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As a sort of practice on solving word problems on atmospheric dynamics, I attempted to solve this problem that I found in page 22 of the book Mid-latitude Atmospheric Dynamics: A First Course by Martin (2006):

1.10. An automobile equipped with a thermometer is heading southward at $100 \space km \cdot h^{-1}$ bound for a location $300 \space km$ away. During transit, the temperature drops to $−5°C$ at the origin. If the temperature at departure was measured to be $0°C$ and the temperature tendency measured along the journey is $+5°C \cdot h^{−1}$, what temperature should the travelers expect at their destination?

In the list of answers to selected problems, the book stipulates that:

1.10. The temperature at the destination will be $15°C$.

But in my solution, the answer I got is $20°C$. This is how I got that answer—

Based on the problem and applying the proper sign conventions, the given are:

  • temperature at departure, i.e. initial temperature at the origin, $T_0 = 0°C$
  • duration of travel $\partial{t} = 3 \space h$
  • distance traveled $\partial{y} = -300 \space km$
  • velocity $v = \dfrac{Dy}{Dt} = -100 \space \dfrac{km}{h}$ (since the automobile moved only southward), assumed to be constant during transit
  • temperature change during transit along the automobile's path (Lagrangian) $\dfrac{DT}{Dt} = +5 \space \dfrac{°C}{h}$
  • temperature change during transit at the origin (Eulerian) $\dfrac{\partial{T}}{\partial{t}} = \dfrac{-5°C}{3 \space h} = -\dfrac{5}{3}\dfrac{°C}{h}$

And the value being asked is the temperature at the destination $T_1$.

I applied the concept of the material derivative where the scalar field is temperature $T$, expanded this equation and neglected the terms for the x- and z-components (since the motion is only meridional, i.e. along the y-axis). $$\dfrac{DT}{Dt} = \dfrac{\partial{T}}{\partial{t}} + \mathbf{U} \cdot \nabla T$$ $$\dfrac{DT}{Dt} = \dfrac{\partial{T}}{\partial{t}} + u \dfrac{\partial{T}}{\partial{x}} + v \dfrac{\partial{T}}{\partial{y}} + w \dfrac{\partial{T}}{\partial{z}}$$ $$\dfrac{DT}{Dt} = \dfrac{\partial{T}}{\partial{t}} + v \dfrac{\partial{T}}{\partial{y}} \space \space \space (meridional \space only) $$

Since $\dfrac{\partial{T}}{\partial{y}}$ was not explicitly given and knowing that $\partial{y} = -300 \space km$, I let $\partial{T} = T_1 - T_0$ and rewrote the equation as $$\dfrac{DT}{Dt} = \dfrac{\partial{T}}{\partial{t}} + v \bigg( \dfrac{T_1 - T_0}{-300 \space km} \bigg)$$

Solving for $T_1$ and plugging in the values, $$\dfrac{DT}{Dt} = \dfrac{\partial{T}}{\partial{t}} + \bigg( \dfrac{v T_1 - v T_0}{-300 \space km} \bigg)$$ $$ v T_1 - v T_0 = -300 \space km \bigg( \dfrac{DT}{Dt} - \dfrac{\partial{T}}{\partial{t}} \bigg)$$ $$ T_1 = \dfrac{v T_0 - 300 \space km \bigg( \dfrac{DT}{Dt} - \dfrac{\partial{T}}{\partial{t}} \bigg)}{v} $$ $$ T_1 = \dfrac{\bigg[\bigg( -100 \dfrac{km}{h} \bigg) (0°C) \bigg] - \bigg[ 300 \space km \bigg( 5 \dfrac{°C}{h} - \bigg( -\dfrac{5}{3} \dfrac{°C}{h} \bigg) \bigg) \bigg]}{-100 \space \dfrac{km}{h}} = \dfrac{-2000 \space km \cdot \dfrac{°C}{h}}{-100 \space \dfrac{km}{h}} = 20°C $$

This makes me wonder how the author came up with 15°C. Now, I would like to ask whether or not my solution is theoretically sound and mathematically correct. If not, I would also like to know the mistakes in the solution and ask for your recommendation on the most appropriate strategy for this problem. Thank you!

