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.

A LevelMathsCambridge (CIE)Pure 3FlashcardsDifferentiationDifferentiation of Parametric Equations

Differentiation of Parametric Equations (Cambridge (CIE) A Level Maths: Pure 3): Flashcards

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FrontBasics of Parametric Equations
Define parametric equations.

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Cards in this collection (24)

Define parametric equations.
A pair of equations giving $x$ and $y$ separately, each in terms of a third variable called the parameter.
So instead of one equation linking $x$ and $y$ , there are two: $x = f (t)$ and $y = g (t)$ .
What do $x$ , $y$ and $t$ often represent in a parametric model of motion?
$x$ is the horizontal position of the object, $y$ is its vertical position, and $t$ is the time.
The curve traced out is then the path the object follows, with the parameter recording when it reaches each point.
Complete the parametric equations of a circle of radius $r$ with centre $(a, b)$ :
$x =______$
$y =______$
The completed equations are:
$x = r \cos θ + a$
$y = r \sin θ + b$
With centre $(0, 0)$ the two constants vanish, leaving $x = r \cos θ$ and $y = r \sin θ$ .
True or False?
A circle can be given by parametric equations even though it cannot be written as $y = f (x)$ .
True.
A circle fails the test for a function of $x$ , since most vertical lines cut it twice, so no single $y = f (x)$ describes the whole of it.
Giving $x$ and $y$ separately sidesteps the problem completely, and that freedom is much of the reason parametric equations are worth having.
What are the centre and radius of the circle $x = 3 \cos θ$ , $y = 3 \sin θ - 4$ ?
The centre is $(0, - 4)$ and the radius is $3$ .
Both can be read straight off by comparing with the standard parametric form, and the Cartesian equation $x^{2} + {(y + 4)}^{2} = 9$ confirms them.
True or False?
The equations $x = 2 \cos θ$ and $y = 2 \sin θ$ involve two parameters.
False.
There is only one parameter, $θ$ . The $2$ is a fixed constant, the radius, and it does not vary along the curve.
A parameter is something each point of the curve has its own value of, which is true of $θ$ and false of the $2$ .
Define the Cartesian equation of a curve.
An equation connecting $x$ and $y$ directly, with no parameter left in it.
A curve given parametrically as $x = f (t)$ and $y = g (t)$ still has one, because $x$ and $y$ are linked through $t$ .
When eliminating the parameter from a pair of parametric equations, which of the two should you rearrange?
Rearrange whichever one is easier to make $t$ the subject of.
A linear equation such as $x = 2 t + 3$ is much easier than a quadratic, and a single exponential is also straightforward, since $x = e^{t}$ gives $t = \ln x$ .
Then substitute that expression for $t$ into the other equation.
How do you eliminate $t$ when $x$ and $y$ are given in terms of $\cos t$ and $\sin t$ ?
Rearrange each equation into the form $\cos t = \dots$ and $\sin t = \dots$ , then square both and add them.
The identity $\sin^{2} t + \cos^{2} t \equiv 1$ removes $t$ in one step.
Rearranging for $t$ itself would bring in an inverse trigonometric function, which is far harder to work with.
A curve has parametric equations $x = 3 + \cos t$ and $y = 2 - \sin t$ . Complete the elimination:
${(x - 3)}^{2} + {(2 - y)}^{2} =______$
The completed equation is:
${(x - 3)}^{2} + {(2 - y)}^{2} = 1$
Squaring $\cos t = x - 3$ and $\sin t = 2 - y$ and adding them gives $\cos^{2} t + \sin^{2} t$ on the left, which is $1$ .
The curve is therefore a circle of radius $1$ centred on $(3, 2)$ .
True or False?
Eliminating the parameter always produces an equation of the form $y = f (x)$ .
False.
The Cartesian equation can be a relation that is not a function, such as the circle ${(x - 3)}^{2} + {(y - 2)}^{2} = 1$ .
A vertical line meets that curve twice, so it cannot be written as a single $y = f (x)$ .
When is it worth converting parametric equations to Cartesian form before differentiating?
When the parameter comes out easily and the resulting Cartesian equation is simple to differentiate.
For $x = \tan t$ , $y = \cos^{2} t$ the conversion gives $y = \frac{1}{1 + x^{2}}$ , and mark schemes can award full marks for that route.
For most pairs neither condition holds, so differentiating with respect to $t$ is the better route.
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 $y = - 9$ .
Complete the rule for differentiating parametric equations:
$\frac{d y}{d x} = \frac{d y}{d t} \div______$
The completed rule is:
$\frac{d y}{d x} = \frac{d y}{d t} \div \frac{d x}{d t}$
It follows from the chain rule with the reciprocal property, since $\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x}$ .
Why is it acceptable for $\frac{d y}{d x}$ to come out in terms of $t$ ?
Because every point on the curve is identified by its parameter value rather than by its $x$ coordinate.
To get a numerical gradient you find the value of $t$ at the point you want and substitute that, which is one step earlier than in ordinary differentiation.
How do you find the gradient of a parametric curve at a given point?
Differentiate both equations to get $\frac{d y}{d t}$ and $\frac{d x}{d t}$ , then divide to get $\frac{d y}{d x}$ in terms of $t$ .
Find the value of $t$ at that point and substitute it in.
What condition gives a stationary point on a parametric curve?
$\frac{d y}{d x} = 0$ , which happens exactly when $\frac{d y}{d t} = 0$ .
It is the numerator of the quotient that has to vanish, so it is the $y$ equation you differentiate and set to zero.
What happens where $\frac{d x}{d t} = 0$ ?
The gradient $\frac{d y}{d x}$ is undefined, since the quotient would be dividing by zero.
Geometrically the tangent there is vertical, which is something an ordinary $y = f (x)$ curve can never show.
True or False?
To find $\frac{d y}{d x}$ you must first eliminate the parameter.
False.
Differentiating each equation with respect to $t$ and dividing is quicker, and it still works when the parameter cannot be eliminated neatly.
Eliminating first is extra work that usually makes the differentiation harder rather than easier.

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