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Writing an Expression for dρ in Terms of ρ and dT.

  1. Sep 21, 2016 #1

    Drakkith

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    Staff: Mentor

    1. The problem statement, all variables and given/known data
    The coefficient, β, of thermal expansion of a liquid relates the change in the volume V (in m3) of a fixed quantity of a liquid to an increase in its temperature T (in °C):
    dV = βV dT
    (a) Let ρ be the density (in kg/m3) of water as a function of temperature. (For a mass m of liquid, we have ρ = m/V.)
    Write an expression for dρ in terms of ρ and dT.


    2. Relevant equations
    dV = βV dT
    ρ = m/V

    3. The attempt at a solution
    I know the answer is dρ = -βρ dT thanks to the back of the book, but I can't figure out how to get there. I thought the answer would have been something like dρ = m/dV, where dV = βV dT, making it dρ = ρ/β dT, but apparently that's all wrong.

    What's really bugging me is that this problem is in the section on "Local Linearity and the Differential" and for the life of me I can't figure out how it even relates to the section except that I have a changing quantity.
     
  2. jcsd
  3. Sep 21, 2016 #2

    Mark44

    Staff: Mentor

    Starting from ρ = m/V, take differentials of both sides. Keep in mind that d(m/V) = ##\frac d {dV}(\frac m V) \cdot dV##, which will require the use of the quotient rule (or writing 1/V as V-1 and using the power rule). From there it's pretty straightforward getting to the book's answer.
     
  4. Sep 21, 2016 #3

    fresh_42

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    I have ##\frac{dV}{dT}= \beta V ## and therefore ##V=\exp(\beta T)= \frac{m}{\rho}##. This gives me ##\rho## and its derivative w.r.t. ##T## the result.
     
  5. Sep 21, 2016 #4

    Drakkith

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    I'll try that, but I don't really understand what differentials have to do with this question. Is ρ(T) even a multi-variable function?

    @fresh_42, I feel stupid for asking, but how did you get from ## \frac{dV}{dT}= \beta V ## to ## V=\exp(\beta T)= \frac{m}{\rho} ##?
     
  6. Sep 21, 2016 #5

    Mark44

    Staff: Mentor

    From your post:
    Have you studied the total derivative yet?
    @fresh_42 was assuming you know how to solve differential equations, which I believe is not a reasonable assumption.
     
  7. Sep 21, 2016 #6

    fresh_42

    Staff: Mentor

    As the solution of the differential equation ##y(t)' = c \cdot y(t)## which is ##y=e^{ct}##. Ok, I left out the constant. ##V=\frac{m}{\rho}## was given.
     
  8. Sep 21, 2016 #7

    fresh_42

    Staff: Mentor

    Yes. And, sorry, if not.

    Edit: Or for short: the exponential function is the one that doesn't change when differentiated.

    Edit: Edit: And to be honest. I took the solution and and played with it and then wrote my answer backwards. :sorry:
     
    Last edited: Sep 21, 2016
  9. Sep 21, 2016 #8

    Mark44

    Staff: Mentor

    No. It's given that ρ is a function of T, and they're asking for dρ (the differential of ρ) in terms of ρ and T.
     
  10. Sep 21, 2016 #9

    Drakkith

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    I don't think so. That term doesn't sound familiar.

    I touched on them at the end of Calc 2 this past spring, but that's it. I haven't taken Diff EQ yet.

    Oh boy, I'm confused. Everything taught in this section is in the form of F(x,y) where the differential is: dF = Fx dx + Fy dy.
     
  11. Sep 21, 2016 #10

    fresh_42

    Staff: Mentor

    I've read so many valuable and profound posts from you, that I automatically assumed you were just blinkered (?, don't know if this is the correct term, had to look it up).
    But my edit above is an idea you may keep in mind: If you know the solution, try a way backwards. How do you think I got rid of the ##e^{-}/3 ## charges I measured in the Millikan experiment at school ... At least I knew it couldn't had been quarks.

    Edit: My personal Kobayashi-Maru.
     
    Last edited: Sep 21, 2016
  12. Sep 21, 2016 #11

    Mark44

    Staff: Mentor

    This is easier than that. You have ρ = m/V. You can assume that the mass stays constant, but that the volume V changes with temperature T.
    After taking the differentials of both sides of ρ = m/V, the rest is just an algebra exercise.

    Again, ##dρ = \frac d {dV} (\frac m V ) \cdot dV##, so you need to take the derivative of m/V, with respect to V.
     
  13. Sep 21, 2016 #12

    Drakkith

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    Blinkered? Never heard that before.

    Okay, so that's -m/v2. If ρ= m/v, then that's dρ = -ρ/v.

