Parametric equations in modellingEdexcel A-Level Maths: Revision notes

In modelling, the parameter is often time $t$, and the pair $x=f(t)$, $y=g(t)$ gives the position of an object at time $t$. This describes a path and also how fast the object moves along it, which a single equation $y=f(x)$ cannot do. Typical questions ask you to:

• build $x(t)$ and $y(t)$ from information in words,
• find the position at a given time (substitute $t$),
• find the time at which a condition holds (solve for $t$, then substitute back),
• find the Cartesian path by eliminating $t$. Always state units, and check that your $t$ lies in the range where the model applies (for example $t\ge0$, or until the ball lands).

If an object starts at $(x_0,y_0)$ and moves with constant velocity $(u,v)$, then $x=x_0+ut,\qquad y=y_0+vt.$ Given two positions at two times, find the velocity by dividing the displacement by the time taken. Example: from $(1,8)$ at $t=0$ to $(6,20)$ at $t=5$: the velocity is $\left(\frac55,\frac{12}{5}\right)=(1,2.4)$, so $x=1+t$, $y=8+2.4t$. The speed is the size of the velocity, $\sqrt{u^2+v^2}$. For $(3,6)$ the speed is $\sqrt{45}=3\sqrt5$. Constant velocity gives a straight line: eliminating $t$ gives $y$ as a linear function of $x$.

When the horizontal motion is steady and the vertical motion is accelerated, the model is of the form $x=ut$, $y=vt-\frac12gt^2$ (a projectile).

• The ball lands when $y=0$: factorise to get $t=0$ (launch) and the landing time.
• The greatest height occurs at the midpoint of the two times when $y=0$, or by completing the square in $t$. For $y=9t-5t^2=-5(t-0.9)^2+4.05$ the maximum is $4.05$ at $t=0.9$.
• The range is the value of $x$ at the landing time.
• Eliminating $t$ gives the Cartesian path, here a parabola $y=\frac34x-\frac{5}{144}x^2$. This model ignores air resistance and spin, so real paths fall slightly short.

Motion around a circle is modelled with sine and cosine: $x=a+r\sin(\omega t)$, $y=b-r\cos(\omega t)$ describes a point on a circle of radius $r$ with centre $(a,b)$. The period (time for one full revolution) is $\frac{2\pi}{\omega}$. For a Ferris wheel seat $x=15\sin\left(\frac{\pi t}{2}\right)$, $y=16-15\cos\left(\frac{\pi t}{2}\right)$: the radius is $15$ m, the centre is $16$ m above the ground, the period is $4$ minutes, the lowest height is $1$ m and the greatest is $31$ m. The Cartesian path is $x^2+(y-16)^2=225$. Use the greatest and least values of $\sin$ and $\cos$ ($\pm1$) to read off extreme values.

To decide whether two objects collide, they must be at the same place at the same time. Equate the $x$-coordinates and solve for $t$, then check whether the $y$-coordinates agree at that same $t$. Two paths can cross without a collision if the objects reach the crossing point at different times. Tug $A$ reaches $(6,5)$ at $t=2$ while tug $B$ reaches it at $t=4$. To find the closest approach, write the vector between the objects, $(x_B-x_A,\ y_B-y_A)$, form $d^2=(x_B-x_A)^2+(y_B-y_A)^2$, which is quadratic in $t$, and find its minimum by completing the square or differentiating. Then square root. For $d^2=26t^2-154t+245$ the minimum is at $t=\frac{77}{26}$. Criticise models too: constant velocity ignores acceleration and turning, and treating tugs as points ignores their size.