Subfield: Mechanics


Question: On a flat playground, choose a Cartesian Oxy coordinate system (in unit of meters). A child running at a constant velocity $V=1 \mathrm{~m} / \mathrm{s}$ around a heart-shaped path satisfies the following order- 6 algebraic equation:
$$
\left(x^{2}+y^{2}-L^{2}\right)^{3}-L x^{2} y^{3}=0, L=10 \text {. }
$$
When the child is at the position $(x, y)=(L, 0)$, what is the magnitude of their acceleration?

![](https://cdn.mathpix.com/cropped/2023_12_21_23e8f9f667608506c947g-1.jpg?height=688&width=1282&top_left_y=526&top_left_x=408)

Solution:
The acceleration can be found from the local geometry of the curves, thus let us study small deviations around the position of interests $(x, y)=(L, 0)$ :
$$
x=L+\delta_{x} \quad, \quad y=0+\delta_{y} \quad, \quad\left|\delta_{x}\right|,\left|\delta_{y}\right| \ll L
$$
Consider the 2nd-order approximation in $\delta_{x}$ of $\delta_{y}$ with quadratic coefficients $\alpha$ and $\beta$ :
$$
\delta_{y} \approx \alpha \delta_{x}+\frac{\beta}{L} \delta_{x}^{2} \sim \delta_{x}
$$
To find these coefficients, we look at the algebraic equation of our heart-shape path up to the two lowest-orders of expansions (which are the 3rd and 4th):
$$
\begin{aligned}
0 & =\left(x^{2}+y^{2}-L^{2}\right)^{3}-L x^{2} y^{3} \approx L^{2}\left[8 L \delta_{x}^{3}+12 \delta_{x}^{4}+12 \delta_{x}^{2} \delta_{y}^{2}-2 \delta_{x} \delta_{y}^{3}-L \delta_{y}^{3}+\mathcal{O}\left(\delta_{x}^{5}\right)\right] \\
& \approx L^{2}\left[8 L \delta_{x}^{3}+12 \delta_{x}^{4}+12 \alpha^{2} \delta_{x}^{4}-2 \alpha^{3} \delta_{x}^{4}-\left(\alpha^{3} L \delta_{x}^{3}+3 \alpha^{2} \beta \delta_{x}^{4}\right)+\mathcal{O}\left(\delta_{x}^{5}\right)\right] \\
& \propto\left(8-\alpha^{3}\right) L \delta_{x}^{3}+\left(12+12 \alpha^{2}-2 \alpha^{3}-3 \alpha^{2} \beta\right) \delta_{x}^{4}+\mathcal{O}\left(\delta_{x}^{5}\right)
\end{aligned}
$$
Thus, $\alpha$ and $\beta$ can be found by solving:
$$
8-\alpha^{3}=0 \quad, \quad 12+12 \alpha^{2}-2 \alpha^{3}-3 \alpha^{2} \beta=0 \Longrightarrow \alpha=2, \beta=\frac{11}{3} .
\tag{3}
$$
We can find the relations between velocities $\left.(\dot{x}, \dot{y})=\dot{\delta}_{x}, \dot{\delta}_{y}\right)$ and accelerations $(\ddot{x}, \ddot{y})=\ddot{\delta}_{x}, \ddot{\delta}_{y}$ ) evaluated at the position $(x, y)=(1,0) \rightarrow\left(\delta_{x}, \delta_{y}\right)=(0,0)$ by taking the time-derivatives:
$$
\dot{\delta}_{y}=\alpha \dot{\delta}_{x}+2 \frac{\beta}{L} \delta_{x} \dot{\delta}_{x}=\left(\alpha+2 \frac{\beta}{L} \delta_{x}\right) \dot{\delta}_{x}=\alpha \dot{\delta}_{x}
\tag{4}
$$

$$
\ddot{\delta}_{y}=\alpha \ddot{\delta}_{x}+2 \frac{\beta}{L} \dot{\delta}_{x}^{2}+2 \frac{\beta}{L} \delta_{x} \ddot{\delta}_{x}=\left(\alpha+2 \frac{\beta}{L} \delta_{x}\right) \ddot{\delta}_{x}+2 \frac{\beta}{L} \dot{\delta}_{x}^{2}=\alpha \ddot{\delta}_{x}+2 \frac{\beta}{L} \dot{\delta}_{x}^{2}
\tag{5}
$$
For a constant running speed $V$, we get:
$$
V=\left(\dot{x}^{2}+\dot{y}^{2}\right)^{1 / 2} \Longrightarrow \quad \dot{\delta}_{x}=\left(1+\alpha^{2}\right)^{-1 / 2} V, \dot{\delta}_{y}=\alpha\left(1+\alpha^{2}\right)^{-1 / 2} V
$$
which we obtain by applying Eq. (4). Also, the temporal-constraint of constant speed means that the acceleration vector (if non-zero) should be perpendicular to the velocity vector:
$$
\frac{d}{d t} V=0=\frac{d}{d t}\left(\dot{x}^{2}+\dot{y}^{2}\right)^{1 / 2} \propto \dot{\delta}_{x} \ddot{\delta}_{x}+\dot{\delta}_{y} \ddot{\delta}_{y}=0 \quad \Longrightarrow \quad \ddot{\delta}_{x}+\alpha \ddot{\delta}_{y}=0
$$
Using Eq. (5), we can arrive at:
$$
\begin{aligned}
\ddot{\delta}_{x}+\alpha\left(\alpha \ddot{\delta}_{x}+2 \frac{\beta}{L} \dot{\delta}_{x}^{2}\right)=0 \Longrightarrow \ddot{\delta}_{x} & =-2 \frac{\beta}{L} \alpha\left(1+\alpha^{2}\right)^{-1} \dot{\delta}_{x}^{2}=-2 \beta \alpha\left(1+\alpha^{2}\right)^{-2} \frac{V^{2}}{L}, \\
\ddot{\delta}_{y} & =-\alpha^{-1} \dot{\delta}_{x}=2 \beta\left(1+\alpha^{2}\right)^{-2} \frac{V^{2}}{L} .
\end{aligned}
$$
The quadratic coefficients are found in Eq. (3), and given that $V=1 \mathrm{~m} / \mathrm{s}, L=10 \mathrm{~m}$, the magnitude of the total acceleration can be calculated:
$$
a=\left(\ddot{\delta}_{x}^{2}+\ddot{\delta}_{y}^{2}\right)^{1 / 2}=2 \beta\left(1+\alpha^{2}\right)^{3 / 2} \frac{V^{2}}{L}=\frac{22}{15 \sqrt{5}} \frac{V^{2}}{L} \approx 0.066591 \mathrm{~m} / \mathrm{s}^{2} .
$$

Final answer: $0.066591$ $\mathrm{~m} / \mathrm{s}^{2}$