Let's pick up on vertical motion with quadratic drag from last time. We identified the general equation of motion to be
\[ \begin{aligned} \dot{v}_y = -g - {\rm sgn}(v_y) \frac{c}{m} v_y^2 \end{aligned} \]
Although this is valid, it's also not especially useful - we'll have to solve the two cases where \( v_y \) is positive and negative separately anyway. Let's begin with \( v_y < 0 \) (the only case that Taylor does), so that
\[ \begin{aligned} \dot{v}_y = -g + \frac{c}{m} v_y^2\ \ (v_y < 0). \end{aligned} \]
Here we see that as before, there is a terminal velocity at which the forces balance,
\[ \begin{aligned} v_{{\rm ter}, c} = \sqrt{\frac{mg}{c}}. \end{aligned} \]
Before we continue with the solution, let's think a bit about the opposite-sign case: if \(v_y > 0\), then
\[ \begin{aligned} \dot{v}_y = -g - \frac{c}{m} v_y^2\ \ (v_y > 0). \end{aligned} \]
In this case, there is no terminal velocity. Physically, this makes sense because if our object is moving upwards, gravity and drag are both pointing down, and the forces can't balance. Eventually our object will turn around and start to fall, and we'll end up in the \(v_y < 0\) case and approach the terminal velocity \(v_{{\rm ter},c}\). (However, we can't use our quadratic solution to describe this motion, because the object has to come to a stop before it turns around, and the linear air resistance will start to matter!)
Now that we've thought carefully about signs, we'll assume \( v_y < 0 \) from here forward. Once again, we have an inhomogeneous ODE. Even worse, this is actually a non-linear ODE because of the square! That means that we can't attack it using the complementary + particular solution method - we'll get the wrong answer if we try.
The good news is that we can still get a general solution because the ODE is still separable. Dividing the right-hand side out,
\[ \begin{aligned} \frac{\dot{v}_y}{-g+\frac{c}{m} v_y^2} = 1 \\ \frac{dv_y}{1 - v_y^2 / v_{{\rm ter},c}^2 } = -g dt \end{aligned} \]
where I plugged in the terminal velocity to make things look nicer. Now integrating and doing a quick variable change,
\[ \begin{aligned} v_{{\rm ter}, c} \int \frac{du}{1-u^2} = -\int g dt \end{aligned} \]
where \(u = v_y / v_{{\rm ter},c}\). This is a very standard integral, and it's easy to look up the result - in fact, it's in the front cover of Taylor:
\[ \begin{aligned} \int \frac{du}{1-u^2} = \tanh^{-1}(u). \end{aligned} \]
(the inverse hyperbolic tangent function, pronounced "tanch-inverse" or "arc-tanch"). Instead of relying on this result, I'm actually going to go through doing the integral by hand, because it shows off a little bit of useful math and it might give you better intuition for what the hyperbolic tangent function is.
You may remember from your calculus class that a good trick to deal with integrals of rational functions is to split them up into separate terms. Here, we notice that
\[ \begin{aligned} \frac{1}{1-u^2} = \frac{1}{(1+u)(1-u)} = \frac{1/2}{1+u} + \frac{1/2}{1-u}. \end{aligned} \]
With the integrand split apart, we're left with two integrals that we know how to do:
\[ \begin{aligned} \int du\ \left( \frac{1/2}{1+u} + \frac{1/2}{1-u} \right) = \frac{1}{2} \ln (1+u) - \frac{1}{2} \ln (1-u) \\ = \frac{1}{2} \ln \left( \frac{1+u}{1-u} \right). \end{aligned} \]
Back to our full equation, we have
\[ \begin{aligned} \frac{1}{2} \ln \left( \frac{1+u}{1 - u} \right) = -\frac{gt}{v_{{\rm ter},c}} + A\\ \frac{1+u}{1 - u} = B e^{-2gt / v_{{\rm ter},c}} \end{aligned} \]
keeping track of our arbitrary constant of integration. Before we continue, notice that we can once again define a natural time in freefall from the terminal velocity in the same way as the linear case,
\[ \begin{aligned} \tau \equiv \frac{v_{\rm ter},c}{g}. \end{aligned} \]
(not the same as \( \tau_c \) above, beware!) I'll use \( \tau \) from this point on to save some writing. The last step is to solve for \( v_y \), which means solving for \( u \). If we begin by multiplying through to get rid of the fraction,
\[ \begin{aligned} 1+u = Be^{-2t/\tau} (1-u) \\ u (1 + Be^{-2t/\tau}) = Be^{-2t/\tau} - 1 \end{aligned} \]
so plugging in \(u = v_y / v_{{\rm ter}, c}\),
\[ \begin{aligned} v_y(t) = v_{{\rm ter},c} \frac{Be^{-2t/\tau} - 1}{Be^{-2t/\tau} + 1}. \end{aligned} \]
Finally, we need to fix the constant of integration. Let's take \( v_y(0) = -v_0 \), where \( v0 \) like \( v{{\rm ter},c} \) is a positive speed. Then we have
\[ \begin{aligned} -v_0 = v_{{\rm ter},c} \frac{B-1}{B+1} \Rightarrow B = \frac{v_{{\rm ter},c} + v_0}{v_{{\rm ter},c} - v_0}. \end{aligned} \]
In the special case that we start from rest, \( v_0 = 0 \), then \( B = 1 \) and our solution is
\[ \begin{aligned} v_y(t) = v_{{\rm ter},c} \frac{e^{-2t/\tau} - 1}{e^{-2t/\tau} + 1}. \end{aligned} \]
First, notice that the exponential pieces are always less than 1, so the fraction is always negative, which it had better be because we assumed \( v_y < 0 \)! As \( t \rightarrow \infty \), the exponentials vanish, and we approach the terminal velocity asymptotically, like in the linear case.
