9Fparametric

Parametric Vectors

So far we have studied vectors with constant x and y components. We will now investigate vectors where the x and y components vary with time. Thus the vector traces out a path over time.

math \text{Introduce }t\text{ for time, where }t\geqslant 0 \qquad. math

t is a **parameter**. We introduced **parametric form** back in Chapter 1.

Now the coefficients of **__i__** and **__j__** become expressions **x(t)** and **y(t)** so the position vector becomes:

math . \qquad \underline{r}(t)=x(t)\underline{i}+y(t)\underline{j} \qquad. math

This is called a **parametric vector equation**.

The **cartesian path** traced out by the parametric vector equation can be found by solving **x(t)** and **y(t)** simultaneously to eliminate **t**.
 * This is the same process that was described back in Chapter 1.

The domain of the cartesian path can be found from **x(t)** given: math . \qquad t\geqslant 0 \qquad. math

Example 1

Find the cartesian equation of the path of a particle with position given by:

math . \qquad \underline{r}(t)=2t\underline{i}+ \big( t^2+t \big) \underline{j} \qquad. math

From the parametric equation, we get that: math . \qquad x=2t \text{ and } y = t^2+t \qquad. math
 * Solution:**

rearrange **x = 2t** to make **t** the subject math . \qquad t=\dfrac{x}{2} \qquad. math

Substitute into **y** math \\ . \qquad y=t^2+t \qquad. \\ . \\ . \qquad y=\left( \dfrac{x}{2} \right)^2 + \left( \dfrac{x}{2} \right) \qquad. \\ . \\ . \qquad y=\dfrac{x^2}{4}+\dfrac{x}{2} math

Hence the path is a parabola (see right)

__Domain__ math \\ \text{Since } t\geqslant 0 \text{ and } t=\dfrac{x}{2} \qquad. \\ . \\ . \qquad \Rightarrow \dfrac{x}{2}\geqslant 0 \qquad. \\ . \\ . \qquad \Rightarrow x\geqslant 0 math

math . \qquad t\geqslant 0 \textit{ doesn't always mean }x\geqslant 0 \qquad. math
 * Be Careful:**

The Classpad can be used to draw parametric equations (see here).

Example 2

Find the cartesian equation for these parametric equations:

... ... ... x = 2 cos(t) + 1 and ... ... ... y = 2 sin(t) + 3 ... ... t Î [0, 2 p ]

This can be done by isolating the trig functions, squaring each equation, then adding the two equations.

Using Pythagoras math .\qquad \cos^2(t)+\sin^2(t)=1 \qquad. math

We get math . \qquad \dfrac{(x-1)^2}{4}+\dfrac{(y-3)^2}{4}=1 \qquad. math

Multiply both sides by 4 math . \qquad (x-1)^2+(y-3)^2=4 \qquad. math

This is the cartesian equation of a circle with centre at (1, 3) and radius = 2

math \\ . \qquad \text {Domain: } x \in \lbrack 1-2, 1+2 \rbrack \Longrightarrow x \in \lbrack -1, 3 \rbrack \qquad. \\ . \\ . \qquad \text {Range: } y \in \lbrack 3-2, 3+2 \rbrack \Longrightarrow y \in \lbrack 1, 5 \rbrack math

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