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Finding an acceleration of the line with tangent.

 
 
Reply Tue 26 Jul, 2005 10:49 pm
How do you find acceleratiion of the graph at any given point when you draw a tangent line to that point? What do you do first? Do you need a slope formula or can you just use acceleration formula?
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Type: Discussion • Score: 1 • Views: 798 • Replies: 5
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Grand Duke
 
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Reply Tue 26 Jul, 2005 11:03 pm
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.
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raprap
 
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Reply Wed 27 Jul, 2005 05:04 am
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∫;?/
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babylonian
 
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Reply Wed 27 Jul, 2005 03:51 pm
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?
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raprap
 
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Reply Wed 27 Jul, 2005 07:05 pm
Too many fuzzy definitions.

I'm not sure what a trend line is. Is the output of the 'trend line' the units being sold or the rate of the units being sold? But in my usual bumfooted manner I'll try to muddle on.

If a ?'trend line' is the former (unit sales), the slope (rise over run) of that line is the rate of sales. This slope is effectively the 'first' derivative.

If the 'trend line' is a rate of unit sales it is already the first derivative (of unit sales) and the slope of that 'line' is the acceleration (second derivative) of unit sales.

I like Physics and Newton (who came up with all of this mess) and how he thought about the behavior of an apple. The distance of an apple is how far it could fall from a tree. This distance is analogous to the number of units sold. But like the apple these sales started slow and got faster (velocity) with time. So the velocity wasn't constant. This velocity is equivalent to the rate of unit sales.

This rate of change of the velocity of the apple is the acceleration, and Newton found this acceleration to be constant. Now if the velocity is increasing the velocity is normally called acceleration, and if decreasing it's called decelerating, but this is different names for the same thing. The difference between acceleration and deceleration is the 'sign' on the rate slope (velocity).

Rap c∫;?/
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babylonian
 
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Reply Sat 30 Jul, 2005 09:36 am
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 Very Happy
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