PARAMETRIC FUNCTIONS
Use Autograph throughout sections 1 to 4 of this topic! Sadly, it won’t find areas enclosed by curves define parametrically, so it’s not much use for section 5.
1. Sketching Parametric Functions
We usually express a curve as an equation connecting the x and y coordinates of any point on the line e.g. . It can sometimes be useful to involve a third variable t, and to express both x and y as functions of t. The two equations are called parametric equations, and t is called the parameter.
The methods for sketching parametric functions are
•Draw up a table of values of t, x and y.
•Convert the parametric equations into Cartesian form by making t the subject of one of the equations (usually the x equation) and substituting into the other equation.
Example 1 : Sketch .
First, with a table of values.
t / –3 / –2 / –1 / 0 / 1 / 2 / 3x / 9 / 4 / 1 / 0 / 1 / 4 / 9
y / –2 / –1 / 0 / 1 / 2 / 3 / 4
Alternatively, we can convert the parametric equations into Cartesian form like this...
To sketch this, we draw the standard square root graph and translate it up one unit.
Either way, we get the graph shown above.
Example2:Sketch the curve with parametric equations .
This is the parametric equation of a circle, centre (0, 0), and radius a.We can see this because any point on the circle at an angle t to the x-axis will have the coordinates described by the parametric equations.
Alternatively, we can use a Pythagorean identity to generate the Cartesian equation....
…which from unit C2 we know to be the equation of a circle, centre (0, 0), and radius a.
Activity1:What is the parametric form for a circle with radius r and centre (a, b)? Check on Autograph.
Example3:Use a table of values to sketch the curve with parametric equations . What is the Cartesian equation of the curve?
The centre of the curve will be the origin. Whilst the y values complete one period, the x values will complete two periods. A table of values helps here.
t / 0 / / / / / / / /x / 0 / 3 / 0 / –3 / 0 / 3 / 0 / –3 / 0
y / 0 / / 2 / / 0 / / –2 / / 0
Using trigonometric identities,
…and this is probably as tidy as we can get it.
Example 4 : Find the Cartesian equation of the curve . Hence sketch the curve.
This graph is fairly straightforward to sketch, although note the restriction on the domain. Since , we have , and hence an abrupt end to the graph at (−1, −1) and (1, −1).
C4 p10 Ex 2A, p15 Ex 2C
2. Solving Problems With Parametric Functions
Example 1 : A curve has parametric equations .
Find the coordinates of the points where the curve crosses the axes.
The curve crosses the x-axis when .
The curve crosses the x-axis at (−3, 0).
The curve crosses the y-axis when .
The curve crosses the y-axis at and (0, 3).
Example 2 : A curve has parametric equations , where a is a constant. The curve meets the x-axis at (−2, 0) and the y-axis at A. Find the value of a and the corresponding coordinates of A.
Substituting
Substituting these and into the other equation,
The curve crosses the y-axis when
A is (0, 104).
Example3:Find the coordinates of the points of intersection of the curve with parametric equations and the line with equation .
Substituting the parametric equations in to the linear equation,
Substituting into the parametric equations, we have the coordinates and (−1, 5).
C4p13 Ex 2B
3. Differentiating Parametric Functions
It is often possible to convert parametric equations into Cartesian form, and then differentiate as usual. If this is impossible or undesirable, we can differentiate parametrically, and we will get an expression for in terms of t. The formula we use is
The second derivative is much more complicated...
Example 1 : Find and in terms of the parameter t for example 1 in section 1.
At this stage stop and consider whether this is sensible as a gradient function for the graph.
Example2:The graph of , is shown opposite.
Find and in terms of the parameter t. Hence find the stationary points on the curve and determine their natures.
There are stationary points when
The stationary points are therefore and.
When t is positive, this expression is positive, and so is a minimum point.
When t is negative, the expression is negative, and so is a maximum point.
Also notice that when , is undefined, and so the graph is vertical at the origin.
4. Tangents and Normals to Parametric Functions
Example 1:Find the equations of the tangent and the normal to the curve at the point (–4, –2).
At the point (–4, –2), it is clear (from the y equation) that t= –1, and so
The equation of the tangent is of the form
The equation of the normal is of the form
Example 2 : Sketch the curve given by the parametric equations.
Find the equation of the tangent to the curve at the point .
We need only consider the range , since the parametric equations are periodic.
t / 0 / / / /x / 2 / 0 / –2 / 0 / 2
y / 0 / 1 / 0 / –1 / 0
Plotting these points gives us an ellipse.
Normally we would have to find the value of t at the point in question (easily done anyway) but in this particular example we don’t need to...
At the point in question, we see from the parametric equations that and .
So the equation of the tangent is
Example 3 : A curve C is given by the parametric equations
Find the equation of the normal to the curve at the point where . Find also the coordinates of the point where the normal meets the curve again.
At , the gradient of the tangent is , and so the gradient of the normal is 4. The coordinates are . So the equation of the normal is
To find where the normal meets the curve again, we substitute the parametric equations into the normal equation.
We know immediately that is a solution of the equation, since this is certainly a point where the normal meets the curve, and so that must be a factor. (This is very useful when the equation is more awkward to factorise e.g. a cubic).
The normal meets C again when The coordinates are therefore .
C4 p34 Ex 4A
5. Areas and Parametric Functions
There are two methods for finding areas under curves : the equation of the curve may be expressed in Cartesian form, eliminating the parameter, or the integration can be written solely in terms of the parameter, eliminating x and y…
Note that the limits of the integral will have to be changed too.
Example1:Find the area between the curve and the x-axis between and .
We will do the two methods side-by-side for comparison.
Cartesian / Parametric/ We begin by working out the new limits of the integral using the parametric equation .
This parametric method is essential if the parametric equations will not easily convert to Cartesian form, and is often simpler anyway.
Example 2 : Sketch the curve with parametric equations
Find the equation of the normal to the curve at the point where and find the area of the finite region enclosed by the curve, the x-axis, and the normal.
The point where is (4, 9).
When , the gradient of the tangent is 3, and so the gradient of the normal is .
The equation of the normal is
To find the area of the region we divide it into two parts, one of which needs integration, the other of which is a triangle.
Activity2:Now try this example again by writing the equation in Cartesian form.
Example3:The graph of, is shown opposite. Find the area of the region bounded by the loop of the curve.
We could find the area by integrating from to , thereby circumnavigating the loop, but it is simpler to integrate from to, and then double the result.
The area is therefore square units. The minus sign arose because we are going along the bottom of the loop as we integrate from to.
Activity3:Now try this example again by writing the equation in Cartesian form.
Example 4 : Prove that the area of a circle is .
We know a circle with radius r and centre (0, 0) has parametric equations .
To find the area we can find the area in the positive quadrant, and then multiply by 4...
Activity4:Find a formula for the area of an ellipse.
Example5:The diagram shows a sketch of the curve with parametric equations .
a) Find the coordinates of the point where the curve crosses the x-axis.
b) Find .
c) Show that the area A of the finite region bounded by the curve and the y-axis is given by…
…and hence evaluate A.
a) On the x-axis, y is zero. Therefore and . So the coordinates are (1, 0).
b) Using the chain rule / / /c) Using parametric integration,
C4p18 Ex 2D Topic Review : Parametric Functions