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Age 16 to 18

Challenge Level

- Problem
- Getting Started
- Student Solutions
- Teachers' Resources

Particle A

Qualitative behaviour: moves in positive direction with an
increasing velocity. $$\frac{dx}{dt} = x \Rightarrow \int
\frac{1}{x}dx = \int dt \Rightarrow x = Ae^t$$ $x(0) = 1$ so $A =
1$ and the solution is $x(t) = e^t$.

Particle B

Qualitative behaviour: moves in negative
direction with a decreasing velocity, gradually approching origin.
$$\frac{dx}{dt} = x \Rightarrow x = Ae^{-t}$$ $x(0) = 1$ so $A = 1$
and the solution is $x(t) = e^{-t}$.

Particle C

Qualitative behaviour: moves in
positive direction with an increasing velocity. $$\frac{dv}{dt} = x
\Rightarrow \frac{d^2x}{dt^2} - x = 0 \Rightarrow x = Ae^t +
Be^{-t}$$ If particle starts at origin with velocity $v$ then $x(t)
= \frac{v}{2}(e^t - e^{-t})$.

Particle D

Qualitative behaviour: oscillates
around origin. $$\frac{dv}{dt} = -x \Rightarrow \frac{d^2x}{dt^2} +
x = 0 \Rightarrow x = Ae^{it} + Be^{-it} = C\sin{t} + D\cos{t}$$ If
particle starts at origin with velocity $v$ then $x(t) =
v\sin{t}$.

Particles C and D have symmetrical
equations of motion, in the sense that starting at origin with
negative velocity would result in the same motion but in the
opposite direction.

Observe that if we start particle C
at $+1$ with velocity $-1$ then our equation of motion is $x(t) =
e^{-t}$, so the particle eventially stops at the origin. Are there
any other initial conditions that can be imposed on particles C or
D which force the particle to stop eventually?

A circle is inscribed in an equilateral triangle. Smaller circles touch it and the sides of the triangle, the process continuing indefinitely. What is the sum of the areas of all the circles?

$2\wedge 3\wedge 4$ could be $(2^3)^4$ or $2^{(3^4)}$. Does it make any difference? For both definitions, which is bigger: $r\wedge r\wedge r\wedge r\dots$ where the powers of $r$ go on for ever, or $(r^r)^r$, where $r$ is $\sqrt{2}$?