# Differentiation with units- Related Rates problem

• opus
In summary: I guess I still don't understand what is happening when we put that into the context of the original problem.The original problem was to find the rate at which the radius is increasing when the radius is 3 cm. So how do we get from the derivative of the volume to the rate of change of the radius?In summary, the problem involves a spherical balloon being filled with air at a constant rate of 2 cm^3/sec. To find the rate of change of the radius when the radius is 3 cm, we use the derivative of the volume function with respect to time. By differentiating the volume function, we can find the relationship between the derivatives of volume and radius. This allows us to solve for the rate of change
opus
Gold Member

## Homework Statement

A spherical balloon is being filled with air at a constant rate of ##2cm^3/sec##.
How fast is the radius increasing when the radius is 3 cm?

## The Attempt at a Solution

This is a problem given and spelled out in my text. However, it seems to be assuming that the reader knows what to do with the units and that is unfortunately not the case with me.
If you'll have a look at the posted image, I have highlighted where units have been added into the RHS of the equation. I understand the substitution on the LHS, but I have a couple of questions in regard to the RHS.

First, we had an original volume function written as ##V(t) = \frac{4}{3}π[r(t)]^3cm^3## and we differentiated both sides, giving ## \frac{dV}{dt} = 4π[r(t)]^2\frac{dr}{dt} ## Notice that the cm units are missing. Why were they taken out?

Second, after we made the substitution (the line which is highlighted in the image), we brought two separate units back in- one was ##cm^2## and one was ##\frac{cm}{s}##.

Thirdly, after we solved for ##\frac{dr}{dt}##, the units didn't cancel out, even though we had a ##cm^3## and ##sec## in both the numerator and the denominator.

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opus said:
Notice that the cm units are missing
If you differentiate cm3 wrt time, you get cm3/s, so that is the dimension of ##V'(t)##, as it says on the next line: ##V'(t) = 2 ## cm3/s.

On the line after 'therefore' the dimensions left and right are also ok: both cm3/s

In the equation $$2\ \rm { cm}^3/\rm s = 4\pi \bigl (r(t)\bigr )^2\ \rm{cm^2}\ \ r'(t) \ \rm{cm/s}$$ left and right are divided by ##4\pi \bigl (r(t)\bigr )^2\ \rm{cm^2}## to get $$r'(t) \ \rm{cm/s} = {2\ \ \ \ \rm { cm}^3/\rm s \over 4\pi \bigl (r(t)\bigr )^2\ \rm{cm^2} } = {2\over 4\pi \bigl (r(t)\bigr )^2}\ \rm{cm/s}$$ again with the same units on both sides.

Does this help ?

opus
opus said:
First, we had an original volume function written as ##V(t) = \frac{4}{3}π[r(t)]^3cm^3## and we differentiated both sides, giving ## \frac{dV}{dt} = 4π[r(t)]^2\frac{dr}{dt}## Notice that the cm units are missing. Why were they taken out?
If the volume is given in ##cm^3## and you take its derivative w.r.t. time, then the units that go with the derivative would be ##\frac{cm^3}{s}## (volume divided by time). I don't know why they were omitted. Probably an oversight.
opus said:
Second, after we made the substitution (the line which is highlighted in the image), we brought two separate units back in- one was ##cm^2## and one was ##\frac{cm}{s}##.
That's easy to see: ##\frac{cm^3}{s}=cm^2 \times \frac{cm}{s}## were the omitted units.
opus said:
Thirdly, after we solved for ##\frac{dr}{dt}##, the units didn't cancel out, even though we had a ##cm^3## and ##sec## in both the numerator and the denominator.
I am not sure I understand what you mean by this. Please post the full problem with the question that is being asked. Then the solution will make more sense.

opus
With regards to the
opus said:
Why were they taken out?
In physics it is custom to assign a symbol to a value plus its units

You can see that happen in your example when the author writes
##V'(t) = ## 2 cm3/s​
##V'(t)## cm3/s = 2 cm3/s

I agree with you that this is somewhat inconsistent when the third equation is compared to the equation just before ( V(t) = ... cm3 )​

opus
BvU said:
If you differentiate cm3 wrt time, you get cm3/s, so that is the dimension of ##V'(t)##, as it says on the next line: ##V'(t) = 2 ## cm3/s.

