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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FrontEliminating the Parameter of Parametric Equations
What does eliminating the parameter from a pair of parametric equations give you?

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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 $y = - 9$ .
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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