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Digging into your long expansion, if you want to go that way... the nuance that led you astray is the fact that $\dfrac{\partial{T}}{\partial{y}}$ is the difference in temperature over y... at one instance. In other words, being more explicit, it's $${\partial{T}} = T_{y1} - T_{y0}$$.

That isn't to say you did bad math... you indeed found $T_1$... but that $T$ must match the instant in time of your other value. In other words:
at $t=0$... your origin was at $0°C$... and your final destination was at $20°C$.
at $t=3hr$... your origin is at $-5°C$... and your final destination at $15°C$.

You validly found there's a $20°C$ difference meridionally between your two sites at any picked instance.

Want some convincing that you have to hold $t$ (and $x$ and $z$) constant for $T_1-T_0 = {\partial{T}}$ when in $\dfrac{\partial{T}}{\partial{y}}$?

Well, if you wanted to calculate the local change, $\dfrac{\partial{T}}{\partial{t}}$... you couldn't just subtract the measurement three hours apart from different locations. You have to keep the other variables constant. If want how the sun's position in the sky varies over time locally, you can't compare the views in two different spots. And if you want how the sun's position in the sky varies by latitude, you can't just compare Shanghai's 10 AM sky to Luzon's 5 PM sky.

Now, there is a second entirely valid way of calculating it that shortcuts the whole problem. Look it from "your" perspective in the automobile. The material derivative is how things are changing from "your" (the car's) perspective. Where will "you" be in 3 hours? You'll just happen to be... at the destination city. So if you apply your perspective's rate of change, which you are directly given... you can get it easy.

Or put another way $$\dfrac{DT}{Dt} = \dfrac{T_1-T_0}{t_1-t_0}$$

Where nothing ($x, y, z, t$) is being held constant... in other words, we're working along your path (where all those independent variables may well be changing [x and z happen not to be in this case, so aren't part of the contributions... but on your path y and t certainly are changing, and the fact you're moving through both time and latitude certainly are both contributing to the temperature change you are getting].

That's another way of expressing the age old concept that total change = rate * time, a concept you learned well before worrying about what different contributions there were to the change.

But anyways... for your equation, as you note: $\dfrac{DT}{Dt} = +5 \space \dfrac{°C}{h}$

Well $t_1 = 3hr$ and $t_0 = 0hr$.
And $T_0 = 0°C$.

So if you plug those and solve for $T_1$, you get:

$+5 \space \dfrac{°C}{h} \cdot 3h = 15°C$

And if you stop and think about it, that just makes sense. If it's 0°C and changing for you 5°C per hour... it'll be 5°C at your location when you hit the one hour mark... 10°C at your location when you hit the two hour mark... and 15°C at your location when you hit the three hour mark.

And that just happens to be where they've asked about!

So you either need to keep emphasizing to yourself that each partial derivative itself is the change in your variable where the other variables are kept constant... or you have to realize that the material derivatives are often giving you what you want if you're interested in the changing path. For example, if there's a 980 mb tropical depression over Guam and it's supposed to deepen (intensify) 20 mb each day and make landfall in the Philippines in four days... you don't need to take a deep dive into local and spatial rates of pressure changes... they've just told you all you need to know, you're expecting a 900 mb typhoon in the Philippines in four days!

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  • $\begingroup$ The first path, though long, is the more universally useful. Because what if they asked what the temperature would be in 4 hours 150 km to the north of the origin? Well having both rates of change, you could just calculate there's a -20°C time contribution and a 15°C meridional contribution from the origin's starting 0°C and you'd have the -5°C answer. But if you're taking a test on this topic, they're more likely to ask simpler questions where there's a basic key to getting the answer... and it's usually in having a firm understanding of just what the derivatives mean and how that's useful $\endgroup$ Commented Jul 10 at 10:37

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