Many curves we integrate come in the form \(y = f (x)\). For most curves, this is fine, but it is not always possible or convenient to write it like this. It is in this scenario where parametric coordinates are useful.
Recap of parametric coordinates
In this scenario, let us introduce a 'dummy' variable, usually denoted as t. We call this a dummy variable as it is an abstract concept that assigns a value to either an x or y coordinate, and isn't plotted.
This means that instead of having a function of the form \( y = f (x)\), we represent a curve by \(y(t) = g(t)\), \(x (t) = h (t)\), where h and t are functions that describe the change of the x and y coordinates respectively.
A curve is described by \(y (t) = 2 (t)\), \(x (t) = 2 (t)\), \(0 < t < 2\pi\).
Expressing the parametric curve as \( (x (t))^2 + (y (t))^2 = (2\cos{t})^2 + (2\sin{t})^2 = 4\) \(\cos^2{t} + \sin^2{t} = 4\), we see that it actually describes a circle of radius 4, or \(x^2 + y^2 = 4\).
Why does parametric integration work?
Normally, we expect to evaluate an integral of the form \( \int y (x) dx\); however, we need to change this because our curve is not in the form \(y (x)\). We use a modified version of the Chain Rule. We can replace dx with \(\frac{dx}{dt}dt\) (you can think of this as the dt's cancelling. While this is not technically how they work, as \(\frac{dx}{dt}\) is not strictly a fraction, we can treat it as one for operational purposes). This gives an integral of the form \(\int{y(t)\frac{dx(t)}{dt}dt}\).
We must also remember to do with parametric integrals is switch limits. Suppose we have an integral of the form \(\int^b_a{f(x)dx}\). We must also switch the limits, which results in the integral being given as\(\int^d_c{f(t)\frac{dx}{dt}dt}\), where \(c = x^{-1}(a)\) and \(d = x^{-1}(b)\).
Examples of parametric integration
At first glance, this can be a tricky topic to get your head around, so let's walk through a couple of examples to try and consolidate what we have said so far.
A curve is defined parametrically with \( x(t) = 2 -t\) and \(y(t) = e^t - 1\). Find the area enclosed by the x-axis, the line x = 0 and the curve.
First thing to do to work out where the curve crosses de x-axis and where the line x = 0 crosses the curve.
If the line crosses de x-axis, then the y-value will be zero. Solving this, we have \(e^t -1 = 0\) which implies \(e^t = 1\) and in turn t = 0. When x = 0, then \(2 - t = 0\) which implies t = 2.
This means that we now have our limits, and we can start the integral. We have:
\[\int^0_2{(e^t -1)} \cdot \frac{d}{dt}(2 - t) \cdot dt = - \int^0_2{(e^t - 1)} dt = \int^0_2{(e^t - 1) dt}\]
where we switch the limits to change the sign.
This then equals \([e^t - t]_{t=0}^{t=2} = [(e^2 - 2) - (1-0)] = e^2 - 3\).
Using parametric integration, find the area of the circle defined as \(x(t) = -3\cos(t), y(t) = 3\sin(t), 0 < t < 2\pi\).
By the formula for parametric integration, we have:
\[\int^{2\pi}_0 {3\sin(t) \cdot \frac {d}{dt} (-3 \cos (t))dt} = 9 \int^{2\pi}_0 \sin^2(t)dt\].
We now need to use a double angle formula here, and we can use the result \(2(t) = \frac{1}{2}(1 - \cos(2t))\).
Filling this in, we get \(\frac{9}{2} \int^{2\pi}_0{(1-\cos(2t))dt} = \frac{9}{2}[t - \frac{1}{2} \sin(2t)]^{t = 2\pi}_{t = 0} = 9\pi\), which is what we would expect of a circle with a radius of 3.
Exam style question
Suppose we have a curve that has been defined parametrically, with \(x(t) = 3\cos(4t)\) and \(y(t) = 6 \sin(8t)\), with \(0 < t < \frac{\pi}{8}\).
i) Find any turning points of the curve.
ii) Find the area under the curve.
i) For a turning point, then \(\frac{dy}{dx}\) must be equal. By the chain rule,
\[\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{dy}{dt}(\frac{dx}{dt})^{-1}\]
We can now use the standard formulae for derivatives of trigonometric functions to find these results.
\(\frac{dy}{dt} = 48 \cos(8t)\) and \(\frac{dx}{dt} = -12 \sin(4t)\), which in turn will give \(\frac{dy}{dx} = \frac{48 \cos(8t)}{-12 \sin(4t)}.
We can then solve this equal to zero to find the t value of the turning point. For this to equal zero, then the numerator must equal zero, which implies that \(\cos(8t) = 0\).
This means that \(8t = \frac{\pi}{2} + n\pi, n \epsilon N\), which we further reduce to \(t = \frac{\pi}{16} + \frac{n\pi}{8}, n \epsilon N\).
The only value of t here which satisfies the \(0 < t < 8\pi\) is \(t = \frac{\pi}{16}\).
This gives the x coordinate of the turning point as \(3\cos(\frac{\pi}{4}) = \frac{3\sqrt{2}}{2}\), and the y coordinate as \(6\sin(\frac{\pi}{2}) = 6\).
First, let's work out. the direction of our limits. \(x(0) = 3\) and \(x(\frac{\pi}{8}) = 0\), which means the area under the curve is given by
\[\int^0_{\frac{\pi}{8}}{y(t) \cdot \frac{dx}{dt}\cdot dt} = \int^0_{\frac{\pi}{8}}{6 \sin(8t)(-12\sin(4t))dt} = 72\int^{\frac{\pi}{8}}_0{\sin(8t) \sin(4t)dt}\]
where we "flip the limits" to get rid of the negative sign. We can use a double angle formula to help us solve this integral.
We know \(\sin(2x) = 2\sin(x)\cos(x)\). This implies that \(\sin(8x) = 2 \cdot \sin(4x) \cos(4x)\).
Filling this in, we get the integral \(144 \int^{\frac{\pi}{8}}_0 \sin^2(4t)\cos(4t)dt\).
Since \(\int \sin(t) = \cos(t)\), it seems intuitive that this is best suited to an integration by substitution.
Let us take \(u = \sin(4t)\) which implies \(\frac{du}{dt} = 4\cos(4t)\), so \(dt = \frac{du}{4\cos(4t)}\).
As this is a definite integral, we also need to change the limits.
\(u_1 = \sin(4 \cdot 0) = 0\) and \(u_2 = \sin(4 \cdot \frac{\pi}{8}) = 1\).
Filling this in, we can say that \(144 \int_0^{\frac{\pi}{8}} \sin^2(4t) \cos(4t) dt = 36 \int^1_0 u^2 du\).
This is a straightforward integral and can be evaluated directly.
\(36 \int^1_0 u^2 du = 12[u^3]^{u = 1}_{u = 0} = 12[1-0] = 12\).
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