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Homework Help: Differentiation with units- Related Rates problem

  1. Oct 10, 2018 #1

    opus

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    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations


    3. 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.

    Thank you for any help you can provide.
     

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  3. Oct 10, 2018 #2

    BvU

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    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 ?
     
  4. Oct 10, 2018 #3

    kuruman

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    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.
    That's easy to see: ##\frac{cm^3}{s}=cm^2 \times \frac{cm}{s}## were the omitted units.
    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.
     
  5. Oct 10, 2018 #4

    BvU

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    With regards to the
    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​
    instead of
    ##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 )​
     
  6. Oct 10, 2018 #5

    opus

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    Thank you for your reply.

    So I guess I'm still a little hung of on the initial differentiation.
    So I'll start with the function ##V(t) = \frac{4}{3}π[r(t)]^3cm^3##
    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.
     
  7. Oct 10, 2018 #6

    opus

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    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.
    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?
     
  8. Oct 10, 2018 #7

    opus

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    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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  9. Oct 10, 2018 #8

    opus

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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.
     
  10. Oct 10, 2018 #9

    kuruman

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    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}##.
     
  11. Oct 10, 2018 #10

    Mark44

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    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.
     
  12. Oct 10, 2018 #11

    Orodruin

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    I disagree. I would consider this a bad habit. The kosher approach is to realise 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.
     
  13. Oct 11, 2018 #12

    Mark44

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    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.
    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.
     
  14. Oct 11, 2018 #13

    Orodruin

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    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.

    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.

    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.
     
  15. Oct 11, 2018 #14

    verty

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    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.
     
  16. Oct 11, 2018 #15

    opus

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    Man I'm real confused :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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  17. Oct 11, 2018 #16

    BvU

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    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 :wink: )

    Same with line 2: it gives a relationship that is correct in any sytem 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 :-p )

    .
     
  18. Oct 11, 2018 #17

    kuruman

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    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.
     
    Last edited: Oct 11, 2018
  19. Oct 12, 2018 #18

    opus

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    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.
    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?


    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.
     
  20. Oct 12, 2018 #19

    opus

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    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.
     
  21. Oct 12, 2018 #20

    Orodruin

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    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}##.
     
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