Some plane curves are not the graphs of functions \(y = f(x).\) In particular, graphs of functions cannot fail the vertical line test: for each \( a \) there can be at most one point on the curve with \(x\)-coordinate \( a.\) In order to describe more curves, it is convenient to consider \(x\) and \(y\) as functions of a separate variable \(t\) (called a parameter), i.e.
This is known as a parametric equation for the curve that is traced out by varying the values of the parameter \( t.\)
Show that the parametric equation \( x=\cos t\) and \(y=\sin t\) \((0 \leqslant t\leqslant 2\pi)\) traces out a circle.
Eliminating \(t\) gives
\[x^2+y^2= \cos^2 t+\sin^2 t=1,\]
which in fact is the equation of a circle with radius \(1\). \(_\square\)
A circle centered at \( (h,k)\) with radius \( r\) can be described by the parametric equation
\[x=h+r\cos t, \quad y=k+r\sin t.\]
Eliminating \(t\) as above leads to the familiar formula
What are the radius \(r\) and center \((h,k)\) of
\[\begin &x=3+8\cos 4t, &y=-2+8\sin 4t, &0 \leq t\leq 2\pi? \end\]
From the general equation above, we have
\[r=8, \quad c(h,k) = (3,-2). \ _\square\]
In the parametric equation
\[\begin &x=8\cos At, &y=8\sin At, &0 \leqslant t\leqslant 2\pi, \end\]
how does \(A\) affect the circle as \(A\) changes?
Eliminating \(t\) gives
\[ < x >^< 2 >+< y >^< 2 >=< \cos >^ < 2 >At+< \sin >^ < 2 >At=1,\]
which is still a circle with radius \(1\) and center at the origin. If we have \(A=\frac\), \((x,y)=\left(\cos\fract, \sin \fract \right)\), i.e. as \(t\) ranges from \(0\) to \(2\pi,\) the equation starts at \((1,0)\) and stops at \((-1,0)\). This means that it goes halfway through the circle. So \(A\) governs the rate at which the equation traces out a circle.
Similarly, if \(A=2,\) the equation moves twice around the circle. \(_\square\)
A line that passes through point \((h,k)\) with slope \(m\) can be described by the parametric equation
\[x = h + t, \quad y = k + mt.\]
More generally, let \(m = \tan \alpha,\) where \(\alpha\) is the tilt angle. Changing \(t\) to \(t\cos\alpha,\) the parametric equation will become
\[x = h+t\cos \alpha, \quad y = k+t\sin \alpha.\]
Let \(P, Q\) be two points of interception between line \(L:x-y+2=0\) and parabola \(y=x^2\). What is the length \(|PQ|?\)
\(L : x-y+2=0\) passes through point \((0,2)\) and has a tilt angle of \(\alpha=\frac<\pi>\) and hence the parametric equations
\[x=t\frac>, \quad y = 2 + t\frac>.\]
Substituting into the parabola gives
\[\begin y &= x^2 \\ 2+t\frac>&=t^2\frac \\ t^2-\sqrtt-4 &= 0\\\\ |PQ| &= \sqrt \\ &= \sqrt<(t_1 - t_2)^2>\\ &= \sqrt \\ &= \sqrt \\ &= 3\sqrt.\ _\square \end\]
Converting from a parametric equation to an equation in terms of Cartesian coordinates involves eliminating \( t\):
What does the parametric equation
\[\begin &x=t^+t, &y =2t-1\end\]
describe?
Plugging the value of \(t\) in \(y,\) which is \(t=\frac (y+1),\) into \(x\) gives
\[x=\fracy^+y+\frac.\]
This is the equation of a parabola opening to the right. \(_\square\)