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This is a contribution question I'm making in hopes that others may benefit. I will provide my answer underneath. Initially I wanted to ask this question, but I solved it myself and I'd like to give back for the question I asked earlier. I will wait a day before selecting an answer in case anyone else wants a crack at it.

I asked a question earlier here about why I was getting a different answer when calculating the tangent slope of a line by using the secant method compared to using the Calculus method. I found out that the Calculus method always produces an answer relative to radian measure.

So the two answers I had were actually the same, except one was slope per radians and one was slope per degrees.

Why is it that when we convert radians to degrees we multiply radians $\ *\frac{180}{\pi}\ $, but when we convert slope per radians to slope per degrees we have to multiply the inverse conversion formula slope per radians$\ *\frac{\pi}{180}\ $?

I should give an example to explain better.

Given r radians, we get degrees by $\ r * \frac{180}{\pi}\ $ Given d degrees, we get radians by $\ d * \frac{\pi}{180}\ $

However, the opposite is true when converting slope per radians and slope per degrees:

Given s slope per radians, we get slope per degrees by $\ s * \frac{\pi}{180}\ $ Given t slope per degrees, we get slope per radians by $\ t * \frac{180}{\pi}\ $

It seems illogical that we convert slope per radians to slope per degrees by using the formula which converts degrees to radians. Since it's already in radian form?

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    $\begingroup$ Same reason an hour is 60 minutes, but one mile per hour is one-sixtieth of a mile per minute. $\endgroup$ – Gerry Myerson Jun 8 '13 at 4:13
  • $\begingroup$ And my question is regarding the peculiarity of using the inverse formula of conversion to convert miles per hour to miles per minute, when compared to just converting regular hours to regular minutes. My apologies if my answer is too concise. $\endgroup$ – Gerry Myerson Jun 8 '13 at 4:23
  • $\begingroup$ @GerryMyerson I must apologize for not understanding your comment initially. Obviously I have a lot to learn and you are much wiser than I. Nonetheless, I put this question not to have it answered, but rather to provide an answer in case someone else isn't as wise and looks for an answer. I did not visualize slope per radians as a unit conversion with radians in the denominator. You are definitely correct in your comment and I retract my silly responses. $\endgroup$ – Klik Jun 8 '13 at 5:10
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    $\begingroup$ The question reminds me about the story of someone who, on an exam for a pilot license was asked a question where the distance between two cities was given, the plane's mileage (in miles per gallon) was given, and he was to compute the fuel needed to travel between those two cities. He gave the correct answer, and the examiner asked how he knew whether to multiply or divide. He answered that he did both and chose the reasonable answer. A similar strategy handles many issues of the present sort. $\endgroup$ – Andreas Blass Jun 8 '13 at 15:50
  • $\begingroup$ That's pretty much how I came up with the answer at first, I investigated as to why the unusual conversion afterwards. $\endgroup$ – Klik Jun 8 '13 at 20:29
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Why is it that when we convert radians to degrees we multiply radians $\times \frac{180}π$ , but when we convert slope per radians to slope per degrees we have to multiply the inverse conversion formula slope per radians $\times \frac{π}{180}$

If we want to know an hour in terms of minutes, we multiply 1 hour $\times \frac{60}{1}$, given the result in minutes. If we want to know how convert 180 minutes, we divide $180$ minutes by $60$, i.e., multiply $180 \times \frac 1{60}$.

You'll find this phenomenon in any conversion: To convert temperature in degrees Celsius to temp in Fahrenheit, we have $F = \frac 95 C + 32$. To convert to from F to C, we need to invert this: $C = \frac 59(F-32)$

$R \text{ radians}\;\times \dfrac{180^\circ}{\pi \;\text{radians}} = \dfrac{R\times 180^\circ}{\pi}$.

"Radians" cancels as the unit, leaving a numeric value expressed in degrees.

$D \text{ degrees}\; \times \dfrac{\pi \;\text{radians}}{180\; \text{degrees}} = \dfrac{D\pi\;}{180}\;\text{ radians}$.

"Degrees" cancels as the unit, leaving the value expressed in radians.


Moved from comments:

Note that slope per radian is a ratio: $\;\dfrac{\text{slope}}{\text{radians}}.\;$ So to obtain "slope per degree", you need to have $π$ radians in the numerator, to cancel the unit "radians" from the denominator, and $180$ degrees in the denominator, to end with slope/degrees.

