8. Motion in a Plane

Video coming soon3 worked examples

Theory

Parametric equations describe motion naturally when the parameter is time. If a point $P$ moves in the $x$ – $y$ plane with position given by $x = f(t)$ and $y = g(t)$ , then each derivative is a velocity in one direction:

\frac{dx}{dt} = \text{velocity in the } x \text{ direction}, \qquad \frac{dy}{dt} = \text{velocity in the } y \text{ direction}

These are the two components of the velocity. The speed is the magnitude of that velocity, found by Pythagoras:

|v| = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}

This gives the instantaneous speed of the particle at time $t$ . Note that speed is always positive, even when one or both components are negative.

The Golden Rule: most questions do not give you $t$ directly — they describe an event (“when it hits the ground”, “when it reaches the target”). Translate that event into an equation, solve it for $t$ first, and only then substitute into the derivatives.

⚠️ Common Examiner Traps

Rejecting the wrong root: “hits the ground” usually gives $t = 0$ as well as the answer — but $t=0$ is the moment of launch, so discard it.
Squaring away the sign: a downward velocity is negative, but $(-20)^2 = 400$ . Keep the sign while differentiating; it only disappears at the squaring stage.
Speed is not $\frac{dy}{dx}$ : the gradient tells you the direction of travel; the speed needs both $t$ -derivatives combined.
Units: if $x$ and $y$ are in metres and $t$ in seconds, the speed is in m/s — state it.

Worked examples

Example 1

The position of a ball $t$ seconds after being struck is given by $x = 12t$ , $y = 20t - 5t^2$ , with distances in metres. Find the speed of the ball when it first hits the ground.

Step 1: The ball is on the ground when $y = 0$ :

20t - 5t^2 = 0 \implies 5t(4 - t) = 0 \implies t = 0 \ \text{ or } \ t = 4

Step 2: $t = 0$ is the moment it was struck, so the ball lands at $t = 4$ .

Step 3: Differentiate each coordinate with respect to $t$ :

\frac{dx}{dt} = 12, \qquad \frac{dy}{dt} = 20 - 10t

Step 4: Substitute $t = 4$ . The vertical component is negative because the ball is falling:

\frac{dx}{dt} = 12, \qquad \frac{dy}{dt} = 20 - 40 = -20

Step 5: Combine using Pythagoras:

|v| = \sqrt{12^2 + (-20)^2} = \sqrt{144 + 400} = \sqrt{544} = 4\sqrt{34} \approx 23.3\ \text{m/s}

Example 2

At time $t$ , the position of a moving point is given by $x = e^{2t}$ , $y = e^{-t}$ . Find its speed when $t = \ln 2$ .

Step 1: Differentiate each coordinate, using the chain rule on both:

\frac{dx}{dt} = 2e^{2t}, \qquad \frac{dy}{dt} = -e^{-t}

Step 2: Evaluate at $t = \ln 2$ , using $e^{2\ln 2} = e^{\ln 4} = 4$ and $e^{-\ln 2} = \frac{1}{2}$ :

\frac{dx}{dt} = 2 \times 4 = 8, \qquad \frac{dy}{dt} = -\frac{1}{2}

Step 3: Combine the components:

|v| = \sqrt{8^2 + \left(-\tfrac{1}{2}\right)^2} = \sqrt{64 + \tfrac{1}{4}} = \sqrt{\tfrac{257}{4}} = \frac{\sqrt{257}}{2} \approx 8.02

Example 3

A particle moves so that its position at time $t$ is $x = 5\cos t$ , $y = 2\sin t$ . Show that its speed is $\sqrt{4 + 21\sin^2 t}$ , and hence find its maximum and minimum speeds and the positions at which they occur.

Step 1: Differentiate each coordinate:

\frac{dx}{dt} = -5\sin t, \qquad \frac{dy}{dt} = 2\cos t

Step 2: Form the sum of the squares:

|v|^2 = 25\sin^2 t + 4\cos^2 t

Step 3: Write everything in terms of $\sin^2 t$ using $\cos^2 t = 1 - \sin^2 t$ :

|v|^2 = 25\sin^2 t + 4(1 - \sin^2 t) = 4 + 21\sin^2 t \implies |v| = \sqrt{4 + 21\sin^2 t}

Step 4: Since $0 \le \sin^2 t \le 1$ , the expression under the root runs from $4$ to $25$ :

|v|_{\min} = \sqrt{4} = 2, \qquad |v|_{\max} = \sqrt{25} = 5

Step 5: Identify the positions. The minimum occurs when $\sin t = 0$ , so $\cos t = \pm 1$ and the particle is at $(5, 0)$ or $(-5, 0)$ . The maximum occurs when $\sin t = \pm 1$ , so $\cos t = 0$ and the particle is at $(0, 2)$ or $(0, -2)$ .

The path is an ellipse, and the particle travels fastest at the ends of the minor axis and slowest at the ends of the major axis.