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$F(t)$ is a rate function measuring widgets per minute, its average rate of change goes to the next level derivative ... hence widgets per min$^2$

In related rates, the area is (unit$^2$) and volume is (unit$^3$). I did not think of those as derivatives, but just 2D vs 3D calculations.

Let me think about the standard Distance/Velocity/Acc. relationship.

  • $S(t)$ = distance traveled (in miles), $t$ = hours
  • $S(1)$ = 50 miles.
  • $S(3)$ = 150 miles.

So, Avg. Rate of Change over the interval $[1,3]$ is $$ \frac{100 \text{ miles}}{2 \text{ hours}} = 50 \frac{\text{miles}}{\text{hour}}.$$ In this case, I have not squared the hours units b/c I started with an amount function (not a rate function) Units make intuitive sense: "50 miles for every hour."

But, let's say we consider the acceleration function, the derivative of the rate function.

  • $v(1) = 40$
  • $v(3) = 60$

The Avg. Rate of Change of velocity from [1,3] is 20 miles / 2 hours Is this why the Avg. Rate of Change of velocity should be written as 10 miles/hour^2?

In a nutshell,

  • $S = \text{miles}$
  • $S' = V = \text{miles/hour}$
  • $S'' = V' = A = \text{miles/hour}^2$
  • $S''' = V'' = A' = \text{miles/hour}^3$ ?

I do see the connection b/w the hour units and derivative level of S. But, is it better to think about this as fractions?

$$S'' = V' = A = \text{miles/hour}^2$$

Does some dimensional analysis thing ever happen? $$\frac{\text{miles}}{\text{hour}} \times \frac{1}{\text{hour}} = \frac{\text{miles}}{\text{hour}^2}$$

Here is where I arrived: If you have a rate function, for example "miles/hour". Then you take the rate of change of that? "mile per hour, per hour" or $$\frac{\frac{miles}{hour}}{hour} = miles/hour^2$$. This also plays out in the average rate of change calculation: $$\frac{f(b) - f(a)}{(b-a)} = \frac{(miles/hour - miles/hour)}{(hour - hour )} = \frac{(miles/hour)}{hour} = \frac{miles}{hour^2}$$

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  • $\begingroup$ I recommend you reading this. $\endgroup$ – user132181 Apr 3 '14 at 20:48
  • $\begingroup$ WAY out of my league $\endgroup$ – JackOfAll Apr 4 '14 at 0:06
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Well it's really the only way to make sense of it. Velocity (I will use m/s) and acceleration ($m/s^2$). The reason the square goes in acceleration is to literally say it's gaining a velocity x m/s every second. Or, $x*(m/s)/s=x*m/s^2$.

All the next level does is add a "per unit" aspect.

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  • $\begingroup$ Here is where I arrived: If you have a rate function, for example "miles/hour". Then you take the rate of change of that? "mile per hour, per hour" or $$\frac{\frac{miles}{hour}}{hour} = miles/hour^2$$. This also plays out in the average rate of change calculation: $$\frac{f(b) - f(a)}{(b-a)} = \frac{(miles/hour - miles/hour)}{(hour - hour )} = \frac{(miles/hour)}{hour} = \frac{miles}{hour^2}$$ $\endgroup$ – JackOfAll Apr 4 '14 at 0:13
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The derivative of a function of one variable measures instantaneous rate of change, which has the same units as average rate of change: $$ \frac{dy}{dt} = \lim_{\Delta t \to 0} \frac{\Delta y}{\Delta t}. $$ Hence, the units of $\frac{dy}{dt}$ are simply the units of $y$ divided by the units of $t$.

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  • $\begingroup$ Precisely. Remember that the definition of derivative always involves a division. If you’re differentiating twice with respect to time, there’ll be a “per second$^2$” in the units. $\endgroup$ – Lubin Apr 3 '14 at 21:05
  • $\begingroup$ @Sammy_Black Here is where I arrived: If you have a rate function, for example "miles/hour". Then you take the rate of change of that? "mile per hour, per hour" or $$\frac{\frac{miles}{hour}}{hour} = miles/hour^2$$. This also plays out in the average rate of change calculation: $$\frac{f(b) - f(a)}{(b-a)} = \frac{(miles/hour - miles/hour)}{(hour - hour )} = \frac{(miles/hour)}{hour} = \frac{miles}{hour^2}$$ $\endgroup$ – JackOfAll Apr 4 '14 at 0:09
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The notation sort of suggests the units of the parameter should be exponentiated (and divided):

$$ f^{(n)}(x) = \frac{d^n}{dx^n}f(x) $$

$dx^n$ will have the units of the $x$ to the $n^\text{th}$ power. But $d^n$ is unit-less so, the final units would be the units of $f$ divided by the units of $x$ to the $n^\text{th}$ power. A better way would be to understand what the derivatives are:

$$ \frac{df}{dx} \tilde{} \frac{\Delta f}{\Delta x} \\ \frac{d^2f}{dx^2} = \frac{d}{dx}\frac{df}{dx} \tilde{} \frac{\Delta \left(\frac{\Delta f}{\Delta x}\right)}{\Delta x} \tilde{} \frac{\Delta(\Delta f)}{\Delta(\Delta x)\Delta x} $$

You can keep going and see that each time, you'll get more and more $\Delta x$ terms on the bottom, but the top just remains $\Delta(\Delta(\Delta ... (f)))$.

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Here is where I arrived: If you have a rate function, for example "miles/hour". Then you take the rate of change of that? "mile per hour, per hour" or $$\frac{\frac{miles}{hour}}{hour} = miles/hour^2$$. This also plays out in the average rate of change calculation: $$\frac{f(b) - f(a)}{(b-a)} = \frac{(miles/hour - miles/hour)}{(hour - hour )} = \frac{(miles/hour)}{hour} = \frac{miles}{hour^2}$$

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