Things you might have noticed

Particle \(P\) is moving anti-clockwise round a circle of radius \(\quantity{1}{m}\) at speed \(\quantity{1}{m\,s^{-1}}\).

The particle is at point \(A\) when \(t=0\) and its position at time \(\quantity{t}{s}\) is shown in the diagram below.

Unit circle with point P(cos t, sin t) marked. Tangent to circle at P drawn and labelled v.

Combine some of the following diagrams to explain why the velocity vector \(\mathbf{v}\) is \((-\sin t, \cos t)\). You may be able to do this in more than one way.

  • How have you used the fact that \(P\) is moving at unit speed on a unit circle?

We’ll use the diagrams with the prompts from the suggestion section.

Why is \(P\) in this position at time \(\quantity{t}{s}\)?

What is the magnitude and direction of \(\mathbf{v}\)?

\(P\) is moving with speed \(\quantity{1}{m\,s^{-1}}\) round a circle of radius \(\quantity{1}{m}\). Therefore, after time \(\quantity{t}{s}\), it has travelled a distance of \(\quantity{t}{m}\) along a circular arc of radius \(\quantity{1}{m}\). This arc subtends an angle \(\quantity{t}{rads}\) at the centre of the circle, and hence the coordinates of \(P\) at time \(\quantity{t}{s}\).

At any time \(\quantity{t}{s}\) the velocity of \(P\) has magnitude \(\quantity{1}{m\,s^{-1}}\) and is directed along the tangent to the circle at the point \((\cos t, \sin t)\).

There are a couple of ways to approach the rest of the problem. We can think about triangles in the plane, or we can think about transforming vectors.

How are the two triangles related?

Why is \(\mathbf{v}\) given by the vector shown?

Both triangles are right-angled, with one angle \(t\) and hypotenuse \(1\). Therefore they are congruent.

The hypotenuses of the two triangles are perpendicular, so by considering the angles around \(P\) we can see that the sides of the red triangle are parallel to the axes.

The length of the horizontal side of the red triangle is \(\sin t\) and the length of the vertical side is \(\cos t\). It follows that \(\mathbf{v}\) can be expressed as \((-\sin t, \cos t)\), as shown.

  • Why is the horizontal component of \(\mathbf{v}\) given by \(-\sin t\) rather than \(\sin t\)?

  • What would happen if \(P\) were in a different quadrant?

How are vectors \(\mathbf{v}\) and \(\overrightarrow{OQ}\) related?

Why are these the coordinates of \(Q\)?

Why is \(\mathbf{v}\) given by the vector shown?

As \(\angle POQ\) is a right-angle, \(\overrightarrow{OQ}\) is parallel to \(\mathbf{v}\) and they have the same direction. Furthermore, \(\overrightarrow{OQ}\) has length \(\quantity{1}{m}\) and \(\mathbf{v}\) has magnitude \(\quantity{1}{m\,s^{-1}}\), so geometrically \(\overrightarrow{OQ}=\mathbf{v}\). Therefore we can think of \(\mathbf{v}\) as a translation of \(\overrightarrow{OQ}\). We know how to find the coordinates of \(Q\), because it is a point on the circle, so this will help us to express \(\mathbf{v}\) in terms of \(t\).

\(Q\) has coordinates \((\cos (t+\tfrac{\pi}{2}), \sin(t+\tfrac{\pi}{2}))\) because \(OQ\) is a rotation of \(OP\) anticlockwise through \(\quantity{\tfrac{\pi}{2}}{rad}\). Therefore we can write \(\mathbf{v}= (\cos (t+\tfrac{\pi}{2}), \sin(t+\tfrac{\pi}{2}))\).

What does the graph of \(\cos (t+\tfrac{\pi}{2})\) look like? What about the graph of \(\sin(t+\tfrac{\pi}{2})\)?

We chose to separate thinking about triangles from thinking about transforming vectors. You may have used a combination of these approaches. What do the approaches have in common? What different things do they tell you?

  • What does this tell you about the derivatives of \(\cos t\) and \(\sin t\)?
In the problem we were told that

Velocity is the rate of change of displacement, so its components are given by the rates at which its \(x\) and \(y\) coordinates are changing with respect to time. Therefore \[\mathbf{v}=\left( \dfrac{dx}{dt}, \dfrac{dy}{dt}\right).\]

The diagrams helped us to establish that at time \(\quantity{t}{s}\), \(x=\cos t\) and \(y=\sin t\) and therefore \(\mathbf{v}=\left( \dfrac{d}{dt}(\cos t), \dfrac{d}{dt}(\sin t)\right)\).

From the diagrams we also saw that \(\mathbf{v}\) can be written as \((-\sin t, \cos t)\). By equating the components in the two expressions for \(\mathbf{v},\) we obtain

\[\frac{d}{dt}(\cos t)= -\sin t \quad \text{and}\quad \frac{d}{dt}(\sin t)=\cos t.\]

But from the diagrams we also saw that \(\mathbf{v}\) could be written as \((\cos (t+\tfrac{\pi}{2}), \sin(t+\tfrac{\pi}{2}))\), which gives a way of thinking about the derivatives of \(\cos t\) and \(\sin t\) as rotations through \(\tfrac{\pi}{2}\),

\[\frac{d}{dt}(\cos t)= \cos \left(t+\frac{\pi}{2}\right) \quad \text{and}\quad \frac{d}{dt}(\sin t)= \sin\left(t+\frac{\pi}{2}\right).\]

What do these ideas tell you about the acceleration of \(P\) at time \(\quantity{t}{s}\)?