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## A Calculus Primer Part VI- Parametric and Polar Equations

### #1 macosxnerd101

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Posted 21 May 2011 - 08:31 PM

This tutorial will introduce Calculus in the parametric and polar coordinate systems.

Parametric Equations
Parametric equations are used to break equations down into their components based on a parameter. Time is a very common example of a parameter. Another name for a set of parametric equations is a vector.

Some common ways of representing parametric equations include:
```x = 3t
y = 6t^2

//Or
a(t) = <x(t), y(t)>

```

So how is a vector or set of parametric equations differentiated? Quite easily- simply differentiate the component equations with respect to the parameter.
```da/dt = <dx/dt, dy/dt>

```

When dealing with cartesian functions, derivatives provide the rate of change for a function at a point. While differentiated parametric equations provide information on the rates of change for the various components, it does not provide the total rate of change at a point. When dealing with cartesian functions, the notation is dy/dx. Since the derivative of the vector form provides dy/dt and dx/dt, dy/dx can be obtained simply by manipulating notation. In other words, treat dy/dt and dx/dt as fractions, letting the dt terms cancel:
```dy/dx = (dy/dt)/(dx/dt)

```

Given the differentiated parametric equations, it is possible to find the speed instead of the velocity by finding the magnitude of the differentiated vector. Simply apply the parametric equation:
```speed = sqrt[(dx/dt)^2 + (dy/dt)^2]

```

Then by integrating the speed based on the parameter, the arc length of the curve can be determined:
```length = Integral(a, b, speed dt)

```

This is equivalent to the cartesian form for arc length:
```length = Integral(a, b, sqrt[1 + (dy/dx)^2] dx)

```

Polar Coordinates
Polar coordinates are used to represent functions in terms of distances with respect to angles rather. Think of it as an expanded unit circle. Remember that the distance is a vector magnitude, equivalent to sqrt(x^2 + y^2). Since the unit circle can be found in the polar coordinate system at r = 1, x and y can be described as x = r cos(theta) and y = r sin(theta). As shown, polar equations are quite easy to parameterize. Given these parametric equations, the arc length of a polar function is found the same way as shown in the last section.

The big application with polar coordinates and calculus is finding the area under the curve. Because the polar coordinate system deals with circles, the formula for area is pi * r^2. Since the angles go from 0 to 2pi, the integral comes out to 1/2 * Integral(0, 2pi, r(theta)^2 dtheta). The 1/2 and 2pi simplifies to pi.

Different polar equations may have different limits of integration based on their periods. Once the period is determined, the limits of integration are usually from 0 to the angle for the period. A good strategy for determining the period is to set up a table, finding the r values for various angles along the unit circle. When values begin to repeat, then you have found the period. Familiarity with the trig functions and identities is also helpful in determining the period, as shown below.

As an example, let's find the area for sin(3theta). Simply by being familiar with sinusoidal functions, it is possible to quickly determine the period as 2pi/3. Therefore, the area for sin(3theta) is 1/2 Integral(0, 2pi/3, sin(3theta) dtheta), which comes out to: -1/6 cos(3theta) evaluated from 0 to 2pi/3, which comes out to 1/2 - 1/6 = 1/3.

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