Slope itself is not a "unit" per se, meaning it isn't a degree, or radian, a mm, or foot. It is unit-free: even if we assign distance units (meters, say) to displacement: e.g. $Δ\,y\text{m}=y_2\,\text{m}−y_1\,\text{m}$ and $Δ\,x\,\text{m}=x_2\text{m}−x_1\,\text{m},$ then we have $$\text{slope}\,= \dfrac{Δy\,\text{ m}}{Δx\,\text{m}}=\frac{Δy}{Δx},$$ you see that the units "m: meter" attached to "change in y" and "change in x" cancel in the ratio defining slope, leaving us with a unit-free scalar which slope really is.

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  • $\begingroup$ Yes, I understand this method of conversion. This is essentially the Stoichiometry method of unit conversion (well at least what I learned). I've elaborated more in my question. My issue was that the formula for converting slope per radians to degrees is actually the formula used to convert degrees to radians. It seemed illogical since it was already in radian form. This is related to my last question actually since the two answers were the same. But the formula to convert the slope per radians to slope per degrees is not the formula you'd think it is, it is the inverse. $\endgroup$ – Klik Jun 8 '13 at 4:46
  • $\begingroup$ Yes, that is exactly it and very concisely written. I just went through a more in depth explanation. It was something I wanted to ask earlier so I decided to write out a good (debatable) explanation for the site since I couldn't find this question on here. On a side note, I didn't initially think of the slope as a unit over radians like you wrote, since I understood slope as delta y over delta x. $\endgroup$ – Klik Jun 8 '13 at 4:59
  • $\begingroup$ Slope itself is not a "unit" per se, meaning it isn't a degree, or radian, or mm, or foot. It is unit-free: even if we assign distance units (meters, say) to displacement: e.g. $\Delta y \;\text{m} = y_2\; \text{m} -\Delta y_1 \;\text{m}$ and $\;\Delta x \;\text{m} = x_2 \;\text{m} - x_1 \;\text{m},\;$ then we have $\;\text{slope} = \dfrac{\Delta y\;\text{m}}{\Delta x \; \text{m}} = \dfrac{\Delta y}{\Delta x},\;$ you see that the units "m: meter" attached to "change in y" and "change in x" cancel in the ratio defining slope, leaving us with a unit-free scalar which slope really is. $\endgroup$ – Namaste Jun 8 '13 at 15:18
  • $\begingroup$ @amWhy: nice write-up +1 $\endgroup$ – Amzoti Jun 9 '13 at 0:17
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First consider what "slope" actually is.

$\ slope = \frac{\Delta y}{\Delta x}\ $

Now when we have trigonometric functions such as y=sin(x) or y=cos(x) the values given by these are neither in radians nor degrees. They only use radians or degrees as input and produce an output which is actually a ratio. Hence SOH CAH TOA. $\ sin(\theta)=\frac{Opp}{Hyp}\ $ etc. These are just ratios of lengths.

That being said, we can agree that $\ \Delta y \ $ is not a factor in the unit measurement of slope (in terms of radians and degrees that is). Now consider $\ \Delta x\ $, it will be a unit of either degrees or radians. Thus we determine if the slope is measured in slope per radians versus slope per degrees by this term specifically. Thus, if $\ \Delta x \ $ is in radians, then the slope is in slope per radians.

As for why we use the inverse conversion formula when compared to converting regular radians and regular degrees. Consider the following series of equations:

First, let's state our formulas for conversion:

For degrees to radians it is $\ degrees * \frac{\pi}{180}\ $ For radians to degrees it is $\ radians * \frac{180}{\pi}\ $

The slope formula is $\ slope = \frac{\Delta y}{\Delta x}\ $

If this were in slope per radians than it is in slope per radians. Therefore, to change to degrees we have to multiply the radians unit by $\ *\frac{180}{\pi}\ $

Given a slope per radians

$\ slope = \frac{\Delta y}{\Delta x}\ $

$\ slope = \frac{\Delta y}{\Delta x * \frac{180}{\pi}}\ $ Multiply the radians unit by $\ \frac{180}{\pi}\ $

$\ slope = \frac{\Delta y}{\Delta x} * \frac{1}{\frac{180}{\pi}}\ $

$\ slope = \frac{\Delta y}{\Delta x} * \frac{\pi}{180}\ $

$\ = slope\ $ (per degrees)

Thus it can be seen why we use the inverse conversion formula to convert slope per radians to slope per degrees. The same is true for slope per degrees to slope per radians, except using the opposite conversion equation.

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