# Mathematical method in school physics

I have a simple physical task: there is a car, riding on an ellipse ($a=500$, $b=250$), with a constant absolute velocity ($|\vec{v}|$ constant). I need to find the maximum and minimum acceleration $\vec{a}$ in each point.

I represent the position vector $\vec{r}$ as a vector function $\vec{r}(t)$. Also, I know that velocity is the first derivative, $\vec{v}=\vec r\,'$, and acceleration is the second derivative, $\vec{a}=\vec r\,''$.

How should I approach this problem?

## 2 Answers

Acceleration is given by the formula $$\vec{a}=\frac{dv}{dt}\vec{u}_T+\frac{v^2}{R}\vec{u}_{\perp}$$ where $\frac{dv}{dt}\vec{u}_T$ is the linear acceleration ($\vec u_T$ is the unit vector tangent to the trajectory) and $\frac{v^2}{R}\vec{u}_{\perp}$ the centripetal/normal acceleration ($\vec u_{\perp}$ is the unit vector perpendicular to $\vec u_T$). $R$ is the radius of curvature, that is the radius of the circumference that best approximates the car's trajectory in every point.

So, knowing that $v$ is constant, linear acceleration is always zero, so only the centripetal one counts, and it is greater when $R$ is smaller, which happens at the vertices of the major axis, so you have maximum acceleration at these two points. On the other hand you have minimum acceleration where $R$ is greater, that is at the vertices of the minor axis.

• You are quit right, but I meant general solution (some function). So, it's my mistake and I will edit the question. – Vanzef Dec 22 '13 at 18:54
• Uh, well, then acceleration would be just $a=\frac{v^2}{R}$. You can drop the vector notation as it all happens on one direction, the one perpendicular to the trajectory at each point. You should then find an equation for the curvature of the ellipse (which I don't know), then put it in that equation, knowing that the radius $R$ is the inverse of the curvature. – yellowquark Dec 22 '13 at 19:08
• @Vanzef: in this find parametric and curvature . – Tony Piccolo Dec 22 '13 at 22:27

There is a nice discussion of the relevant computations at http://www2.math.umd.edu/~jcooper/math241/accel.pdf. The following is a quick summary of the relevant parts:

Suppose $r$ is a $C^2$ curve. Suppose $v(t) = \dot{r}(t) \neq 0$, and let $\tau(t) = { v(t) \over \|v(t)\|}$. Notice that $\|\tau(t)\| = 1$, and so $\langle \dot{\tau}(t), \tau(t) \rangle = 0$. Further, suppose $\dot{\tau}(t) \neq 0$ and let $\nu(t) = { \dot{\tau}(t) \over \| \dot{\tau}(t) \| }$. Note that $v(t) = \|v(t)\| \tau(t)$, and so $a(t) = \dot{v}(t) = a_t(t) \tau(t) + a_n \nu(t)$, where $a_t(t) = \dot{s}(t)$, with $s(t) = \|v(t)\|$ and $a_n(t) = \|v(t)\| \| \dot{\tau}(t) \|$.

Now we need to parametrize the ellipse appropriately. If we let $p(\theta) = (a \cos \theta, b \sin \theta)$ we see that $p$ traces out the appropriate ellipse, but the speed is not constant. We need to find a curve $t \mapsto \theta(t)$ so that $r(t) = p(\theta(t))$ satisfies $\|v(t)\| = \sigma$, where $\sigma$ is the constant speed.

We have $v(t) = (-a\sin \theta(t), b \cos \theta(t) ) \dot{\theta}(t)$, and so $\sqrt{(a \sin \theta(t))^2 + (b \cos \theta(t))^2}|\dot{\theta}(t)| = \sigma$. Hence if we let $\theta$ solve the differential equation $\dot{\theta}(t) = { \sigma \over \sqrt{(a \sin \theta(t))^2 + (b \cos \theta(t))^2} }$ subject to $\theta(0) = 0$, then $r$ traces out the ellipse with constant speed.

Since the speed is constant, we see that $a_t(t) = 0$. We have $\tau(t) = { (-a\sin \theta(t), b \cos \theta(t) ) \over \sqrt{(a \sin \theta(t))^2 + (b \cos \theta(t))^2} }$. Differentiating gives $\dot{\tau}(t) = { -( a b^2 \cos \theta(t), a^2 b \sin \theta(t)) \over ({(a \sin \theta(t))^2 + (b \cos \theta(t))^2})^{3\over 2} } {\sigma \over \sqrt{(a \sin \theta(t))^2 + (b \cos \theta(t))^2} }$.

Hence $\| \dot{\tau}(t) \| = {\sigma |ab| \over ({(a \sin \theta(t))^2 + (b \cos \theta(t))^2})^{3\over 2} }$, and so $\|a(t)\| = a_n(t) = {\sigma^2 |ab| \over ({(a \sin \theta(t))^2 + (b \cos \theta(t))^2})^{3\over 2} }$.