| Question asksโฆ | What to do |
|---|---|
| Where is it at time $t = 2$? | Substitute $t = 2$ into both equations |
| When does it reach the ground? | Solve $y(t) = 0$ (taking the sensible root) |
| When is it highest? | Find the maximum of $y(t)$ โ complete the square, or use calculus |
| What shape is the path? | Eliminate $t$ to get a Cartesian equation |
| For what values is the model valid? | Restrict $t$ so that the situation makes sense (e.g. $y \geqslant 0$, $t \geqslant 0$) |
A ball thrown through the air moves at a steady speed horizontally while its height follows a quadratic in $t$. That gives parametric equations of the form $x = at$, $y = bt - ct^2$. (Where these come from is covered in Year 2 mechanics, on Projectiles.)
A ball is thrown from ground level. Its position $t$ seconds later is $x = 12t$, $y = 16t - 5t^2$, where $x$ and $y$ are the horizontal and vertical distances in metres.
(a) When and where does it land? (b) Find its greatest height. (c) Find the Cartesian equation of its path, with the domain for which the model is valid.
One revolution takes $\dfrac{2\pi}{\omega}$ (with the angle in radians)
A capsule on an observation wheel has position $x = 20\sin\left(\dfrac{\pi t}{15}\right)$, $y = 22 - 20\cos\left(\dfrac{\pi t}{15}\right)$, where $t$ is in minutes, $y$ is the height above the ground in metres and $x$ is the horizontal displacement from the vertical line through the centre.
(a) Find the radius, the height of the centre, and the time for one revolution. (b) Find the lowest and greatest heights. (c) For how long in each revolution is the capsule more than $32$ m above the ground?
A wheel has radius $0.35$ m. Using the cycloid model, find (a) the greatest height of the stone and where it occurs in the first revolution, (b) how far the bicycle moves in one revolution, and (c) the first value of $\theta$ at which the stone is $0.525$ m above the ground.
Two boats move so that, $t$ hours after noon, boat A is at $x = t$, $y = 2t$ and boat B is at $x = 6 - t$, $y = t + 3$ (distances in km). Do their paths cross? Do the boats collide?
Give three limitations of the model $x = 12t$, $y = 16t - 5t^2$ for the ball in Worked Example 1.
Position at time $t$
Substitute into both equations.
Lands
Solve $y(t) = 0$, take $t > 0$.
Highest point
Complete the square on $y(t)$.
Path
Eliminate $t$; state the domain.
Circle
$x = a + r\sin\omega t$, $y = b - r\cos\omega t$; period $\tfrac{2\pi}{\omega}$.
Cycloid
$x = r(\theta - \sin\theta)$, $y = r(1 - \cos\theta)$.
Collision
Same point at the same $t$.
Validity
Restrict $t$: e.g. $t \geqslant 0$, $y \geqslant 0$.
A cyclist's position $t$ hours after setting off is $x = 2t$, $y = 3t + 1$ (km). (a) Where is she when $t = 2$? (b) Find the Cartesian equation of her path. (c) Find her speed.
โถ Show solution
(a) $(4, 7)$.
(b) $t = \tfrac{x}{2}$, so $y = \tfrac32 x + 1$ โ a straight line.
(c) Each hour $x$ increases by $2$ and $y$ by $3$, so she travels $\sqrt{2^2 + 3^2} = \sqrt{13} \approx 3.61$ km per hour.
A particle moves so that $x = t^2 - 4$, $y = 2t$, for $0 \leqslant t \leqslant 3$. Find its start and end points, and when and where it crosses the $y$-axis.
โถ Show solution
$t = 0$: $(-4, 0)$. $t = 3$: $(5, 6)$.
$x = 0$: $t^2 = 4$, $t = 2$ (as $t \geqslant 0$). Position $(0, 4)$.
A ball is hit from a height of $1.5$ m. Its position after $t$ seconds is $x = 15t$, $y = 1.5 + 10t - 4.9t^2$ (metres). Find (a) its greatest height, (b) the horizontal distance travelled before it hits the ground. Give answers to 3 significant figures.
โถ Show solution
(a) $y = -4.9\left(t - \tfrac{10}{9.8}\right)^2 + 1.5 + \dfrac{10^2}{4 \times 4.9}$, so the greatest height is $1.5 + \dfrac{100}{19.6} = 6.60$ m (3 s.f.), at $t \approx 1.02$ s.
(b) $4.9t^2 - 10t - 1.5 = 0 \Rightarrow t = \dfrac{10 + \sqrt{100 + 29.4}}{9.8} = 2.181\ldots$ (the other root is negative).
$x = 15 \times 2.181\ldots = 32.7$ m (3 s.f.).
A car drives round a roundabout with $x = 15\cos\left(\dfrac{t}{3}\right)$, $y = 15\sin\left(\dfrac{t}{3}\right)$ (metres, seconds). Find the Cartesian equation of the path and the time for one full circuit.
โถ Show solution
$x^2 + y^2 = 225$: a circle of radius $15$ m centred on the origin.
One circuit when $\tfrac{t}{3} = 2\pi$: $t = 6\pi \approx 18.8$ seconds.
