A cricket analyst lays a grid over the ground, in metres, with its origin at the middle stump. A fielder starts at the point \((10, 5)\). He chases the ball along a curve and gathers it at the point \((22, 14)\). The analyst wants his displacement, the straight-line distance between the two points, and she has only the four numbers.
The distance between two points
Mark the two points \(P_1(x_1, y_1)\) and \(P_2(x_2, y_2)\) on a grid with equal scales on the two axes. Go from \(P_1\) straight across by \(x_2 - x_1\), and then straight up by \(y_2 - y_1\). You reach \(P_2\), and the two steps make the two shorter sides of a right triangle. The distance \(P_1P_2\) is the longest side. So Pythagoras' theorem gives the distance.
\[ \Delta x = x_2 - x_1, \qquad \Delta y = y_2 - y_1 \]
Why. the step across and the step up from \(P_1\) to \(P_2\); a sign only says the direction of each step.
\[ d^2 = (\Delta x)^2 + (\Delta y)^2 \]
Why. the two steps are at right angles, because the axes are, so Pythagoras' theorem applies.
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Read it. take the positive root, since a distance is never negative. Squaring removes the signs, so the order of the points does not matter.
In space a point needs three coordinates, \((x, y, z)\). The distance gains a third term, \((\Delta z)^2\), under the root. For example, the points \((1, 0, 2)\) and \((3, 3, 8)\) are \(\sqrt{4 + 9 + 36} = 7\) units apart. The chapter Gravitational Force and Field uses exactly this distance for a force between two masses.
The distance from an axis
A spinning body needs a different distance: the distance of a point from the axis of spin, measured at right angles to it. For the \(z\) axis, that distance is \(r_z = \sqrt{x^2 + y^2}\). The coordinate \(z\) does not enter, because moving along the axis does not change the distance from it. In the same way, \(r_x = \sqrt{y^2 + z^2}\).
What is the distance between the points \((1, 2)\) and \((4, 6)\), both in metres?
A \(0.5\,\text{kg}\) bead sits at the point \((6\,\text{m}, 8\,\text{m}, 3\,\text{m})\). The body turns about the \(z\) axis, and the moment of inertia of the bead is \(mr^2\), with \(r\) its distance from that axis. Find the moment of inertia, in kg m².
The point that divides a line
Take the line from \(A(x_1, y_1)\) to \(B(x_2, y_2)\). A point P on it divides it in the ratio \(m : n\) when \(AP : PB = m : n\). Then P lies a fraction \(m/(m + n)\) of the way from A to B. So its \(x\) coordinate is \(x_1 + \tfrac{m}{m + n}(x_2 - x_1)\), which simplifies to \((nx_1 + mx_2)/(m + n)\). The \(y\) coordinate follows the same rule. This is the section formula.
Notice that each end is weighted by the part of the ratio next to the other end. With \(m = n\) the point is the midpoint, \(\left(\tfrac{x_1 + x_2}{2}, \tfrac{y_1 + y_2}{2}\right)\). This formula is the centre of mass in disguise. For masses \(m_1\) at A and \(m_2\) at B, the centre of mass is \((m_1x_1 + m_2x_2)/(m_1 + m_2)\). That is the section formula with the ratio \(AP : PB = m_2 : m_1\).
A \(1\,\text{kg}\) ball sits at \((0, 0)\) and a \(3\,\text{kg}\) ball at \((4\,\text{m}, 0)\), on a light rod. Where is their centre of mass?
The circle
A circle is the set of all points at one distance \(r\), the radius, from one point, the centre. Let the centre be \((a, b)\). A point \((x, y)\) lies on the circle when its distance from the centre is \(r\). Write that with the distance formula and square both sides. The result is the equation of the circle, \((x - a)^2 + (y - b)^2 = r^2\).
An equation such as \(x^2 + y^2 - 6x = 0\) is also a circle. To see its centre, complete the square in \(x\): \(x^2 - 6x = (x - 3)^2 - 9\). The equation becomes \((x - 3)^2 + y^2 = 9\). So the centre is \((3, 0)\) and the radius is \(3\).
Perpendicular lines and the tangent to a circle
Take a line of slope \(m\): a run of \(1\) with a rise of \(m\). Turn the little slope triangle through a right angle. The run of \(1\) becomes a rise of \(1\), and the rise of \(m\) becomes a run of \(-m\). So the perpendicular line has slope \(1/(-m) = -1/m\). Two lines are at right angles when their slopes multiply to \(-1\).
The tangent to a circle at a point P touches the circle only at P, and it is perpendicular to the radius drawn to P. So its slope is \(-1/m\), where \(m\) is the slope of the radius. The line through P with that slope is the tangent, in the point-slope form \(y - y_1 = m_t(x - x_1)\). For a curve that is not a circle, the slope of the tangent comes from the derivative, which the chapter Differentiation teaches.
What is the slope of the tangent to the circle \(x^2 + y^2 = 25\) at the point \((3, 4)\)?
What are the centre and the radius of the circle \(x^2 + y^2 - 6x = 0\)?
Back to physics: the fielder's displacement
The fielder's run along the curve is long, but his displacement is a straight line from start to finish. The distance formula gives its size from the four coordinates alone, whatever path he ran.
A fielder starts at the point \((10, 5)\,\text{m}\) and, after running along a curve to collect the ball, ends at \((22, 14)\,\text{m}\). Find the displacement vector and its magnitude.
The steps are \(\Delta x = 22 - 10 = 12\,\text{m}\) and \(\Delta y = 14 - 5 = 9\,\text{m}\), so the displacement is \(12\hat i + 9\hat j\) metres.
The distance formula gives its size: \(\sqrt{12^2 + 9^2} = \sqrt{144 + 81} = \sqrt{225} = 15\,\text{m}\).
The fielder may have run \(20\,\text{m}\) or \(30\,\text{m}\) along the curve; that length does not affect the displacement. The midpoint of the displacement is \((16, 9.5)\,\text{m}\).
Two speakers sit at \((0, 1.5)\) and \((0, -1.5)\), in metres. A listener stands at \((8.0, 0.82)\). Find the difference of the two path lengths, \(L_2 - L_1\), from the lower speaker and the upper speaker, in metres.
The distance formula needs perpendicular axes that measure the same quantity in the same unit. On a v-t graph a "distance" between two points mixes seconds with metres per second and means nothing. Perpendicular slopes and circles also need equal scales on the two axes. With unequal scales a circle looks like an oval, and a right angle no longer looks square.
Motion in Two Dimensions writes positions as coordinates and finds displacements with the distance formula. Rigid Bodies and Moment of Inertia measures each mass from the axis by \(\sqrt{x^2 + y^2}\), and the centre of mass is the section formula. Interference and Reflection of Sound finds the two path lengths from two speakers to a listener. The mirror's sign convention puts every distance on one axis. The velocity against position of a body in SHM traces an ellipse, and interference fringes of two point sources are hyperbolas.
- Adding the steps instead of squaring them.The distance is \(\sqrt{(\Delta x)^2 + (\Delta y)^2}\), not \(\Delta x + \Delta y\). Steps of 3 and 4 give 5, not 7.
- Using the distance from the origin for the distance from an axis.From the \(z\) axis the distance is \(\sqrt{x^2 + y^2}\). The coordinate along the axis never enters.
- Weighting each end of the section formula by its own part of the ratio.For \(AP : PB = m : n\), A is weighted by \(n\) and B by \(m\). The point is nearer the end with the larger weight.