Physics Lab Propagation of Error Issue

[Browntown]

Homework Statement

Physics Lab Propagation of Error Issue

Relevant Equations

∂f/∂b=∂/∂b (tan^(-1)⁡(a/4b))= 1/(1+(a/4b)^2 )×(-a)/(4b^2 )=1/(1+((5.922 cm)/(4×1.766 cm))^2 )×(-5.922 cm)/(4×(1.766 cm)^2 )

Homework Statement: Physics Lab Propagation of Error Issue
Homework Equations: ∂f/∂b=∂/∂b (tan^(-1)⁡(a/4b))= 1/(1+(a/4b)^2 )×(-a)/(4b^2 )=1/(1+((5.922 cm)/(4×1.766 cm))^2 )×(-5.922 cm)/(4×(1.766 cm)^2 )

Hello,

I'm trying to find the uncertainty in a function from the lab manual for the critical angle of refraction and since this is my first time doing such a thing I'm a bit confused.
When I take the derivative of the inverse tan function provided and input my values, the units of cm don't seem to cancel and I can't figure out how to fix that

 

[tnich]

It seems to me that you want $\Delta f = \Delta \tan^{-1}(\frac{AB}{4t})$ to be unitless. What happens when you multiply $\frac{\partial f}{\partial a}$ by $\Delta a$?

 

[Browntown]

Δa is in units of cm so technically when they're multiplied they cancel out (cm^-1 x cm) but I thought anytime you take the inverse of a trig function you get an angle?

 

[tnich]

That is true. Then you differentiated that angle with respect to a variable with units of cm. What would you expect to be the units of the derivative?

 

[Browntown]

Would the derivative of a variable with units in cm be then cm/s? instantaneous velocity?

 

[tnich]

Consider the definition of derivative:

$$\frac {df}{da} = \lim_{\Delta a \rightarrow 0} {\frac {f(a +\Delta a)- f(a)}{\Delta a}}$$

With $f(a)$ an angle and $a$ a distance, what would be the units of $\frac {df}{da}$?

 

[Browntown]

I'm sorry but I honestly have no clue..

 

[tnich]

I'm sorry but I honestly have no clue..

OK. Let's back up one step. if $f(a)$ is unitless and $a$ has units of cm, what are the units of $$\frac {f(a +\Delta a)- f(a)}{\Delta a}$$

 

[Browntown]

Would they just cancel?

 

[Browntown]

I'm sorry, my lab report is due tomorrow and I think the stress of trying to get it done is just not letting me think properly.

 

[tnich]

Here's the bottom line: for the purpose of determining its units, you can treat a derivative like a fraction. So if $f$ is unitless and $a$ has units of cm, then $\frac {df}{da}$ has units of $\frac {\text {unitless}}{\text {cm}} = \text {cm}^{-1}$

 

[Browntown]

So that means that even though its the inverse tangent function, because it's the derivative of it, it can have units of cm^-1?

 

[tnich]

So that means that even though its the inverse tangent function, because it's the derivative of it, it can have units of cm^-1?

Yes.

 

[Browntown]

Awesome! Thank you so much!