Turning an exponential graph into a line graph

**iamdenis** · February 8th 2008, 03:50 PM

Hi, in physics we did an experiment which would determine the acceleration due to gravity. we dropped a golf ball from a table and made a d-t graph and all the good stuff. My teacher said that as one of the methods we must manipulate the graph so that the outcome is a straight line from which we can find the slope ( not a secant or line of best fit). Can anyone please tell me the best way to change an exponential graph into a straight line graph?

- Thanks

**ecMathGeek** · February 8th 2008, 04:13 PM

Originally Posted by iamdenis

Hi, in physics we did an experiment which would determine the acceleration due to gravity. we dropped a golf ball from a table and made a d-t graph and all the good stuff. My teacher said that as one of the methods we must manipulate the graph so that the outcome is a straight line from which we can find the slope ( not a secant or line of best fit). Can anyone please tell me the best way to change an exponential graph into a straight line graph?

- Thanks

I will venture a guess to this: Try doing some "algibraic" operation to every data point to make them all appear in a straight line. You said it's an "exponential" graph (to be clear, this does not necessarily mean it's actually an exponential graph, it just has that type of shape). So, think of different types of "exponential" functions, $e^t,~10^t,~t^2,~t^3$ , that your graph might represent and then "de-exponentialize" (I'm fairly certain that's not a word) your graph by using various inverse-"exponential" functions on each of your data points.

I hope that's somewhat clear. My main point is, if you think it looks like an $e^t$ graph, then use the opposite of $e^t$ , the $\ln$ function, to make the points a straight line. Do you know what the inverse functions to the ones I listed are?

**mr fantastic** · February 8th 2008, 05:03 PM

Originally Posted by iamdenis

Hi, in physics we did an experiment which would determine the acceleration due to gravity. we dropped a golf ball from a table and made a d-t graph and all the good stuff. My teacher said that as one of the methods we must manipulate the graph so that the outcome is a straight line from which we can find the slope ( not a secant or line of best fit). Can anyone please tell me the best way to change an exponential graph into a straight line graph?

- Thanks

First of all, plot the data. What curve seems to fit it best? The simplest curve. Probably parabolic (quadratic model) ..... Does it look like it goes through the origin?

So assume a power rule, something of the form $d = k t^m$ .

Now take the log of both sides (doesn't matter what base).

$\log d = \log (k t^m) = \log k + \log t^m = \log k + m \log t$ .

In other words, $\log d = m \log t + \log k$ .

This has the form $y = m x + c$ where y is log d, x is log t and c is log k. In other words, a line.

So here's what you do:

Take the log of all your data. Do a plot of log d versus log t. Draw the line of best fit.

The gradient gives you m. The log d intercept lets you calculate k.

Crystal ball gazing: You should find the value of m is very close to 2 and the value of k is close to 5. Depending on the accuaracy of the data, you might even be able to do better than 5 .....

Note: This data alone will not let you calculate the acceleration due to gravity ......

**iamdenis** · February 9th 2008, 04:28 PM

thanks for the help, i ended up makign the graph a fairly straight line. but it also says that we have to use the equation of the line along with one of the basic d/a/t ( d = v1t+0.5at^2..) formulas to find the acceleration

any ideas lol?
thanks again

**mr fantastic** · February 9th 2008, 04:55 PM

Originally Posted by iamdenis

thanks for the help, i ended up makign the graph a fairly straight line. but it also says that we have to use the equation of the line along with one of the basic d/a/t ( d = v1t+0.5at^2..) formulas to find the acceleration

any ideas lol?
thanks again

Well, on the one hand you have your model for the data, namely $d = k t^n$ . From the graph of log d versus log t you have calculated the values of k and n .....

On the other hand, there's a formula that says that for constant acceleration, $d = v_1 t + 0.5 a t^2$ . In this formula $v_1$ is the initial velcocity ( = 0 in your experiment, I assume?) and a is acceleration. In your experiment, a = g .....

Compare $d = k t^n$ with $d = 0.5 g t^2$ : k = 0.5 g therefore g = .....

**iamdenis** · February 9th 2008, 05:16 PM

Originally Posted by mr fantastic

Well, on the one hand you have your model for the data, namely $d = k t^n$ . From the graph of log d versus log t you have calculated the values of k and n .....

On the other hand, there's a formula that says that for constant acceleration, $d = v_1 t + 0.5 a t^2$ . In this formula $v_1$ is the initial velcocity ( = 0 in your experiment, I assume?) and a is acceleration. In your experiment, a = g .....

Compare $d = k t^n$ with $d = 0.5 g t^2$ : k = 0.5 g therefore g = .....

that helped alot man. wasnt too hard, but im just not thinking straight today. thanks for the descriptive information

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