Functions are often written in terms of two variables that can be plotted as a single equation.
For example, the function \(f(x)\) is drawn as the graph \(y = f(x)\).
In projectile motion, an object accelerates vertically due to gravity while travelling horizontally at constant velocity.
The horizontal and vertical positions can be independently calculated at any given time t.
If \(x = f(t)\) and \(y = g(t)\) are functions of time \(t\), then each value of \(t\) gives a point \((x,y)\) on a curve.
The equations \(x = f(t)\) and \(y = g(t)\) are called
parametric equations, and \(t\) is the parameter.
Example
The path of an object is defined by:
\[
x = 3(1+t), \qquad y = 6(1 - t)
\]
Values at different times:
| t | x | y |
|---|
| -5 | -12 | 36 |
| -4 | -9 | 30 |
| -3 | -6 | 24 |
| -2 | -3 | 18 |
| -1 | 0 | 12 |
| 0 | 3 | 6 |
| 1 | 6 | 0 |
| 2 | 9 | -6 |
| 3 | 12 | -12 |
| 4 | 15 | -18 |
| 5 | 18 | -24 |
| 6 | 21 | -30 |
| 7 | 24 | -36 |
| 8 | 27 | -42 |
| 9 | 30 | -48 |
| 10 | 33 | -54 |
At \(t = 0\), the object is at \((3,6)\).
But how can the two parametric equations be written as one equation in terms of y and x ?
Rearrange one of the equations .
\[
x = 3(1+t)
\]
\[
so
\]
\[
x = 3 + 3t
\]
\[
\frac{x - 3}{3} = t
\]
now substitute into the other
\[
\; y = 6(1 - t)
\]
\[
\; y = 6\left(1 - \frac{x - 3}{3}\right)
\]
\[
\; y = 6 - 2(x - 3)
\]
\[
\; y = 6 - 2x + 6
\]
\[
\; y = 12 - 2x
\]
\(
y = 12 - 2x
\) is the constraint equation.
how to try to get back to y = f(x)
\[
\text{Given } x = f(t),\quad t = f^{-1}(x)
\]
\[
\text{so } y = g(t) \text{ becomes } y = g\!\left(f^{-1}(x)\right)
\]
Example
Find the constraint equation
of the parametric equations
\(
x = 4 - t
\)
,
\(
y = 1 + 3t
\)
\[
x = 4 - t,\qquad y = 1 + 3t
\]
\[
\Rightarrow\; t = 4 - x
\]
\[
\Rightarrow\; y = 1 + 3(4 - x)
\]
\[
\Rightarrow\; y = 1 + 12 - 3x
\]
\[
\Rightarrow\; y = 13 - 3x
\]
\[
\Rightarrow\; y + 3x = 13
\]
This is a straight line.
Example
Find the constraint equation of the locus defined by:
x = 3 + 4 sinθ
y = 2 − 5 cosθ
\[
x = 3 + 4\sin\theta,\qquad y = 2 - 5\cos\theta
\]
\[
\sin\theta = \frac{x - 3}{4},\qquad \cos\theta = \frac{2 - y}{5}
\]
Using sin²θ + cos²θ = 1:
\[
\left(\frac{x - 3}{4}\right)^2 + \left(\frac{2 - y}{5}\right)^2 = 1
\]
This is an ellipse , centre \((3,2)\), major axis \(8\), minor axis \(10\).
\[
x = r\cos t,\qquad y = r\sin t
\]
Circle of radius \(r\), centre \((0,0)\).
Example
\[
x = 3\cos t,\qquad y = 3\sin t
\]
Shifted circle:
\[
x = 3 + 3\cos t,\qquad y = 5 + 3\sin t
\]
Differentiating Parametric Equations
First derivative
When x = f(t) and y = g(t) and y = f(x)
\[
f'(x) = \frac{dy}{dx}
\]
\[
\quad = \frac{dy}{dt} \times \frac{dt}{dx}
\]
\[
\quad = \frac{dy}{dt} \div \frac{dx}{dt}
\]
\[
= \frac{y'(t)}{x'(t)}
\]
\[
\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}
\]
Example
Find the gradient of the
curve defined by :
\(
x = t^2,\qquad y = 4t
\)
\[
x = t^2 \qquad\qquad\qquad y = 4t
\]
\[
x'(t) = 2t \qquad\qquad\qquad y'(t) = 4
\]
\[
\frac{dy}{dx} = \frac{y'(t)}{x'(t)}
\]
\[
\quad = \frac{4}{2t}
\]
\[
\quad = \frac{2}{t}
\]
Example
The radius ,\( r \) cm, of a circular pool of water is increasing at a constant rate of 2cm per minute.
