EXAMPLE 6.2 We can use the calculus of variations to solve a classic problem in the history of ". physics: the brachistochrone. \( { }^{\dagger} \) Consider a particle moving in a constant force field starting at rest from some point \( \left(x_{1}, y_{1}\right) \) to some lower point \( \left(x_{2}, y_{2}\right) \). Find the path that allows the particle to accomplish the transit in the least possible time.
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Define the problem: We want to find the path that minimizes the time taken for a particle to move from point \( (x_{1}, y_{1}) \) to point \( (x_{2}, y_{2}) \) in a constant force field. Show more…
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(a) Let us differentiate twice the path equation $y(x)$ with respect to time. $$ \frac{d y}{d t}=2 a x \frac{d x}{d t} ; \frac{d^{2} y}{d t^{2}}=2 a\left[\left(\frac{d x}{d t}\right)^{2}+x \frac{d^{2} x}{d t^{2}}\right] $$ Since the particle moves uniformly, its acceleration at all points of the path is normal and at the point $x=0$ it coincides with the direction of derivative $d^{2} y / d t^{2}$. Keeping in mind that at the point $x=0,\left|\frac{d x}{d t}\right|=v$ We get $$ w=\left|\frac{d^{2} y}{d t^{2}}\right|_{x=0}=2 a v^{2}=w_{n} $$ So, $w_{n}=2 a v^{2}=\frac{v^{2}}{R}$, or $R=\frac{1}{2 a}$ Note that we can also calculate it from the formula of problem $(1.35 b)$ (b) Differentiating the equation of the trajectory with respect to time we see that $$ b^{2} x \frac{d x}{d t}+a^{2} y \frac{d y}{d t}=0 $$ which implies that the vector $\left(b^{2} x \vec{i}+a^{2} y \vec{j}\right)$ is normal to the velocity vector $\vec{v}=\frac{d x}{d t} i+\frac{d y}{d t} \vec{j}$ which, of course, is along the tangent. Thus the former vactor is along the normal and the normal component of acceleration is clearly $$ w_{n}=\frac{b^{2} x \frac{d^{2} x}{d t^{2}}+a^{2} y \frac{d^{2} y}{d t^{2}}}{\left(b^{4} x^{2}+a^{4} y^{2}\right)^{1 / 2}} $$
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A curve $\mathbf{r}(t)$ is parameterised by a scalar variable $t .$ Show that the length of the curve between two points, $A$ and $B$, is given by $$ L=\int_{A}^{B} \sqrt{g_{i j} \frac{d u^{i}}{d t} \frac{d u^{\prime}}{d t}} d t $$ Using the calculus of variations (see chapter 22 ), show that the curve $\mathbf{r}(t)$ that minimises $L$ satisfies the equation $$ \frac{d^{2} u^{i}}{d t^{2}}+\Gamma^{i}{ }_{j k} \frac{d u^{j}}{d t} \frac{d u^{k}}{d t}=\frac{\ddot{s}}{s} \frac{d u^{i}}{d t} $$ where $s$ is the arc length along the curve, $s=d s / d t$ and $\mathfrak{s}=d^{2} s / d t^{2} .$ Hence, show that if the parameter $t$ is of the form $t=a s+b$, where $a$ and $b$ are constants, then we recover the equation for a geodesic (26.101). [A parameter which, like $t$, is the sum of a linear transformation of $s$ and a translation is called an affine parameter.]
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