    And that's all I can even think about right now. I'll try to tackle this again tomorrow. I'm too tired and stressed out right now to think straight. Thanks guys.
     
  14. Sep 21, 2016 #13

    Mark44

    Staff: Mentor

    You're skipping steps.
    The -m/V2 is this part: ## \frac d {dV} (\frac m V )##
    So where you are so far is ##dρ = \frac{-m}{V^2} \cdot dV##
    Now substitute in for dV, which you're given.
     
  15. Sep 22, 2016 #14

    Drakkith

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    Where did dV come from? Is it the same idea as the dx and dy in dF = Fx dx + Fy dy?
     
  16. Sep 22, 2016 #15

    Mark44

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    Yes, but you're coming at this from the more complex (a function F(x, y) of two variables) whereas here ρ is a function of only one variable, V.

    If y = f(x), then dy = f'(x) * dx

    If z = g(x, y), then dz = gx * dx + gy * dy, something you're already familiar with, but which doesn't apply in this problem. In the above gx is the partial (derivative) of g with respect to x, and gy if the partial of g w.r.t. y.

    There's a similar formula for a function of three variables, etc.
     
  17. Sep 22, 2016 #16

    Drakkith

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    I see. This is one of those rare times that you actually have to treat dx or dy (or dV in this case) as something other than... nothing... I guess. (I know they represent small changes in the variable, but we usually just ignore them for the most part.)

    Ok. So, if ρ = m/V, then dρ = -m/V2 dV.
    Then, since dV = βV dT, we have dρ = -m/V2 * βV dT
    Simplifying gives: dρ = -mβ/V dT.
    Since p = m/v, dρ = -βρ/ dT.

    Look good?
     
  18. Sep 22, 2016 #17

    Mark44

    Staff: Mentor

    "rare times" -- No.
    It's a mistake to think of the differentials as "nothing."
    In integrals such as ##\int x^2 dx##, the dx indicates the variable of integration. It might seem that you can ignore it, but for integrals that require the use of techniques such as integration by parts or trig substitution, the differential plays a very important and nontrivial role. Ignoring it will cause you to get incorrect answers.

    In the section you're working in, the differentials also play an important role, by allowing you to approximate infinitesimally small quantities (like dρ and dT in this example) by measured nonzero values. If you're given numeric values for ρ, V, m, and ##\Delta T## (a small change in temperature), you can calculate ##\Delta \rho##, which can be a reasonably good approximation for dρ.
    Yes, that's exactly what I had in mind.
     
  19. Sep 22, 2016 #18

    Drakkith

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    Of course. They aren't "nothing". My point was that when learning calculus, the dx and dy is almost always just ignored by students since you rarely do anything with them. So when you run into a situation where you do have to use them, such as this problem, you get lost.

    Maybe I shouldn't say "rarely". As you point out below, you often have to use the differential in integration. I guess it's just something I always forgot about.

    Absolutely.

    Why is this merely an approximation?
     
  20. Sep 22, 2016 #19

    Mark44

    Staff: Mentor

    Let's take a simple example, with y = f(x) = x2. We know that f(1) = 1. In the following work I will approximate f(1.01) by using the differential.

    ##f(1.01) = f(1) + \Delta y##, where ##\Delta y## represents the exact change in y values between the point (1, 1) and (1.01, f(1.01)).
    Since I'm going to pretend that I don't know what ##\Delta y## is, I'm going to approximate ##\Delta y## by dy, which equals f'(1)dx

    For this problem ##\Delta x = .01##, which I will use in place of dx. This is reasonable provided that ##\Delta x## is reasonably small in comparison to x.

    So ##f(1.01) = f(1) + \Delta y \approx f(1) + dy = f(1) + f'(1) dx \approx f(1) + f'(1) \Delta x##
    In the last expression f(1) = 1, f'(1) = 2, and ##\Delta x = .01##, so we have ##f(1.01) \approx 1 + 2 \cdot .01 = 1.02##

    This isn't too bad, as the actual value of 1.012 is 1.0201.

    I'll try to put together a drawing of this sometime this afternoon or evening. Having an image to look at should make things a little clearer.
     
  21. Sep 22, 2016 #20

    Mark44

    Staff: Mentor

    Here's the picture I promised, that goes with the work of my example. We're using the tangent line to approximate the y value at (1.01, (1.01)2). The vertical distance between the points (1, 1) and (1.01, 1.012) is ##\Delta y##. The estimate, dy, is equal to f'(1) * dx, (but I'm approximating dx by using ##\Delta x##, or .01).

    Because the graph of y = x2 is concave up, the tangent line lies below the graph, so our approximation will be less than the actual function value.
    Graph.png
     
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