What about the hyperbolic trig function? Let's review those very quickly. The definitions of hyperbolic sine and cosine are
\[ \begin{aligned} \sinh(x) = \frac{1}{2} (e^x - e^{-x}) \\ \cosh(x) = \frac{1}{2} (e^x + e^{-x}) \end{aligned} \]
At large negative and positive \( x \), these are dominated by one or the other exponential. So the whole functions resemble two exponential curves stuck together, with or without a minus sign:
The reason these are called hyperbolic trig functions is that they satisfy a number of identities that are similar to the regular trig identities: for example, \( \cosh^2 x - \sinh^2 x = 1 \). (Aside: the real reason is because you can define them by trying to do trigonometry using a unit hyperbola instead of a unit circle, and the deeper reason is that they're actually related to ordinary trig functions once you start using complex numbers.)
Finally, the hyperbolic tangent is defined in analogy to the regular tangent:
\[ \begin{aligned} \tanh(x) = \frac{\sinh(x)}{\cosh(x)} =\frac{e^x - e^{-x}}{e^x + e^{-x}} = \frac{e^{2x} - 1}{e^{2x} + 1}. \end{aligned} \]
Here's a sketch of that function, showing the asymptotes at \( \pm 1 \):
This function comes up frequently in physics - we've just seen an example - but the fact that it smoothly goes from \( -1 \) to \( +1 \) makes it useful in a variety of engineering applications as well, because it maps an arbitrary input signal into a finite range. One example is in machine learning, where the response of individual neurons in a neural net is often modeled with \( \tanh \).
Thus, for the special case of starting at rest, we can write our solution simply as a hyperbolic tangent:
\[ \begin{aligned} v_y(t) = -v_{{\rm ter},c} \tanh \left( \frac{t}{\tau} \right) = -v_{{\rm ter},c} \tanh \left( \frac{gt}{v_{{\rm ter},c}} \right). \end{aligned} \]
This is nice and simple, but it's also only true if we start at rest! For \( v_0 \neq 0 \), the general solution is
\[ \begin{aligned} v_y(t) = v_{{\rm ter},c} \frac{(v_{{\rm ter},c} + v_0) e^{-2t/\tau} - (v_{{\rm ter},c} - v_0)}{(v_{{\rm ter},c} + v_0) e^{-2t/\tau} + (v_{{\rm ter},c} - v_0)} \end{aligned} \]
which can't be written as a hyperbolic tangent in any particularly nice way. (This is a good reminder that it's important to carry your integration constants along when solving ODEs! If you just solved at rest, you might be tempted to think that the general solution for a moving initial condition would just be to add \( v_0 \) to the \( \tanh(...) \) solution above - but that would be wrong!)
One final thing to point out, linking back to our general discussion of ODEs. Remember that this is a non-linear ODE, for which we were warned that we can't necessarily find a general solution. What we just found looks general, but don't forget that we had to assume \( v_y < 0 \). If instead we had taken \( v_y > 0 \), a very important minus sign would be flipped, and the key integral above would become
\[ \begin{aligned} \int \frac{du}{1+u^2} = \tan^{-1}(u). \end{aligned} \]
So solutions for positive initial speed look like regular tangent instead of hyperbolic tangent. In other words, the functional form of our solution changes depending on initial conditions! This is always a possibility if our ODE is non-linear.
Example: oops!
