. The acceleration is the second derivative of this dependant variable with respect to the independent variable. The acceleration is then rate of change of the velocity at that point.

If I were to graphically determine the acceleration. I would take tangents of the graph around the point in question and determine the slope of those tangent lines. Then I'd plot those slopes against the independent variable values corresponding to those slopes. Then I take the tangent of those rates of changes at the point I was interested in. The slope of that tangent would be the acceleration at the independent point of interest.

Rap c∫;?/

rap...let me ask mate.....
if i did a trend line in excel of such a graph, and made it a second order, then got the equation of the line, could i equate this roughly to accel in this case?

spot on buddy...also,, just out of interest as i did go ahead and try it...
in excel, the trend utility is the rate and when i graphed some numbers, the trend yeilded the velocity slope and displayed the equation from which i could draw the factor....
increasing the order only increased the 'accuracy' depending on the visual fit....
to yeild the acceleration, you have to take a trend of the original trend line.....
hahahha

silly but fun to know