Find the rate at which the area of the pool is increasing
with time, when the radius is 8cm.
\[
\frac{dr}{dt} = 2\,\text{cm per minute}
\]
\[
\text{Area of a circle } = \pi r^{2}
\]
\[
\;\;\Rightarrow\; \frac{dA}{dr} = 2\pi r
\]
\[
\;\;\;\;\frac{dA}{dr}
= \frac{dA}{dt} \times \frac{dt}{dr}
\]
\[
\Rightarrow\; 2\pi r
= \frac{dA}{dt} \times \frac{1}{2}
\]
\[
\Rightarrow\; 4\pi r
= \frac{dA}{dt}
\]
when \( r=8 \)
\[
\frac{dA}{dt} = 32\pi\text{ cm}^2\text{/min}
\]
\[
f''(x)
=
\frac{d^2 y}{dx^2}
\]
\[
\quad =
\frac{d}{dx}
\left(
\frac{dy}{dx}
\right)
\]
\[
\quad =
\frac{d}{dt}
\left(
\frac{dy}{dx}
\right)
\cdot
\frac{dt}{dx}
\]
Substituting the equation for dy/dx
\[
=
\frac{d}{dt}
\left(
\frac{y'(t)}{x'(t)}
\right)
\cdot
\frac{1}{\dfrac{dx}{dt}}
\]
\[
=
\frac{d}{dt}
\left(
\frac{y'(t)}{x'(t)}
\right)
\cdot
\frac{1}{x'(t)}
\]
now apply the quotient rule
\[
=
\left(
\frac{
y''(t)\,x'(t)\;-\;x''(t)\,y'(t)
}{
\left(x'(t)\right)^{2}
}
\right)
\cdot
\frac{1}{x'(t)}
\]
\[
=
\frac{
y''(t)\,x'(t)\;-\;x''(t)\,y'(t)
}{
\left(x'(t)\right)^{3}
}
\]
\[
f''(x)
=
\frac{
x'(t)\,y''(t)\;-\;x''(t)\,y'(t)
}{
\left(x'(t)\right)^{3}
}
\]
\[
\frac{d^2y}{dx^2}
=
\frac{
\left(\frac{d^2y}{dt^2}\right)\left(\frac{dx}{dt}\right)
-
\left(\frac{dy}{dt}\right)\left(\frac{d^2x}{dt^2}\right)
}{
\left(\frac{dx}{dt}\right)^3
}
\]
Example
Describe the nature of the curve \(x=t^2,\; y=t^3\) at \(t=3\).
\[
x = t^2
\]
\[
x'(t) = 2t
\]
\[
x''(t) = 2
\]
\[
y = t^3
\]
\[
y'(t) = 3t^2
\]
\[
y''(t) = 2t
\]
\[
\frac{d^{2}y}{dx^{2}}
=
\frac{
x'(t)\,y''(t)\;-\;x''(t)\,y'(t)
}{
\left(x'(t)\right)^{3}
}
\]
\[
=
\frac{
2t \cdot 6t \;-\; 2 \cdot 3t^{2}
}{
\left(2t\right)^{3}
}
\]
\[
=
\frac{
12t^{2} - 6t^{2}
}{
8t^{3}
}
\]
\[
=
\frac{
6t^{2}
}{
8t^{3}
}
\]
\[
=
\frac{3}{4t}
\]
\[
\frac{d^2y}{dx^2} = \frac{3}{4t}
\]
\[
\text{at } t = 3
\]
\[
\frac{d^{2}y}{dx^{2}}
=
\frac{3}{12}
\]
\[
\frac{d^{2}y}{dx^{2}} > 0,\;\text{ so the graph is concave up}
\]
At \(t=3\): concave upward.
Projectile Motion Example
A projectile is fired horizontally at 250 m/s from a cliff 45 m high.
- How long after firing does the projectile hit the ground ?
- How far does the projectile travel horizontally ?
