True or False? In the parametric equations of a circle, the radius is a second parameter alongside the angle.

False. For a given circle the radius is a constant . Only the angle varies, so there is just one parameter, and each value of it gives one point on the circle.

Which features should you find before sketching a curve given parametrically?

Where the curve crosses each axis, any asymptotes , and the location of any stationary points . Mark those on the axes and join them up in a way that fits the shape of the curve.

True or False? A sketch of a parametric curve should be built from an accurate table of values.

False. A sketch shows the key features and the general shape: the axis intercepts, any asymptotes and any turning points. Working out a few extra points can help when the shape is unfamiliar, but that is a check, not the method.

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Parametric Equations (AQA A Level Maths: Pure): Flashcards

Exam code: 7357

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FrontBasics of Parametric Equations
Define parameter in the context of parametric equations.

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Cards in this collection (19)

Define parameter in the context of parametric equations.
A third variable that both $x$ and $y$ are given in terms of, instead of $y$ being given directly in terms of $x$ .
Each value of the parameter gives one point on the curve. The letters $t$ and $θ$ are the usual choices.
An object moves in a plane, and its position is modelled by $x = f (t)$ and $y = g (t)$ . What do $x$ , $y$ and $t$ represent?
$x$ is the horizontal position of the object, $y$ is its vertical position, and $t$ is the time.
The two equations together give the position at any moment, and the curve they trace out is the object's path.
Define Cartesian equation of a curve.
An equation connecting $x$ and $y$ directly, with no third variable involved. It can often be rearranged into the form $y = f (x)$ .
$y = 6 x - x^{2} - 5$ and $x^{2} + y^{2} = 1$ are both Cartesian equations.
How do you plot a curve from its parametric equations?
Choose values of the parameter across the range required and work out $x$ and $y$ for each one, setting the results out in a table.
Then plot those $(x, y)$ pairs and join them up. The parameter values themselves never appear on the graph.
True or False?
In the parametric equations of a circle, the radius is a second parameter alongside the angle.
False.
For a given circle the radius is a constant. Only the angle varies, so there is just one parameter, and each value of it gives one point on the circle.
What are the parametric equations of a circle with centre $(0, 0)$ and radius $r$ ?
The angle $θ$ is the parameter:
$x = r \cos θ$
$y = r \sin θ$
For a circle with centre $(a, b)$ , add the coordinates of the centre: $x = r \cos θ + a$ and $y = r \sin θ + b$ .
A circle has parametric equations $x = 5 \cos θ + 2$ and $y = 5 \sin θ - 4$ . Fill in the blanks:
centre $(______,______)$ , radius $______$
Centre $(2, - 4)$ , radius $5$ .
The numbers added to $r \cos θ$ and $r \sin θ$ are the coordinates of the centre, so the Cartesian equation is $(x - 2)^{2} + (y + 4)^{2} = 25$ . Notice the sign change, and that the radius is squared on the right.
What does eliminating the parameter from a pair of parametric equations give you?
Eliminating the parameter from a pair of parametric equations gives you the Cartesian equation of the curve: a single equation connecting $x$ and $y$ directly, with the parameter gone.
Both forms describe the same curve, so the connection between $x$ and $y$ is there all along. Eliminating the parameter is what makes it visible.
A curve has parametric equations $x = \frac{t - 3}{4}$ and $y = e^{2 t}$ . Complete the working that eliminates $t$ , filling in the missing expression and the missing index:
$t =______$
$y = e^{______}$
The completed working is:
$t = 4 x + 3$
$y = e^{2 (4 x + 3)} = e^{8 x + 6}$
Rearranging the equation for $x$ is the easier route here, because it is linear in $t$ .
How do you eliminate the parameter to find the Cartesian equation of a curve?
Make the parameter the subject of one of the two equations, substitute that expression into the other equation, then tidy the result into the form asked for.
So, for example, $x = t + 3$ gives $t = x - 3$ , which goes into $y = 4 - t^{2}$ to give $y = 4 - (x - 3)^{2}$ .
True or False?
A curve given by parametric equations always includes every point of its Cartesian equation.
False.
The range of values the parameter can take may restrict the curve to a section of the Cartesian curve.
So, for example, if $x = 3 + 2 \sin t$ then $- 1 \leq \sin t \leq 1$ forces $1 \leq x \leq 5$ , and no point outside that strip belongs to the curve.
When eliminating a parameter, which of the two parametric equations should you rearrange?
Whichever is easier to make the parameter the subject of.
A linear equation such as $x = 2 t + 3$ is easier than a quadratic such as $y = 3 t^{2} + 3 t - 4$ , and a single exponential is easy too, since $x = e^{t}$ gives $t = \ln x$ .
How do you eliminate the parameter when $x$ and $y$ are given in terms of the sine and cosine of the same angle?
Rearrange each equation so that the sine and the cosine stand alone, then square both and add them. The identity $\sin^{2} θ + \cos^{2} θ \equiv 1$ removes the parameter in one step.
So, for example, $x = 3 \sin 2 t$ and $y = 2 \cos 2 t$ give $\frac{x}{3} = \sin 2 t$ and $\frac{y}{2} = \cos 2 t$ , so ${(\frac{x}{3})}^{2} + {(\frac{y}{2})}^{2} = 1$ .
Which features should you find before sketching a curve given parametrically?
Where the curve crosses each axis, any asymptotes, and the location of any stationary points.
Mark those on the axes and join them up in a way that fits the shape of the curve.
Complete the method for finding where a parametric curve crosses the $x$ -axis:
Set the equation for $______$ equal to zero, solve it for the parameter, then substitute the value(s) into the equation for $______$ .
The completed method is:
Set the equation for $y$ equal to zero, solve it for the parameter, then substitute the value(s) into the equation for $x$ .
For the $y$ -axis it is the other way round: set the equation for $x$ to zero, then substitute into the equation for $y$ .
True or False?
Every curve given by parametric equations crosses the $y$ -axis somewhere.
False.
There is a crossing only if the equation for $x$ can actually equal zero.
So, for example, $x = e^{t}$ is positive for every value of $t$ , so $e^{t} = 0$ has no solutions: the curve never reaches the $y$ -axis, and the $y$ -axis is an asymptote.
Why might a value of the parameter found when looking for an axis intercept have to be rejected?
Because it may lie outside the values the parameter is allowed to take, in which case it gives no point on the curve.
So, for example, $x = \ln t^{3}$ is only defined for $t > 0$ , so a solution of $t = - \frac{3}{2}$ has to be discarded and only the positive root used.
True or False?
A sketch of a parametric curve should be built from an accurate table of values.
False.
A sketch shows the key features and the general shape: the axis intercepts, any asymptotes and any turning points.
Working out a few extra points can help when the shape is unfamiliar, but that is a check, not the method.
When is it worth finding the Cartesian equation before sketching a parametric curve?
When the parametric form leaves the shape unclear and you cannot make progress otherwise. Recognising the Cartesian equation tells you the shape at once.
So, for example, $x = \ln t^{3}$ and $y = 4 t^{2} - 9$ become $y = 4 e^{\frac{2}{3} x} - 9$ , an exponential curve with an asymptote at $y = - 9$ .

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