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## 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.

\(36 \int^1_0 u^2 du = 12[u^3]^{u = 1}_{u = 0} = 12[1-0] = 12\).

## Parametric Integration - Key takeaways

The formula for parametric integration is given as \(\int{y(t)\frac{dx(t)}{dt}dt}\).

We must remember to switch limits when changing from x coordinates to t coordinates.

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##### Frequently Asked Questions about Parametric Integration

How do you integrate parametric equations?

Integrate parametrically using the formula ∫y(t).dx(t/)dt.dt

How do you find the limits of integration for parametric equations?

Use the inverse formula for x(t) to find the t limits from the x limits.

Why does parametric integration work?

Parametric integration works as we introduce a dummy variable which essentially shifts our curve to a different plane where it is easier to integrate.

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