Parametric Equations and Curves — Question 1

Question 1 – Solution

Solution

1. Square both parametric equations: $$x^2=\left(t+\frac1t\right)^2=t^2+2+\frac1{t^2}, \qquad y^2=\left(t-\frac1t\right)^2=t^2-2+\frac1{t^2}.$$

2. Subtract the second equation from the first: $$x^2-y^2=4.$$ Thus the Cartesian curve is the hyperbola $$\boxed{\frac{x^2}{4}-\frac{y^2}{4}=1}.$$

3. Because $t>0$, the arithmetic–geometric mean inequality gives $$x=t+\frac1t\ge 2.$$ Therefore, the parametrization reaches only the right-hand branch of the hyperbola.

4. To determine the direction of travel, note that $$\frac{dx}{dt}=1-\frac1{t^2}, \qquad \frac{dy}{dt}=1+\frac1{t^2}>0.$$ As $t$ increases from $0$ to $1$, the point moves from the lower right toward $(2,0)$. As $t$ increases beyond $1$, it moves from $(2,0)$ toward the upper right.

5. Hence the mechanism traces the entire right branch exactly once, moving upward as $t$ increases.