On the line after 'therefore' the dimensions left and right are also ok: both cm3/s

In the equation $$2\ \rm { cm}^3/\rm s = 4\pi \bigl (r(t)\bigr )^2\ \rm{cm^2}\ \ r'(t) \ \rm{cm/s}$$ left and right are divided by ##4\pi \bigl (r(t)\bigr )^2\ \rm{cm^2}## to get $$r'(t) \ \rm{cm/s} = {2\ \ \ \ \rm { cm}^3/\rm s \over 4\pi \bigl (r(t)\bigr )^2\ \rm{cm^2} } = {2\over 4\pi \bigl (r(t)\bigr )^2}\ \rm{cm/s}$$ again with the same units on both sides.

Does this help ?

So I guess I'm still a little hung of on the initial differentiation.
In differentiating both sides with respect to the independent variable, ##t##,
the LHS would simply be ##\frac{dV}{dt}## which is of course what the text shows.
Now for the RHS, if I were to just forget about the ##cm^3## for a second, it would differentiate into ##4π[r(t)]^2\frac{dr}{dt}##, which is what it shows in the text.
So for the units, are we actually differentiating the units themselves, or are they just "there" and we kind of forget about them while we differentiate the rest?

So now that we have a relationship between the derivatives of both of the different values that are changing (volume and radius), we can look back at the given problem and see that pumping air into the balloon corresponds to a volume change ( in other words, ##\frac{dV}{dt}##. So we can substitute ##\frac{dV}{dt}## for ##\frac{2cm^3}{sec}## and now we can solve for the rate of change for the radius.
However, along with the volume substitution, we added in the ##cm^2## and ##\frac{cm}{sec}## pieces into the RHS.
So, if we "forgot" about the units when we differentiated, is this us bringing them back in? And if so, why are they split up like that? The second term on the RHS makes sense, because the radius is changing in centimeters per second. But the ##cm^2## inside the first time doesn't make sense to me.

I hope this question made sense.

kuruman said:
If the volume is given in cm3cm^3 and you take its derivative w.r.t. time, then the units that go with the derivative would be cm3s\frac{cm^3}{s} (volume divided by time). I don't know why they were omitted. Probably an oversight.
Ok this part I understand. The omission is what is confusing in that particular line. An oversight would make sense. There's a handful every section.
kuruman said:
That's easy to see: cm3s=cm2×cms\frac{cm^3}{s}=cm^2 \times \frac{cm}{s} were the omitted units
Yes but why are they separated as such? I can see why the cm/s is in the second term, because the radius is changing in centimeters per second. But I don't understand why the ##cm^2## is in the first term. Is it because that's just what was left over and it has to go somewhere?

kuruman said:
I am not sure I understand what you mean by this. Please post the full problem with the question that is being asked. Then the solution will make more sense.
The full question in posted verbatim in the OP, but here is what the text spells out. Please see attached images.

What I mean is that, if I've solved for ##\frac{dr}{dt}## correctly, then the units should cancel. I will write my work out in the following post.

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Given: ##\frac{2cm^3}{sec} = (4π[r(t)]^2cm^2)⋅(\frac{dr}{dt}\frac{cm}{sec})##

In solving for ##\frac{dr}{dt}##:
##\frac{\frac{2cm^3}{sec}}{4π[r(t)]^2cm^2} = \frac{dr}{dt} ⋅\frac{cm}{sec}##

##\frac{2cm^3}{sec}⋅\frac{1}{4π[r(t)]^2cm^2}⋅\frac{sec}{cm} = \frac{dr}{dt}##

Now if you'll have a look at the final equation, the centimeters and seconds should cancel because there's a ##cm^3## and ##sec## in both the numerator and denominator on the LHS.

opus said:
Given: ##\frac{2cm^3}{sec} = (4π[r(t)]^2cm^2)⋅(\frac{dr}{dt}\frac{cm}{sec})##

In solving for ##\frac{dr}{dt}##:
##\frac{\frac{2cm^3}{sec}}{4π[r(t)]^2cm^2} = \frac{dr}{dt} ⋅\frac{cm}{sec}##

