Parametric Equations and Curves — Question 3

Question 3 – Solution

Solution

1. Solve the horizontal equation for the parameter: $$x=6t \quad\Longrightarrow\quad t=\frac{x}{6}.$$

2. Substitute $t=x/6$ into the vertical equation: $$\begin{aligned} y&=9\left(\frac{x}{6}\right)-\frac92\left(\frac{x}{6}\right)^2\\ &=\frac32x-\frac18x^2. \end{aligned}$$ Thus the Cartesian path is the downward-opening parabola $$\boxed{y=\frac32x-\frac18x^2}.$$

3. The droplet lands when it returns to ground level, so set $y=0$: $$0=\frac32x-\frac18x^2 =\frac{x}{8}(12-x).$$ The solutions are $x=0$ and $x=12$.

4. The value $x=0$ is the launch point. The nonzero intercept is therefore the landing point $$\boxed{(12,0)}.$$ It occurs at $t=12/6=2$.

5. The Cartesian equation describes only the geometric path. The parameter $t$ also records time, so it identifies the droplet’s position and direction of motion at each instant.