# How to reconcile these two approximations?

I'm working on what should be a simple problem (Taylor Classical Mechanics problem 6.23.), but I'm having a tough time reconciling the many ways it can be tackled.

An aircraft whose speed is $v_0$ has to fly from town $\mathcal{O}$ (at the origin) to town $P$, which is a distance $D$ due east. There is a steady gentle wind shear, such that $v_{\text{wind}}=Vy \hat{x}$, where $x$ and $y$ are measured east and north respectively. Find the path $y(x)$ which the plane should follow to minimize its flight time, as follows: (a) Find the plane’s ground speed in terms of $v_0$, $V$ , $\varphi$ (the angle by which the plane heads to the north of east), and the plane’s position. (b) Write down the time of flight as an integral of the form $\int_{0}^D f \mathrm{d}x$. Show that if we assume that $y'$ and $\varphi$ both remain small (as is certainly reasonable if the wind speed is not too large), then the integrand f takes the approximate form $f =\frac{1+ \frac{1}{2}y'^2}{1+ky}$ (times an uninteresting constant) where $k = V /v_0$.

(there's more to the problem but this is the part I'm interested in)

There are several solutions online (one, another) but I used a different method and can't get the same approximation using it.

The ground velocity is $(vy+v_0 \cos(\varphi),v_0 \sin(\varphi))=(\dot{x},\dot{y})$.

# The Intended Solution

The method used in the linked solutions is to write, where $s$ is the arc length of the curve $y(x)$, $$\int dt=\int \frac{ds}{dx} \left(\frac{ds}{dt}\right)^{-1} dx=\int \frac{\sqrt{1+y'^2}}{\sqrt{(vy+v_0\cos(\varphi))^2+v_0^2\sin^2(\varphi)}} dx$$ Then they write $\sqrt{1+y'^2}\approx 1+\frac{1}{2}y'^2$, $\cos(\varphi)\approx 1$, and $\sin(\varphi)\approx 0$ to get the approximation (with $k=v/v_0$) $$\int dt=\int \frac{1}{v_0} \frac{1+\frac{1}{2}y'^2}{1+ky} dx$$

# My Solution Number One

I didn't do the problem the way above. I wrote $$\int dt=\int \frac{1}{\dot{x}}dx=\int\frac{1}{vy+v_0\cos(\varphi)}dx$$

(And indeed, as shown in my answer below, this integrand is exactly equal to the above integrand in the intended solution, but is simpler)

You can remove $\cos(\varphi)$ from the equation by noting: $$y'=\frac{dy}{dx}=\frac{\dot{y}}{\dot{x}}=\frac{v_0\sin(\varphi)}{vy+v_0\cos(\varphi)}=\frac{\sqrt{1-\cos^2(\varphi)}}{ky+\cos(\varphi)}$$

Squaring/simplifying/quadratic-solving gives you a formula for $\cos(\varphi)$ in terms of $y'$ and $y$, with which you find that $$vy+v_0\cos(\varphi)=v_0\frac{ky+ \sqrt{1+y'^2-y'^2 y^2 k^2}}{1+y'^2}$$

If you say that the square root is $\approx 1$ (which seems to me just as justified as the cosine/sine approximations in the "official answer"), then you get:

$$\int dt=\int \frac{1}{v_0} \frac{1+y'^2}{1+ky} dx$$

The same except for a factor of one half!

# My Solution Number Two

Worse still, if you do a proper taylor series expansion on the exact root expression (so that you can actually tack on a $+O(y'^4)$ on to the approximation formula), you get:

$$\int dt=\int \frac{1}{v_0} \left(\frac{1}{1+ky}+\frac{y'^2}{2}\right) dx$$

Are these just examples of what can happen when you're too relaxed with approximations, or is there something worse going on here? It makes this seem like kind of a poor problem.

There are three approximations. The third one is probably the best because it came about from an actual Taylor expansion. Only the first approximation is what the problem intended, so indeed you have to hunt around through two different methods of solving the problem until you make an approximation that coincides with the intended approximation. I'd say this makes this a bad problem to assign/work on.

$$\int dt\approx \int \frac{1}{v_0} \frac{1+\frac{1}{2}y'^2}{1+ky} dx$$

$$\int dt\approx \int \frac{1}{v_0} \frac{1+y'^2}{1+ky} dx$$

$$\int dt\approx \int \frac{1}{v_0} \left(\frac{1}{1+ky}+\frac{y'^2}{2}\right) dx$$

Checking my algebra through Mathematica:

In[]:= Simplify[k y + cosq /. Solve[yprime == Sqrt[1 - cosq^2]/(k y + cosq), cosq]][[2]]
Out[]:= (k y+Sqrt[1+yprime^2-k^2 y^2 yprime^2])/(1+yprime^2)


Checking the equivalence of both integrands by computing their difference numerically (Mathematica doesn't seem to want to simplify to prove they're equivalent and neither do I. Equality $dt=dx/(dx/dt)=dx (ds/dx)/(ds/dt)$ is enough)

In[]:= cosqsolution=Solve[yprime==Sqrt[1-cosq^2]/(k y+cosq),cosq][[2]];
differenceExpression=(1/(k y+cosq)-Sqrt[1+yprime^2]/Sqrt[k^2 y^2 + 2 k y cosq+1]/.cosqsolution);