##\frac{2cm^3}{sec}⋅\frac{1}{4π[r(t)]^2cm^2}⋅\frac{sec}{cm} = \frac{dr}{dt}##

Now if you'll have a look at the final equation, the centimeters and seconds should cancel because there's a ##cm^3## and ##sec## in both the numerator and denominator on the LHS.
I am not sure what you expect to cancel or not cancel. Starting with ##\frac{\frac{2cm^3}{sec}}{4π[r(t)]^2cm^2} = \frac{dr}{dt} ⋅\frac{cm}{sec}##,
and you simplify units on the left, you get ##\frac{2}{4 \pi r(t)^2}\frac{cm}{sec} = \frac{dr}{dt} ⋅\frac{cm}{sec}##
This equation is dimensionally correct and consistent. That's all you can say. It is understood that when you put a number for ##r(t)##, it should be a quantity expressed in ##cm##. Personally, I would not substitute the ##2 cm^3/sec## in the equation. I would skip the units altogether, write the equation as
$$\frac{\frac{dV}{dt}}{4π[r(t)]^2} = \frac{dr}{dt}$$and put in the numbers at the very end in whatever units are appropriate. Note that this equation is dimensionally correct; both sides have dimensions ##LT^{-1}##.

opus
opus said:
Given: ##\frac{2cm^3}{sec} = (4π[r(t)]^2cm^2)⋅(\frac{dr}{dt}\frac{cm}{sec})##
The units on the left (##\frac {m^3}{sec}##) match those on the right, which are ##cm^2 \cdot \frac {cm}{sec}##, with the units of ##\frac{dr}{dt}## being ##\frac {cm}{sec}##. What's the point of dragging along the units? They only complicate things needlessly.

The near-universal approach in calculus is to do the calculations without the units, and add them into the final answer.
opus said:
In solving for ##\frac{dr}{dt}##:
##\frac{\frac{2cm^3}{sec}}{4π[r(t)]^2cm^2} = \frac{dr}{dt} ⋅\frac{cm}{sec}##

##\frac{2cm^3}{sec}⋅\frac{1}{4π[r(t)]^2cm^2}⋅\frac{sec}{cm} = \frac{dr}{dt}##

Now if you'll have a look at the final equation, the centimeters and seconds should cancel because there's a ##cm^3## and ##sec## in both the numerator and denominator on the LHS.

opus
Mark44 said:
The near-universal approach in calculus is to do the calculations without the units, and add them into the final answer.
I disagree. I would consider this a bad habit. The kosher approach is to realize that the quantities in a formula are dimensionful and carries units. You should never write out units in a formula (unless your formula contains dimensionful constants). That is just confusing and restricting you to use a particular type of unit.

For example, the above practice is particularly useful when your input data have different base units:
$$d = vt$$
is a valid formula that matches dimensionally. If I have v = 5 m/s and t = 3 min, this would give
d = (5 m/s)(3 min) = (15 m)(1 min/1 s) = (15 m)60 = 900 m.

opus and BvU
Orodruin said:
I disagree. I would consider this a bad habit.
I think we're coming at this from different perspectives -- you with more from the physics side, and me more from the mathematics side. In most calculus books that I've seen, there is an assumption in problems of a physical nature that the units are all compatible, so that we aren't mixing, say, ft/sec in one part and mi/hr in another. I don't recall ever seeing any calculus textbook that carried units all the way through a problem. The usual practice that I've seen is to add them in for the reported answer.

I don't recall ever seeing any calculus textbook writing something like ##\frac{2cm^3}{sec} = (4π[r(t)]^2cm^2)⋅(\frac{dr}{dt}\frac{cm}{sec})##, as Opus wrote in an earlier post. In prefatory text, they would have explained that volume of a sphere is ##V(r) = \frac 4 3 \pi r^3##, and that if the radius was in cm., then V(r) would be in ##cm^3##. Differentiating w.r.t. time t would produce the expected units (##\frac{cm^3} {sec}##) on both sides.
Orodruin said:
The kosher approach is to realize that the quantities in a formula are dimensionful and carries units.
Yes, I understand that. My point is that in long calculations, dragging the units along really clutters things up, thereby increasing the chances for making an error.

opus
Mark44 said:
I think we're coming at this from different perspectives -- you with more from the physics side, and me more from the mathematics side. In most calculus books that I've seen, there is an assumption in problems of a physical nature that the units are all compatible, so that we aren't mixing, say, ft/sec in one part and mi/hr in another. I don't recall ever seeing any calculus textbook that carried units all the way through a problem. The usual practice that I've seen is to add them in for the reported answer.
I agree that pure math texts typically are very sloppy on this, mainly because they are concerned with the mathematics and not the physics. That does not make it right. If this approach is taken it needs to be explicitly stated and make no pretention to be generally valid physical examples. Even if this approach is fine for the mathematics part, it later causes a lot of problems for students later on. As evidenced by the very existence of this thread.

