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A parametric curve read as a motion: velocity and acceleration as vectors, speed as the length $\sqrt{x'^2 + y'^2}$ of the velocity, and the circular motion whose speed never changes while its velocity never stops changing.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you can find the velocity and acceleration vectors of a moving point by differentiating its components, and the speed as the length of the velocity, combining the components by Pythagoras rather than by addition. You can say why a particle moving round a circle at a steady rate has constant speed and a velocity that changes at every instant, and what that forces the acceleration to be.
You can differentiate a parametric curve, and in calculus AB you read motion on a line: position, velocity, acceleration, and distance as speed accumulated. All of that survives in two dimensions. The one thing that is genuinely new is that velocity now has a direction, and so velocity and speed stop being the same word.
Position vector: $\langle x(t), y(t) \rangle$, where the particle is.
Velocity vector: $\langle x'(t), y'(t) \rangle$ — its derivative, with a direction.
Speed: $\sqrt{x'(t)^2 + y'(t)^2}$, the length of the velocity, a single non-negative number.
Acceleration vector: $\langle x''(t), y''(t) \rangle$, the derivative of the velocity.
Displacement: the straight-line change in position; not the distance travelled.
Read a parametric curve as a motion. The position is the vector $\langle x(t), y(t)\rangle$; differentiating each component gives the velocity
$$\mathbf{v}(t) = \langle x'(t),\, y'(t) \rangle$$
and differentiating again gives the acceleration. Differentiation is done component by component, with no new rule to learn.
Speed is the length of the velocity. $\|\mathbf{v}\| = \sqrt{x'^2 + y'^2}$: one non-negative number, with no direction in it. The components are perpendicular, so they combine by Pythagoras — never by addition. Velocity $\langle 3, 4\rangle$ is speed $5$, not $7$.
Constant speed does not mean constant velocity. A particle going round a circle at a steady rate has a velocity whose length never changes and whose direction changes every instant. Its acceleration is therefore not zero, and it points at the centre. This is the sentence to hold on to, because it is where the two words stop being interchangeable and start meaning different things.
The velocity points along the curve. It is tangent to the path at every instant, which is why its two components are the same two derivatives that gave the slope in the last lesson: $\dfrac{dy}{dx} = \dfrac{y'(t)}{x'(t)}$ is the direction of $\mathbf{v}$, written as a gradient.
Another way: picture
Picture a car on a roundabout with the speedometer pinned at thirty. The speedometer reads the speed — one number, never changing. A compass on the dashboard reads the direction, and it swings the whole way round. The velocity is both readings together, and it is changing continuously even though one of the two readings never moves. What you feel pushing you sideways in the seat is the acceleration that that change requires.
Another way: steps
| Quantity | What it is | Kind |
|---|---|---|
| Position | $\langle x(t), y(t) \rangle$ | vector |
| Velocity | $\langle x'(t), y'(t) \rangle$ | vector |
| Speed | $\sqrt{x'^2 + y'^2}$ | number, never negative |
| Acceleration | $\langle x''(t), y''(t) \rangle$ | vector |
Three of the four are vectors and one is a number. An answer of the wrong kind — a pair of numbers where a speed was asked for, or a single number offered as a velocity — is wrong before its arithmetic is even checked, and noticing that is the cheapest error check in the topic.
In calculus AB a particle moved along a line, and velocity was a single signed number. Speed was its absolute value, and the only thing the sign carried was which of two directions the particle was going.
In the plane there are infinitely many directions, so the sign has to grow into a whole vector. Everything else is the same sentence:
| On a line | In the plane |
|---|---|
| velocity $v(t)$, a signed number | $\langle x'(t), y'(t) \rangle$, a vector |
| speed $\lvert v(t) \rvert$ | $\sqrt{x'^2 + y'^2}$ |
| distance $\int \lvert v \rvert\,dt$ | $\int \sqrt{x'^2 + y'^2}\,dt$ |
The absolute value has become a square root, because that is what the length of a vector is; in one dimension the two are the same thing. Nothing else in the story of position, velocity and acceleration had to be rewritten.
Adding the components to get the speed. They are perpendicular. $\langle 3, 4 \rangle$ has length $5$; the $7$ is the distance you would go if you travelled east and then north instead of diagonally.
Treating speed and velocity as the same word. A particle can have constant speed and continuously changing velocity, and that is what circular motion is. Velocity carries a direction; speed does not.
Squaring the positions instead of the derivatives. $\sqrt{x^2 + y^2}$ is the distance from the origin, which is a perfectly real quantity and is not the speed. Differentiate first, then square.
$x = t^2$, $y = 2t$: velocity $\langle 2t, 2 \rangle$ by differentiating each component.
Component by component.
Speed $= \sqrt{4t^2 + 4} = 2\sqrt{t^2 + 1}$, which grows with $t$.
Length of the velocity.
Acceleration $\langle 2, 0 \rangle$ — constant, and pointing purely along the $x$-direction even though the path is curved.
$x = 5\cos t$, $y = 5\sin t$: velocity $\langle -5\sin t, 5\cos t \rangle$.
A circle of radius five.
Speed $= \sqrt{25\sin^2 t + 25\cos^2 t} = 5$, the same at every instant.
The identity does the work.
But the velocity is different at every instant, so the acceleration $\langle -5\cos t, -5\sin t \rangle$ is not zero — it points at the centre.
Differentiate each component: the velocity is $\langle 3, 4 \rangle$, the same at every instant.
Constant here.
Speed is its length: $\sqrt{3^2 + 4^2} = \sqrt{25}$.
Square, add, root.
So the speed is $5$, not $7$ — and because the velocity itself is constant, the acceleration is the zero vector and the path is straight.
A particle moves with $x = 4t$ and $y = t^2$. Plot its position at $t = 1$, $t = 2$ and $t = 3$.
Plot your answer on the grid:
A particle moves with $\dfrac{dx}{dt} = 9$ and $\dfrac{dy}{dt} = 12$. What is its speed?
Answer:
A particle moves with $x = 6t$ and $y = 8t$. Fill in its position and its speed at each time.
| $x$ | $y$ | speed | |
|---|---|---|---|
| $t = 1$ | |||
| $t = 2$ | |||
| $t = 3$ |
A particle's velocity at an instant is $\langle 3, 4 \rangle$. What is its speed at that instant?
Put the steps of finding the speed of a particle with $x = 4t^2$ and $y = t^3$ into the order you do them.
Number the steps in order (write the number in the box):
A particle moves with $x = 6\cos t$ and $y = 6\sin t$, once round a circle. Select every statement that is true of it.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A particle moves with $x = t$ and $y = t^2$. Plot its position at $t = 1$, $t = 2$ and $t = 3$.
Plot your answer on the grid:
You can tell velocity from speed and compute both. Without looking: what is the speed of a particle whose velocity is the vector with components three and four, and why is the answer not seven?
10. Your turn: the speed of $x = 3t$, $y = 4t$ at any time, step 3