- What is the horizontal and vertical velocity of the projectile when it hits the ground ?
- What is the horizontal and vertical acceleration of the projectile when it hits the ground ?
1 ) Newton's equation of motion
\[
s = ut + \frac{1}{2}at^{2}
\]
allows the horizontal and vertical displacements to be calculated separately.
\[
\text{When } t = 0,\quad u_x = 250\,\text{m/s},\quad a_x = 0\,\text{m/s}^2
\]
\[
s_x = u_x t + \frac{1}{2} a_x t^{2}
\]
\[
s_x = 250t + 0
\]
\[
s_x = 250t
\]
\[
\text{When } t = 0,\quad u_y = 0\,\text{m/s},\quad a_y = -9.8\,\text{m/s}^2
\]
\[
s_y = u_y t + \frac{1}{2} a_y t^{2}
\]
\[
s_y = 0 + \frac{1}{2} \times (-9.8) \times t^{2}
\]
\[
s_y = -4.9 t^{2}
\]
The original displacement is zero units in the x direction, but 45 units in the y direction, since the cliff is 45 m tall.
\[
x(t) = 250t,\qquad y(t) = 45 - 4.9t^2
\]
The individual positions at t seconds can now be calculated:-
| t |
x |
y |
| 0 |
0 |
45 |
| 1 |
250 |
40.1 |
| 2 |
500 |
25.4 |
| 3 |
750 |
0.9 |
| 4 |
1000 |
-33.4 |
| 5 |
1250 |
-77.5 |
From the table, the projectile hits the ground somewhere between 3 and 4 seconds after being fired.
Plotting these points give a picture of the flight path:
But when exactly does it hit the ground ?
The projectile hits the ground when y = 0
so
\[
y(t) = 45 - 4.9t^{2}
\]
\[
0 = 45 - 4.9t^{2}
\]
\[
4.9t^{2} = 45
\]
\[
t^{2} = \frac{45}{4.9}
\]
\[
t = \sqrt{\frac{45}{4.9}}
\]
\[
t = 3.03\ \text{seconds (2dp)}
\]
The projectile hits the ground 3.03 seconds (2dp) after being fired.
2) Substituting this value of t into the parametric equation for x :-
\[
x(t) = 250t
\]
\[
x(3.03) = 250 \times 3.03
\]
\[
= 757.5\,\text{m}
\]
so the projectile has travelled 757.5 m horizontally.
The constraint equation is
\[
x(t) = 250t
\]
\[
\frac{x}{250} = t
\]
\[
y(t) = 45 - 4.9t^{2}
\]
\[
y = 45 - 4.9\left(\frac{x}{250}\right)^{2}
\]
\[
y = 45 - \frac{4.9x^{2}}{62500}
\]
this can be used to check the horizontal distance travelled, since y = 0 at ground level
\[
y = 45 - \frac{4.9x^{2}}{62500}
\]
\[
0 = 45 - \frac{4.9x^{2}}{62500}
\]
\[
\frac{4.9x^{2}}{62500} = 45
\]
\[
x = \sqrt{\frac{45 \times 62500}{4.9}}
\]
\[
x = 757.614
\]
Why is there a difference ? Because the time was rounded.
3) Finding the horizontal velocity
\[
x(t) = 250t
\]
\[
x'(t) = 250\ \text{m/s}
\]
(which is , of course, as stated in the question!)
and the vertical velocity
\[
y(t) = 45 - 4.9t^{2}
\]
\[
\quad y'(t) = -9.8t
\]
so when the projectile hits the ground, it has a vertical velocity
\[
y(t) = 45 - 4.9t^{2}
\]
\[
\quad y'(3.03) = -9.8 \times 3.03
\]
\[
= -29.694\ \text{m/s}
\]
The negative value shows that it is travelling downwards.
4) Acceleration
This is the second derivative.
Horizontally,
\[
x(t) = 250t
\]
\[
x'(t) = 250\ \text{m/s}
\]
\[
x''(t) = 0\ \text{m/s}^{2}
\]
(as expected since travelling at a constant velovity)
and vertically,
\[
y(t) = 45 - 4.9t^{2}
\]
\[
\quad y'(t) = -9.8t
\]
\[
\quad y''(t) = -9.8\ \text{m/s}^{2}
\]
which is acceleration due to gravity.