My conclusion is: Do not try to teach or learn units and dimensional analysis in a math class.

Mark44 said:
I don't recall ever seeing any calculus textbook writing something like ##\frac{2cm^3}{sec} = (4π[r(t)]^2cm^2)⋅(\frac{dr}{dt}\frac{cm}{sec})##, as Opus wrote in an earlier post. In prefatory text, they would have explained that volume of a sphere is ##V(r) = \frac 4 3 \pi r^3##, and that if the radius was in cm., then V(r) would be in ##cm^3##. Differentiating w.r.t. time t would produce the expected units (##\frac{cm^3} {sec}##) on both sides.

And it should not say anything like that. If you just state that the volume of a sphere is ##V(r) = \frac 4 3 \pi r^3##, then you need no further explanations. The units will follow through from the ##r^3##. Differentiating with respect to time will produce a result which has physical dimension ##\mathsf{L^3/T}##, which is going to be generally valid. The actual units of the result depend on the units of the input.

Yes, I understand that. My point is that in long calculations, dragging the units along really clutters things up, thereby increasing the chances for making an error.
Of course, I do this too. But one has to remember that if one wants to do this, all the input needs to put in the same base units first.

Every time I see a math (or physics) text say something like "Consider a mass of ##m## kg ..." I die a little bit inside.

opus and BvU
opus said:
So for the units, are we actually differentiating the units themselves, or are they just "there" and we kind of forget about them while we differentiate the rest?

If I write formula ##V'(t) = 4\pi r^2 {dr \over dt}##, it's as though the ##4\pi r^2## is a modifier to ##dr \over dt## and we can write ##{dV \over dr} = 4 \pi r^2##. Then ##{dr \over dt} = {dV/dt \over dV/dr}##. Here ##{1 \over 4\pi r^2}## is a modifier to ##dV \over dt##. This should justify that the units aren't being differentiated, they just come out like a normal multiplication or division.

opus
Man I'm real confused
Maybe this pic will help to see what I’m asking. It’s just notes on the actual problem from the text.
Part of my question is on page 2 in orange.

An additional part of my question is, if we do leave the units inside the term from the beginning, do they get differentiated?
This question partly stems from my use of dimensional analysis in chemistry where any manipulation to a constant attached to a unit where both are inside a term, the units get manipulated too. An example to this would be using dimensional analysis to convert something to volume.

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opus said:
Man I'm real confused
I can imagine. Part of the discsourse goes over your head and the didactic quality of the material that is presented to you is less than perfect.

As an example: (i) on the first picture: Volume of a sphere of radius r: $$V = {4\over 3 }\pi r^3 \ {\rm cm^3}$$ I am convinced many of our helpers do not like this at all:
in one interpretation there is only a symbol on the left, so it has a numerical value and a dimension:$$\ V = \Bigl (\ {4\over 3 }\pi r^3 \Bigr ) \ \ {\rm cm^3}$$ another way to read it is $$\Bigl (\ V = {4\over 3 }\pi r^3 \Bigr ) \ \ {\rm cm^3}$$saying the equation is expressed in terms of cm3. But that is a completely unnecessary restriction: the volume has the dimension [length]3 and the unit of length has little to do with that!
Of course you do have to use the same units on both sides, but that isn't made clear at all in this way of presenting.

Bottom line is that everyone (well...) would higly prefer the equation in the form
$$\ V = \ {4\over 3 }\pi r^3$$which is correct in every system of units, even weird ones like some cultures use, or even weirder ones like some branches of science prefer.

In answer to your question in orange: line 1 (##\ {dV\over dt} = 2\ {\rm cm^3/s}\ ##) is acceptable (in the sense: perfectly ok): it gives a numerical value plus a dimension plus a choice of units. Something you can substitute in an expression later on (preferably at the very last - and after a thorough check of the dimensions in this last expression )

Same with line 2: it gives a relationship that is correct in any system of units. No need for any addtions. You write 'no cm', but you could just as well have written 'no inches' or 'no parsecs' .

In line 3 the author tries to be helpful, but in reality throws concientious students like you into utter confusion: why the additions in grey all of a sudden ? The way it is put in the picture the left hand side is replaced by a numerical value plus units in a specific choice of a system of units which again is perfectly ok.

