Exponential Graph: Is Square Root an Exponential?

[jdawg]

Homework Statement

Is a function that has a square root an exponential graph since you can rewrite a square root as x^1/2?

Homework Equations

The Attempt at a Solution

 

[Ray Vickson]

Homework Statement

Is a function that has a square root an exponential graph since you can rewrite a square root as x^1/2?

Homework Equations

The Attempt at a Solution

I cannot figure out what you are asking, since your question is not written as a proper sentence.

However, you seem to be confusing "exponential" with "exponent"; exponent = power n in the formula x^n. This is a "power" function, not an "exponential" function.

 

[jdawg]

... So is a square root function linear?

 

[BvU]

Nope. "function(a times x)" is not equal to "a times function (x)"

 

[jdawg]

How would you classify this equation?: v=A√h

 

[Ray Vickson]

How would you classify this equation?: v=A√h

Its a power law of the form $v = A h^n$ with $n = 1/2$.

 

[jdawg]

Hm... I'm trying to write a lab report and the question asks if the graph is linear or exponential. When I plugged in my values for v (the y axis) and √h (the x axis)to excel the graph came up as looking linear, but with a very slight curve to it. I know I have my axes right because my lab sheet told us which values to put on each axis.

 

[Ray Vickson]

Hm... I'm trying to write a lab report and the question asks if the graph is linear or exponential. When I plugged in my values for v (the y axis) and √h (the x axis)to excel the graph came up as looking linear, but with a very slight curve to it. I know I have my axes right because my lab sheet told us which values to put on each axis.

You could do an excel "least-squares fit" to the data log(v) vs. log(h). If a power law $v = A h^m$ applies then you get a linear equation $\log(v) = \log(A) + m \log(h)$ between the new variables $y = \log(v)$ and $x = \log(h)$. In the straight-line fit, the intercept would be log(A) and the slope would be the exponent, m.

On the other hand, if the relationship is "exponential", that means that $v = A b^h$, and you would get a linear relationship between $y = \log(v)$ and $x = h$ itself; that is, $\log(v) = \log(A) + \log(b)\: h$.

In EXCEL you can just find the best straight line fit to both data sets, and thus determine the best functional form for your data (together with parameter estimates for A, b, m, etc).

 

[BvU]

If you measured v and varied h, you could try a plot of v² as a function of h ?
Error handling will be a bit more difficult then, though.

 

[jdawg]

I'll give it a shot, thanks for your help everybody!