3.5 Parametric equations in modelling — Practise with help

Work each one first, closed book. Before you calculate, decide what the question wants: a position, a time, or the path itself. Write down the domain of the parameter as well.

Warm-up: remember

Solve $12t - 3t^2 = 0$ .
Complete the square on $y = 10t - t^2$ and state its greatest value.
Write the equation of the circle with centre $(3, -1)$ and radius $4$ .

Answers

$3t(4 - t) = 0$ , so $t = 0$ or $t = 4$ .
$y = -(t^2 - 10t) = -((t - 5)^2 - 25) = -(t - 5)^2 + 25$ . The greatest value is $25$ , at $t = 5$ .
$(x - 3)^2 + (y + 1)^2 = 16$ .

If question 1 was hard, revisit solving a quadratic by factorising before you continue.

Practise with help

1. A ball is modelled by $x = 10t$ , $y = 15t - 5t^2$ . Find the greatest height of the ball and the time at which it occurs.

Hint

The height is the $y$ equation. It is a quadratic in $t$ , so complete the square or read off the vertex.

Next step

Write $y = -5(t^2 - 3t)$ , then complete the square inside the bracket.

Full working

y = -5(t^2 - 3t) = -5\left(\left(t - \frac{3}{2}\right)^2 - \frac{9}{4}\right) = -5\left(t - \frac{3}{2}\right)^2 + \frac{45}{4}.

The squared term is subtracted, so the greatest height is $\frac{45}{4} = 11.25$ m, reached at $t = \frac{3}{2} = 1.5$ s.

2. For the model $x = 10t$ , $y = 15t - 5t^2$ , find the Cartesian equation of the path.

Hint

Make $t$ the subject of the simpler equation, $x = 10t$ , then substitute.

Next step

$t = \dfrac{x}{10}$ . Substitute that into $y = 15t - 5t^2$ and simplify.

Full working

y = 15\left(\frac{x}{10}\right) - 5\left(\frac{x}{10}\right)^2 = \frac{3}{2}x - 5 \times \frac{x^2}{100} = \frac{3}{2}x - \frac{x^2}{20}.

The path is a parabola. The equation says where the ball goes; the parameter says when.

3. For the model $x = 10t$ , $y = 15t - 5t^2$ , find where the ball lands.

Hint

The ball lands when its height is $0$ , so solve $y = 0$ for $t$ . Then find $x$ .

Next step

$15t - 5t^2 = 5t(3 - t) = 0$ , so $t = 0$ or $t = 3$ . Reject $t = 0$ as the kick.

Full working

5t(3 - t) = 0, \qquad t = 0 \text{ or } t = 3.

The ball lands at $t = 3$ s. Substitute into $x = 10t$ :

x = 10 \times 3 = 30.

The ball lands $30$ m from the kick. The domain of the model is $0 \le t \le 3$ .

4. A wheel cabin is modelled by $x = 7\cos t$ , $y = 9 + 7\sin t$ . Find the path of the cabin and its greatest height.

Hint

For the path, rearrange each equation so the trigonometric function stands alone, then square and add. For the greatest height, use the largest value of $\sin t$ .

Next step

$\cos t = \dfrac{x}{7}$ and $\sin t = \dfrac{y - 9}{7}$ . The largest value of $\sin t$ is $1$ .

Full working

\left(\frac{x}{7}\right)^2 + \left(\frac{y - 9}{7}\right)^2 = 1,

x^2 + (y - 9)^2 = 49.

The path is a circle with centre $(0, 9)$ and radius $7$ . The greatest height is where $\sin t = 1$ :

y = 9 + 7 \times 1 = 16.

The cabin rises to $16$ m.

5. A point moves on $x = 3 + 2\cos t$ , $y = 1 + 2\sin t$ . Find its position at $t = \frac{\pi}{2}$ and at $t = \pi$ .

Hint

Substitute each value of $t$ into both equations. Use $\cos\frac{\pi}{2} = 0$ , $\sin\frac{\pi}{2} = 1$ , $\cos\pi = -1$ and $\sin\pi = 0$ .

Next step

At $t = \frac{\pi}{2}$ : $x = 3 + 2 \times 0$ and $y = 1 + 2 \times 1$ .

Full working

At $t = \frac{\pi}{2}$ :

x = 3 + 2 \times 0 = 3, \qquad y = 1 + 2 \times 1 = 3.

The point is $(3, 3)$ .

At $t = \pi$ :

x = 3 + 2 \times (-1) = 1, \qquad y = 1 + 2 \times 0 = 1.

The point is $(1, 1)$ .

6. An object moves around the circle $(x - 4)^2 + (y - 6)^2 = 9$ . It starts at the easternmost point and travels anticlockwise at a steady speed, one lap every $20$ seconds. Write parametric equations for its position after $t$ seconds.

Hint

Read the centre and the radius from the circle. The easternmost point is where the angle is $0$ . One lap is a full turn, so the angle after $t$ seconds is $\frac{2\pi t}{20}$ .

Next step

The centre is $(4, 6)$ and the radius is $3$ , so start from $x = 4 + 3\cos(\theta)$ and $y = 6 + 3\sin(\theta)$ with $\theta = \frac{2\pi t}{20}$ .

Full working

The circle has centre $(4, 6)$ and radius $3$ , so

x = 4 + 3\cos\left(\frac{\pi t}{10}\right), \qquad y = 6 + 3\sin\left(\frac{\pi t}{10}\right),

since $\frac{2\pi}{20} = \frac{\pi}{10}$ . At $t = 0$ the point is $(4 + 3, 6) = (7, 6)$ , which is the easternmost point. After $20$ seconds the angle is a full turn and the point returns to $(7, 6)$ , so one lap takes $20$ seconds as required.