But the right hand side? The author wants to make sure you see that the dimensions and units on left match those of the expression on the right. In fact he (she) has changed notation and taken the units and dimensions out of the symbols: all of a sudden the symbol ##r## is no longer a length but a number.

The path to some unpleasant place is paved with good intentions - in this case the attempt to be helpful

Is this clear so far ? Ask away, please: this part of science is very important to grasp thoroughly -- you benefit enormously seeing through and make less errors so higher scores if you get it right.

(and for helpers it's a lot easier to answer questions and correct mistakes than it is to come up with 100% perfect replicas of lecture material )

.

opus
When I write problems, I make sure to use symbols to label any given numerical quantity that has units. This encourages the algebraic manipulation of equations before insertion of the given numerical values. This is how I would present the problem and its solution

Example
A spherical balloon is filled with air at constant rate ##\beta=2~cm/s^3##. How fast is the radius increasing when the radius is ##R=3~cm##?

Solution
The volume of the balloon at any time ##t## is,
##V(t)=\frac{4}{3}\pi r(t)^3##
Taking the time derivative gives its rate of change,
##\beta=\frac{dV}{dt}=4\pi r(t)^2\frac{dr}{dt}~\rightarrow~\frac{dr}{dt}=\frac{\beta}{4\pi r(t)^2}##
Substituting the given values,$$\frac{dr}{dt}=\frac{\beta}{4\pi R^2}=\frac{2~cm^3/s}{4\pi (3~cm)^2}=1.7\times 10^{-2}~cm/s.$$
I think that this is a cleaner presentation, but let's hear what OP thinks.

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opus and BvU
Thank you for taking the time to write all that. Very helpful! This seems like it should be a real easy concept but these dimensions have thrown a wrench into everything and I want to make sure I get it.
BvU said:
I can imagine. Part of the discsourse goes over your head and the didactic quality of the material that is presented to you is less than perfect.

As an example: (i) on the first picture: Volume of a sphere of radius r:
V=43πr3 cm3​
V = {4\over 3 }\pi r^3 \ {\rm cm^3} I am convinced many of our helpers do not like this at all:
in one interpretation there is only a symbol on the left, so it has a numerical value and a dimension:
V=( 43πr3) cm3​
\ V = \Bigl (\ {4\over 3 }\pi r^3 \Bigr ) \ \ {\rm cm^3} another way to read it is
( V=43πr3) cm3​
\Bigl (\ V = {4\over 3 }\pi r^3 \Bigr ) \ \ {\rm cm^3} saying the equation is expressed in terms of cm3. But that is a completely unnecessary restriction: the volume has the dimension [length]3 and the unit of length has little to do with that!
Of course you do have to use the same units on both sides, but that isn't made clear at all in this way of presenting.

Bottom line is that everyone (well...) would higly prefer the equation in the form
V= 43πr3​
\ V = \ {4\over 3 }\pi r^3 which is correct in every system of units, even weird ones like some cultures use, or even weirder ones like some branches of science prefer.
So since the definition of the volume states that we are in units of length cubed, it is redundant to tack on the ##cm^3## at the end? So it is safe to not write the ##cm^3## because I know that in the equation there is some unit of length cubed and it doesn't really matter what unit it is?
BvU said:
In line 3 the author tries to be helpful, but in reality throws concientious students like you into utter confusion: why the additions in grey all of a sudden ? The way it is put in the picture the left hand side is replaced by a numerical value plus units in a specific choice of a system of units which again is perfectly ok.
So I'm okay with the LHS substitution and the way I have viewed it is that ##\frac{dV}{dt}## is change in volume, and change in volume was given in the problem, so it's pretty natural to just be able to swap those two as they're saying the same thing.
But on the RHS, the ##cm^2## and ##cm/sec## came out of nowhere and are actually dimensions used in the final solution. What I mean is that we made the LHS substitution, and just threw in a ##cm^2## and ##cm/sec## into the RHS and solved for ##\frac{dr}{dt}##, and these units that were plugged into the RHS actually were used in manipulations to get ##\frac{dr}{dt}## so they don't seem as though they're notes for the reader to see where units are going.
Maybe I'm way over complicating things. I'm used to either not caring about the any of the units until the final solution like I did in Algebra where it was really just number manipulation, or, always have units and dimensions throughout the entire calculation like in Chemistry because it has a lot of converting back and forth between things. This case seems like a bastardization of both where sometimes they're in, sometimes they're out.

kuruman said:
When I write problems, I make sure to use symbols to label any given numerical quantity that has units. This encourages the algebraic manipulation of equations before insertion of the given numerical values. This is how I would present the problem and its solution

Example
A spherical balloon is filled with air at constant rate ##\beta=2~cm/s^3##. How fast is the radius increasing when the radius is ##R=3~cm##?

