Mathematical Immaturity

14.7 Exercises

• A particle of mass 1 moves in a plane according to the equation $\textbf{r}(t) = x(t)\mathbf{i} + y(t)\mathbf{j}.$ It is attracted toward the origin by a force whose magnitude is four times its distance from the origin. At time $t = 0,$ the initial position is $r(0) = 4\mathbf{i}$ and the initial velocity is $v(0) = 6\mathbf{j}.$ (a) Determine the components $x(t)$ and $y(t)$ explicitly in terms of $t.$ (b) The path of the particle is a conic section. Find a Cartesian equation for this conic, sketch the conic, and indicate the direction of motion along the curve.
• (a) Since the particle is of unit mass, force can be simplified as $F = m\,\mathbf{a} = \mathbf{a}.$ If the particle is attracted towards the origin with magnitude four times its distance from the origin, then it must be the case that $$\mathbf{a}(t) = -4\,\mathbf{r}(t)$$ (b) Set $(x, y)$ as the horizontal and vertical components of the motion.
• (a) We know that $F = m\mathbf{a},$ and since $m = 1,$ we have $F = \mathbf{a}.$ And since we know that $F$ is towards the origin with four times the magnitude of the position vector, we have $$F(t) = \mathbf{a}(t) = -4 \mathbf{r}(t)$$ At time $t = 0,$ we have $\mathbf{r}(t) = 4\mathbf{i}$ and $\mathbf{v}(t) = 6\mathbf{j}.$ If we write write position parametrically $$\mathbf{r}(t) = x(t)\,\mathbf{i} + y(t)\,\mathbf{j}$$ We see that $$\displaylines{x(0) = 4,\quad y(0) = 0\\ x'(0) = 0,\quad y'(0) = 6\\ x''(0) = -16,\quad y''(0) = 0}$$ These values, and the condition $\mathbf{a}(t) = -4\mathbf{r}(t)$ can be satisfied by the components $$x(t) = 4\cos(2t), \quad y(t) = 3\sin(2t) \quad \blacksquare$$ (b) We wish to find values $a^2$ and $b^2$ such that: \begin{align*} \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \end{align*} Let $a = 4,$ $b = 3,$ then we have $$\frac{x^2}{16} + \frac{y^2}{9} = 1\quad \blacksquare$$