- Calculus, Volume 1: One Variable Calculus, with an Introduction to Linear Algebra
- Tom M. Apostol
- Second Edition
- 1967
- 978-1-119-49673-1
14.9 Exercises
-
A particle moves along the ellipse $3x^2 + y^2 = 1$ with position vector
$$
\begin{align*}
\mathbf{r}(t) = f(t)\mathbf{i} + g(t)\mathbf{j}
\end{align*}
$$
The motion is such that the horizontal component of the velocity vector at time $t$ is $-g(t).$
(a) Does the particle move around the ellipse in a clockwise or counterclockwise direction?
(b) Prove that the vertical component of the velocity vector at time $t$ is proportional to $f(t)$ and find the factor of proportionality.
(c) How much time is required for the particle to go once around the ellipse?
-
(a) Horizontal velocity is negative when vertical position is positive. What does this mean about the direction of the motion?
(b) Differentiate the ellipse equation with respect to the parameter $t.$
(c) We can express the components of an ellipse as $x = a\cos \omega t$ and $y = b\sin \omega t$ for some real $a,$ $b.$ What property of the sine and cosine can we use to deduce the time taken to complete a revolution?
-
(a) If horizontal component of velocity is the negative of the vertical component of position, then the particle moves left when $y > 0$ and right when $y < 0,$ making the motion counter-clockwise.
\begin{align*}
\end{align*}
(b) Taking the derivative of the ellipse equation with respect to $t:$
$$
\begin{align*}
\frac{d}{dt}\left(3x^2 + y^2\right) &= 6x\frac{dx}{dt} + 2y\frac{dy}{dt} \\
&= 0
\end{align*}
$$
Substituting the component functions $x = f(t)$ and $y = g(t),$ and recalling that $\frac{dx}{dt} = -g(t),$ the above equation becomes:
$$
\begin{align*}
-6f(t)g(t) + 2g(t)\frac{dy}{dt} &= 0
\end{align*}
$$
Or in other words,
$$
\frac{dy}{dt} = 3f(t) \quad \blacksquare
$$
\begin{align*}
\end{align*}
(c) We know that the parametric component functions for $x$ and $y$ are such that $f'(t) = -g(t),$ $g'(t) = 3f(t),$ and that $3f(t)^2 + g(t)^2 = 1.$ These conditions are satisfied by:
$$
\begin{align*}
f(t) &= \frac{\sqrt{3}}{3}\cos \sqrt{3}\,t \\
g(t) &= \sin \sqrt{3}\,t
\end{align*}
$$
And to complete a full revolution, $\sqrt{3}\,t$ must be $2\pi,$ or $t = \frac{2\pi}{\sqrt{3}}. \quad \blacksquare$