Solution
The volume of the balloon at any time ##t## is,
##V(t)=\frac{4}{3}\pi r(t)^3##
Taking the time derivative gives its rate of change,
##\beta=\frac{dV}{dt}=4\pi r(t)^2\frac{dr}{dt}~\rightarrow~\frac{dr}{dt}=\frac{\beta}{4\pi r(t)^2}##
Substituting the given values,$$\frac{dr}{dt}=\frac{\beta}{4\pi R^2}=\frac{2~cm^3/s}{4\pi (3~cm)^2}=1.7\times 10^{-2}~cm/s.$$
I think that this is a cleaner presentation, but let's hear what OP thinks.

Ok interesting. Now the text does mention to "assign symbols to all variables involved in the problem". I wasn't sure how to take this as either a) give things variables like normal, or b) something new that I haven't done before which looks like the approach that you used. I asked a tutor and they said the symbols are just variables but I took that as being the normal case where a variable is usually just a single value. But what it seems like you've done is assign a symbol to an actual expression(?) that is the whole change in volume itself that was given in the problem. Let me try this approach out and see where I come into questions.

opus said:
So it is safe to not write the cm3cm3cm^3 because I know that in the equation there is some unit of length cubed and it doesn't really matter what unit it is?
Yes. As long as you start by saying "where the radius of the sphere is ##r##" rather than "where the radius of the sphere is ##r## cm". The second of those should be avoided in my opinion. The first makes ##r## a dimensionful number and you can check the dimensions of your formula according to (square parenthesis signifying "the physical dimension of the contained expression")
$$[V] = \left[\frac{4\pi r^3}{3}\right] = \left[\frac{4\pi}{3}\right] [r^3] = 1 [r]^3 = \mathsf{L}^3,$$
just using that the physical dimension of a product is the product of the physical dimensions of the factors, i.e., ##[AB] = [A][ B]##.
Whatever unit you use for ##r##, ##r## will be expressed as ##r = r_u u##, where ##u## is the unit length and ##r_u## a dimensionless number. The volume will then be
$$V = \frac{4\pi r^3}{3} = \frac{4\pi r_u^3}{3} u^3,$$
where ##u^3## is a unit of volume (1 cm3 if your length unit was 1 cm).

When it comes to the units of a derivative, the derivative is a limit of a quotient. As long as you are not changing units (highly recommended to avoid), the units of the derivative will be those of the differentiated quantity divided by those of the parameter you are differentiating with respect to. In particular, for the physical dimensions of some function ##f(\lambda)##, where ##f## and ##\lambda## are dimensionful quantities, you would have
$$\left[\frac{df}{d\lambda}\right] = \frac{[f]}{[\lambda]}.$$
When it comes to the units, if ##f = f_0 u_f## (where, again ##f_0## is a dimensionless number and ##u_f## is the units you are using for ##f##) and ##\lambda = \lambda_0 u_\lambda##, you would have
$$\frac{df}{d\lambda} = \frac{d(f_0 u_f)}{d(\lambda_0 u_\lambda)} = \frac{df_0}{d\lambda_0} \frac{u_f}{u_\lambda}$$
(again, assuming constant units) so if you are using units ##u_f## for ##f## and units ##u_\lambda## for ##\lambda##, then the units of ##df/d\lambda## would be ##u_f/u_\lambda##. In your case, you would have ##u_r = u_V = 1~{\rm cm}^3## and ##u_t = 1~{\rm s}## and therefore the units of ##dV/dt## would be ##u_V/u_t = 1~{\rm cm}^3/{\rm s}## and those of ##dr/dt## would be ##u_r/u_t = 1~{\rm cm}/{\rm s}##.

opus, SammyS and BvU
I think I don't mind what the author wrote because if you see the units as a kind of tag hanging off the variable or formula, it makes sense. Then one could write something like ##V(t)\ cm^3 = {4 \over 3} \pi r^3## where one has tagged the left side but not the right. Or one could write ##V(t) = {4 \over 3} \pi r^3\ cm^3## or even ##V(t) = {4 \over 3} \pi (r\ cm)^3##, where one has tagged the r variable, or any combination of the above. Or like the author wrote, ##{dV \over dt} = 4 \pi r^2\ cm^2 \cdot {dr \over dt} cm/s##, he or she has tagged two subformulas. If they are just tags and one isn't removing units or anything like that, it isn't a tragedy if they are missing, one can still work out what the units should be, but where they help they can be added.

