Curves by parameter
Tonight's idea
Until now a curve was \(y = f(x)\). Now let a point move, giving each coordinate as a function of time: \(x = x(t)\), \(y = y(t)\). The variable \(t\) is the parameter; the curve is the trail the point leaves. The unit circle is \((\cos t, \sin t)\).
A wheel of radius \(r\) rolls along the ground. Mark the rim point touching the ground at the start. After turning through angle \(t\) the wheel has rolled a distance \(rt\), so its centre is at \((rt, r)\). The marked point sits \(t\) back from the bottom, displaced from the centre by \((-r\sin t, -r\cos t)\). So
\(x = r(t - \sin t), \qquad y = r(1 - \cos t).\) This is the cycloid.
By the chain rule \(\dfrac{dy}{dt} = \dfrac{dy}{dx}\cdot\dfrac{dx}{dt}\), so
\(\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} = \dfrac{r\sin t}{r(1 - \cos t)} = \dfrac{\sin t}{1 - \cos t}.\)
At \(t = \pi\), the top, this is 0. As \(t \to 0\) the point is on the ground, barely moving, and the slope blows up: a cusp.
For the area under one arch, write \(dx = x'(t)\,dt\) and integrate \(t\) from 0 to \(2\pi\):
\[\int y\,dx = \int_0^{2\pi} r(1 - \cos t)\cdot r(1 - \cos t)\,dt = r^2\int_0^{2\pi} \big(1 - 2\cos t + \cos^2 t\big)\,dt = r^2(2\pi - 0 + \pi) = 3\pi r^2.\]
Three times the area of the wheel. Galileo tried to find this by weighing a metal cut-out; the exact answer came a generation later, by integration.
Watch one: the slope, simplified
The half-angle identities \(1 - \cos t = 2\sin^2(t/2)\) and \(\sin t = 2\sin(t/2)\cos(t/2)\) tidy the slope:
\[\frac{dy}{dx} = \frac{2\sin(t/2)\cos(t/2)}{2\sin^2(t/2)} = \cot\frac{t}{2}.\]
At \(t = 2\pi/3\), the wheel in the figure, the point is at \(x = r(2\pi/3 - \tfrac{\sqrt3}{2}) \approx 1.23r\), \(y = r(1 + \tfrac12) = 1.5r\), and the slope is \(\cot(\pi/3) = 1/\sqrt3\): the tangent climbs at \(30^\circ\). At \(t = \pi/2\) the slope is \(\cot(\pi/4) = 1\), a \(45^\circ\) tangent at the point \((r(\pi/2 - 1),\, r)\).
Your turn
- The circle \(x = 3\cos t\), \(y = 3\sin t\). Find \(dy/dx\) in terms of \(t\), and check it agrees with the implicit answer \(-x/y\) from last time. What is the slope at \(t = \pi/4\)?
- The ellipse \(x = a\cos t\), \(y = b\sin t\). Its upper half is traced as \(t\) runs from \(\pi\) down to 0. Use \(\int y\,dx\) to find the area of the upper half, and hence the area of the whole ellipse.
- A projectile: \(x = 20t\), \(y = 15t - 5t^2\) (metres, seconds). (a) Find \(dy/dx\). (b) Find the highest point. (c) Eliminate \(t\) to write \(y\) as a function of \(x\).
- The length of one arch. A small step along the curve has length \(ds = \sqrt{dx^2 + dy^2} = \sqrt{x'(t)^2 + y'(t)^2}\,dt\). Show \(x'^2 + y'^2 = 2r^2(1 - \cos t) = 4r^2\sin^2(t/2)\), then integrate from 0 to \(2\pi\).
Stretch: the tangent and the top of the wheel
Show that the tangent to the cycloid at parameter \(t\) passes through the top of the wheel, the point \((rt, 2r)\). Compute the slope from \(P\) to that point and compare with \(\cot(t/2)\). Where, then, does the normal at \(P\) pass?
Answers
- \(\dfrac{dy}{dx} = \dfrac{3\cos t}{-3\sin t} = -\cot t\). Since \(x/y = \cos t/\sin t = \cot t\), this is \(-x/y\). At \(t = \pi/4\) the slope is \(-1\).
- \(\displaystyle\int_{\pi}^{0} b\sin t\cdot(-a\sin t)\,dt = ab\int_0^{\pi}\sin^2 t\,dt = ab\cdot\dfrac{\pi}{2}\). The whole ellipse has area \(\pi ab\). With \(a = b = r\) this is the circle's \(\pi r^2\).
- (a) \(\dfrac{dy}{dx} = \dfrac{15 - 10t}{20} = \dfrac34 - \dfrac{t}{2}\). (b) Zero at \(t = 1.5\) s: the point \((30, 11.25)\). (c) \(t = x/20\), so \(y = \dfrac{3x}{4} - \dfrac{x^2}{80}\), a parabola, as it should be.
- \(x' = r(1 - \cos t)\), \(y' = r\sin t\), so \(x'^2 + y'^2 = r^2(1 - 2\cos t + \cos^2 t + \sin^2 t) = 2r^2(1 - \cos t) = 4r^2\sin^2(t/2)\). On \([0, 2\pi]\), \(\sin(t/2) \geq 0\), so \(ds = 2r\sin(t/2)\,dt\) and the length is \(\Big[-4r\cos(t/2)\Big]_0^{2\pi} = 4r + 4r = 8r\). Eight radii, with no \(\pi\) in sight.
Stretch. From \(P = (r(t - \sin t),\, r(1 - \cos t))\) to \((rt, 2r)\) the rise is \(2r - r(1 - \cos t) = r(1 + \cos t)\) and the run is \(rt - r(t - \sin t) = r\sin t\). The slope is \(\dfrac{1 + \cos t}{\sin t} = \dfrac{2\cos^2(t/2)}{2\sin(t/2)\cos(t/2)} = \cot\dfrac t2\), the same as the tangent's, so the tangent passes through the top of the wheel. The top, \(P\) and the contact point \((rt, 0)\) lie on a circle with the top-to-contact line as diameter, so the angle at \(P\) is a right angle: the normal at \(P\) passes through the contact point. The rim point is, for an instant, rotating about where the wheel touches the ground.