A capsule on a big wheel has $x = 60\sin\left(\dfrac{\pi t}{15}\right)$, $y = 65 - 60\cos\left(\dfrac{\pi t}{15}\right)$, with $t$ in minutes and $y$ the height in metres. Find the greatest height, and how long in each revolution the capsule is more than $35$ m above the ground.
โถ Show solution
Greatest height $65 + 60 = 125$ m. One revolution takes $30$ minutes.
$65 - 60\cos\left(\dfrac{\pi t}{15}\right) > 35 \Rightarrow \cos\left(\dfrac{\pi t}{15}\right) < \dfrac12$.
For one revolution: $\dfrac{\pi}{3} < \dfrac{\pi t}{15} < \dfrac{5\pi}{3}$, so $5 < t < 25$.
That is $20$ minutes of each revolution.
A stone in a tyre of radius $0.3$ m follows $x = 0.3(\theta - \sin\theta)$, $y = 0.3(1 - \cos\theta)$. Find the horizontal position of the stone when it first reaches a height of $0.45$ m.
โถ Show solution
$0.3(1 - \cos\theta) = 0.45 \Rightarrow \cos\theta = -0.5 \Rightarrow \theta = \dfrac{2\pi}{3}$ (first time).
$x = 0.3\left(\dfrac{2\pi}{3} - \dfrac{\sqrt3}{2}\right) = 0.3(2.0944 - 0.8660) = 0.369$ m (3 s.f.).
A stone is thrown with $x = 20t$, $y = 30t - 5t^2$ (metres, seconds). Find the Cartesian equation of its path, and state the values of $x$ for which the model is valid.
โถ Show solution
$t = \tfrac{x}{20}$, so $y = \dfrac{30x}{20} - \dfrac{5x^2}{400} = \dfrac{3x}{2} - \dfrac{x^2}{80}$.
Valid while $y \geqslant 0$: $5t(6 - t) \geqslant 0$, so $0 \leqslant t \leqslant 6$, i.e. $0 \leqslant x \leqslant 120$.
Drone P is at $x = 2t$, $y = t + 1$ and drone Q at $x = t + 3$, $y = 3t - 4$, $t$ seconds after launch (in metres, at the same height). Show that their paths cross, and determine whether they collide.
โถ Show solution
P's path: $y = \tfrac{x}{2} + 1$. Q's path: $t = x - 3$, so $y = 3(x - 3) - 4 = 3x - 13$.
Paths meet when $\tfrac{x}{2} + 1 = 3x - 13 \Rightarrow \tfrac52 x = 14 \Rightarrow x = 5.6$, $y = 3.8$.
P reaches $x = 5.6$ at $t = 2.8$; Q reaches $x = 5.6$ at $t = 2.6$.
They are at the crossing point at different times, so they do not collide.
A bird flies with $x = 10t$, $y = 20 + 5\sin\left(\dfrac{\pi t}{2}\right)$ (metres, seconds). (a) State its lowest and greatest heights. (b) Find the Cartesian equation of its path. (c) Find when and where it first reaches a height of $22.5$ m.
โถ Show solution
(a) $15$ m and $25$ m.
(b) $t = \tfrac{x}{10}$, so $y = 20 + 5\sin\left(\dfrac{\pi x}{20}\right)$.
(c) $5\sin\left(\dfrac{\pi t}{2}\right) = 2.5 \Rightarrow \sin\left(\dfrac{\pi t}{2}\right) = \tfrac12 \Rightarrow \dfrac{\pi t}{2} = \dfrac{\pi}{6} \Rightarrow t = \tfrac13$ s.
Then $x = \tfrac{10}{3} \approx 3.33$ m.
A golf ball is struck from ground level. Its position after $t$ seconds is modelled by $x = 30t$, $y = 20t - 4.9t^2$, in metres.
(a) Find the Cartesian equation of the path. (b) Find the horizontal distance to where the ball lands. (c) A tree $10$ m tall stands $90$ m away along the line of flight. Does the ball clear it? (d) Find the greatest height. (e) State the domain of $t$ for which the model applies, and give one reason why the actual distance might be shorter.
โถ Show solution
(a) $t = \tfrac{x}{30}$: $y = \dfrac{20x}{30} - \dfrac{4.9x^2}{900} = \dfrac{2x}{3} - \dfrac{49x^2}{9000}$.
(b) $y = 0$: $t(20 - 4.9t) = 0 \Rightarrow t = \dfrac{20}{4.9} = 4.08\ldots$ s. Then $x = 30 \times 4.0816\ldots = 122$ m (3 s.f.).
(c) $x = 90$ when $t = 3$: $y = 60 - 4.9 \times 9 = 15.9$ m. That is more than $10$ m, so the ball clears the tree by about $5.9$ m.
(d) $y = -4.9\left(t - \tfrac{20}{9.8}\right)^2 + \dfrac{20^2}{4 \times 4.9}$, so the greatest height is $\dfrac{400}{19.6} = 20.4$ m (3 s.f.), at $t \approx 2.04$ s.
(e) $0 \leqslant t \leqslant \dfrac{20}{4.9}$ (about $4.08$ s), while the ball is in the air. Air resistance, ignored by the model, would slow the ball and reduce the distance.