I accidentally drop my wedding ring (mass ~ 10 g, width ~ 6 mm) into a glass of dark beer. Beer is more viscous than air, by a factor of about 137 according to Wikipedia; it's also more dense than air, by a factor of about 750.
a) Is linear or quadratic drag dominant? b) What is the terminal velocity? c) How long will it take to hit the bottom of a 15-cm tall pint glass, using a first-order series expansion? Is this result trustworthy?
Starting with part a), I know how the drag coefficients scale with viscosity and density,
\[ \begin{aligned} \beta = 3\pi \eta,\ \ \gamma = \frac{1}{2} C_D \rho (A/D^2). \end{aligned} \]
The important part is that \( \beta \) is linear in the viscosity \( \eta \), and \( \gamma \) is linear in medium density \( \rho \). Comparing to air as a baseline, the density goes up more than the viscosity does, so \( \gamma \) will increase more than \( \beta \). Because we know that everyday objects tend to be quadratic-drag dominated in air, this will probably still be true in the beer, so this should be a quadratic drag problem.
Just to be thorough, I'll calculate both drag coefficients, using Taylor's values for air and \( D = 6 \) mm:
\[ \begin{aligned} b = \beta_{\rm air} \frac{\eta_{\rm beer}}{\eta_{\rm air}} D \\ = (1.6 \times 10^{-4}\ {\rm N} \cdot {\rm s} / {\rm m}^2) (137) (6\times 10^{-3}\ {\rm m}) \\ = 1.3 \times 10^{-4}\ {\rm N} \cdot {\rm s} / {\rm m} \end{aligned} \]
and
\[ \begin{aligned} c = \gamma_{\rm air} \frac{\rho_{\rm beer}}{\rho_{\rm air}} D^2 \\ = (0.25\ {\rm N} \cdot {\rm s}^2 / {\rm m}^4) (750) (6 \times 10^{-3}\ {\rm m})^2 \\ = 6.8 \times 10^{-3}\ {\rm N} \cdot {\rm s}^2 / {\rm m}^2. \end{aligned} \]
So the Reynolds number is
\[ \begin{aligned} R = \frac{cv}{b} = \frac{v}{0.019\ {\rm m}/{\rm s}}. \end{aligned} \]
As long as the terminal velocity for quadratic drag isn't too close to \( 0.019\ {\rm m}/{\rm s} \), then, we expect the quadratic force to dominate.
On to part b), which is simple: we just plug in to the formula for quadratic freefall and find
\[ \begin{aligned} v_{\rm ter,c} = \sqrt{\frac{mg}{c}} = \sqrt{\frac{(10^{-2}\ {\rm kg})(9.8\ {\rm m}/{\rm s}^2)}{6.8 \times 10^{-3}\ {\rm N} \cdot {\rm s}^2 / {\rm m}^2}} \approx 3.8\ {\rm m}/{\rm s}. \end{aligned} \]
This is safely large enough that our Reynolds number will be very large for the majority of the motion, so the quadratic model is okay. For good measure, we should also calculate the natural time:
\[ \begin{aligned} \tau = \frac{g}{v_{\rm ter,c}} = 2.6\ {\rm s}. \end{aligned} \]
Finally, we ask about series expansion of the answer - although we know the full solution in this case, it's still useful to think about to improve our understanding of series expansion use in general. As we just found, the full solution starting from rest is
\[ \begin{aligned} v_y(t) = -v_{\rm ter,c} \tanh \left( \frac{t}{\tau} \right) \end{aligned} \]
We could look up a formula for the series expansion of \( \tanh \), or look up its derivatives and expand ourselves, but I'll use a trick to find the expansion to first order instead. Recall the definition:
\[ \begin{aligned} \tanh(x) = \frac{e^{2x} - 1}{e^{2x} + 1} \end{aligned} \]
Now I'll apply the series expansions for \( e^{2x} \):
\[ \begin{aligned} \tanh(x) = \frac{1 + 2x + 2x^2 + ... - 1}{1 + 2x + 2x^2 + ... + 1} = \frac{x+...}{1+x+...} \end{aligned} \]
cancelling off a 2. We're still not finished, since this isn't a polynomial yet. But we can use the expansion \( 1/(1+x) = 1-x + ... \) to move the denominator up top,
\[ \begin{aligned} \tanh(x) = (x+...) (1 - x + ...) = x + \mathcal{O}(x^2) \end{aligned} \]
Actually, there is no \( \mathcal{O}(x^2) \) term at all, because of the relation \( \tanh(-x) = -\tanh(x) \). So the missing terms start at \( \mathcal{O}(x^3) \).