The only problem I see with these tags is one might want to tag the t variable like so: "V(t s)" or "V(t sec)" which is perhaps going too far, I don't know. But it makes things very clear indeed: ##V(t\ sec) = {4 \over 3} \pi r^3\ cm^3##. I suppose one has to write it somewhere that t is in seconds, so why not there? The tags don't change the formula, the formula is still ##V(t) = {4 \over 3} \pi r^3## and is manipulated like usual. The tags are immutable, they don't get manipulated, they just get replaced with new ones.

Well that's my theory as to how one can place units liberally in formulas in a sensible way. But like I said, if "V(t sec)" is going too far then disallow that use.

opus
verty said:
I think I don't mind what the author wrote because if you see the units as a kind of tag hanging off the variable or formula, it makes sense.
On the contrary, I think this is an appalling practice that makes absolutely no sense when you do dimensional analysis. Putting the units on one variable but not on the others is extremely confusing.
verty said:
If they are just tags and one isn't removing units or anything like that, it isn't a tragedy if they are missing, one can still work out what the units should be, but where they help they can be added.
I think it is just confusing, as shown by the OP's confusion.

opus
Ok thanks so much guys. I'm going to work on some problems to get the hang of it and report back if I run into any issues. Had to put it on the backburner for a moment because I was stuck on a program that I was writing for my computer science class (had to write a program to find perfect numbers in the lowest iterations possible). But now that's done I need to hammer out some of these problems!

A rocket is launched so that it rises vertically. A camera is positioned 5000 ft from the launch pad. When the rocket is 1000 ft above the launch pad, its velocity is 600 ft/sec. Find the necessary rate of change of the camera's angle as a function of time so that it stays focused on the rocket.

Now after doing the leg work, I get to the point where I almost have the rate of change of the angle with respect to ##t##, which is given as
##600 = 5000(\frac{26}{25})(\frac{dθ}{dt})##

Now I plug in the units to get:

Now thinking about the problem, I know that I'm going to get some constant in radians per second. But when I spell everything out, and do a dimensional analysis type of cancellation, it doesn't turn out correct:
$$\frac{600 ft}{sec} = 5000~ft(\frac{26}{25}rad)\frac{dθ}{dt}$$
$$\frac{dθ}{dt} = (\frac{600 ft}{sec})(\frac{1}{5000 ft})(\frac{25}{26 rad})$$

Now in the last equation, the feet cancel, but the radians and seconds are both on the bottom, so I can't really say it's radians/sec.
Thoughts?

Sorry if this seems complex but it seems we can say the following: $${d\theta \over dt} = {|\vec{r} \times {d\vec{y} \over dt}| \over |\vec{r}|^2}$$ where ##\vec{r}## is the vector from the camera to the rocket. This nicely shows the units; ##d\theta \over dt## has units 1/sec. So I think the mistake was calling ##26 \over 25## radians.

-Edit-

So basically, the top is a product with ft and ft/sec and the bottom is ft^2.

Last edited:
opus
Thanks for the reply. The 26/25 rad I got from my book and it seems like it makes sense. I've been thinking about your post and I think I get the idea in the most basic form but I'm not sure about it entirely. For example I can see why you might call the camera line of sight to the rocket a vector but everything else, including the absolute value notation has me lost a bit.
In the way that I have it written out, it units don't cancel properly and I can't seem to figure out why. I know that the answer should be in the form "the camera goes through x radians per second to stay focused on the rocket", but it's just not panning out that way.

Why don’t you show us how you arrived at your expression symbolically? (Ie, denote all the quantities by a letter) It is very hard to understand and track what you have done otherwise.

I haven't quite got the hang of expressing everything symbolically, partly because I feel like I'm going to mess something up, but I will write out all of my work here.

(i) I want the rate of change of θ with respect to time when the rocket is 1000 feet off the ground. In other words, I need ##\frac{dθ}{dt}## when h = 1000 ft.
At this time (when h = 1000 ft), the velocity of the rocket is ##\frac{dh}{dt}= 600 ft/sec##

(ii) So my equations are: ##\frac{dh}{dt}= 600 ft/sec## and ##h = 5000tan(θ)##

(iii) I now want to differentiate the equation that relates my variables, which is the equation ##h = 5000tan(θ)##
This gives me ##\frac{dh}{dt} = 5000sec^2(θ)\frac{dθ}{dt}##

(iv) Now I will substitute all known values into the equation (I didn't before because I would have differentiated a constant which would equal 0). However, first I need to know the value of ##sec^2(θ)## when ##h = 1000##
By the Pythagorean Theorem, when ##h = 1000## and my ground length is 5000, ##sec^2(θ) = \frac{26}{25}## and this is in radians.

Now that I have the value of ##sec^2(θ)## when ##h = 1000##, I can continue on to plug in my values to find ##\frac{dθ}{dt}##:

##\frac{dh}{dt} = (5000)(sec^2(θ))\frac{dθ}{dt}##

Solving for my unknown, ##\frac{dθ}{dt}##:

The reason I wrote the last equation as such was so that I could perform cancellations of units to see what I'm left with. For example, when a ft is on top, and a ft is on bottom, they would cancel, etc. But as you can see here, the feet cancel and I'm left with radians and seconds on the bottom.

opus said:
partly because I feel like I'm going to mess something up, but I will write out all of my work here.
I would say you are far more likely to mess things up without any way of understanding where you went wrong if you do not do things symbolically first.

opus said:
##\sec^2(θ) = \frac{26}{25}## and this is in radians.
This is not correct. Trigonometric functions are dimensionless. In particular, the secant is 1/cos, which is the length of the hypothenuse divided by the adjacent edge so this is a ratio between two lengths - which has what dimension?

Edit: Also, if you want to have your units properly, ##h## is not equal to ##5000 \tan(\theta)##, it is given by ##h = (5000~{\rm ft}) \tan(\theta)##.

opus said:
Thanks for the reply. The 26/25 rad I got from my book and it seems like it makes sense. I've been thinking about your post and I think I get the idea in the most basic form but I'm not sure about it entirely. For example I can see why you might call the camera line of sight to the rocket a vector but everything else, including the absolute value notation has me lost a bit.
In the way that I have it written out, it units don't cancel properly and I can't seem to figure out why. I know that the answer should be in the form "the camera goes through x radians per second to stay focused on the rocket", but it's just not panning out that way.

Sorry, I was just trying to show that there is a formula where the only units involved are ft and ft/sec. I realize you won't know what a cross product is. It's a complex formula but (and you'll have to trust me on this) it simplifies to the following result which I believe shows that radians have no dimension:
$${d\theta \over dt} = {|\vec{r} \times {d\vec{y} \over dt}| \over |\vec{r}|^2} = {(5000 ft) \times (600 ft/sec) \over (5000 ft)^2 + (1000 ft)^2}$$

verty said:
it simplifies to the following result which I believe shows that radians have no dimension
The way of understanding why radians have no dimension is to look at their definition in terms of arc-length (length) divided by radius (length).

## What is differentiation with units in related rates problems?

Differentiation with units in related rates problems is a mathematical technique used to find the rate of change of one variable with respect to another variable, both of which are measured in different units. It involves using the chain rule to find the derivative of the given function and then converting the units to find the final answer.

## Why is it important to use units in related rates problems?

Units are important in related rates problems because they provide context and meaning to the problem. Without units, the answer may be mathematically correct but it may not make sense in the real world. Units also help in converting the final answer to the desired unit of measurement.

## What is the chain rule and how is it used in differentiation with units in related rates problems?

The chain rule is a calculus rule used to find the derivative of a composite function. In differentiation with units in related rates problems, the chain rule is used to find the derivative of the given function with respect to time. This allows us to find the rate of change of one variable with respect to another variable.

## What are some common mistakes to avoid when solving differentiation with units in related rates problems?

Some common mistakes to avoid when solving differentiation with units in related rates problems include forgetting to convert units, using the wrong units, and not using the chain rule correctly. It is important to carefully read the problem and identify the given variables and their units before solving the problem.

## Can differentiation with units in related rates problems be applied to real-world situations?

Yes, differentiation with units in related rates problems can be applied to real-world situations. For example, it can be used to find the rate of change of the volume of a balloon as it is being inflated, or the rate of change of the distance between two cars as they are moving towards each other. It is a useful tool in many fields such as physics, engineering, and economics.

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