10.2 Parametric Equations

Find parametric equations for $f(x)=x^2-x-6$.

SolutionSolution 1: For any choice for $x$ we can determine the corresponding $y$ by substitution. If we choose $x=t-1$ then $y={(t-1)}^2-(t-1)-6=t^2-3t-4$. Thus $f(x)$ can be represented by the parametric equations

On the graph of this parameterization (Figure 10.2.3(a)) the points have been labeled with the corresponding $t$-values and arrows indicate the path of a particle traveling on this curve. The particle would move from the upper left, down to the vertex at $(.5,-6.25)$ and then up to the right.

Solution 2: If we choose $x=3-t$ then $y={(3-t)}^2-(3-t)-6=t^2-5t$. Thus $f(x)$ can also be represented by the parametric equations

On the graph of this parameterization (Figure 10.2.3(b)) the points have been labeled with the corresponding $t-$values and arrows indicate the path of a particle traveling on this curve. The particle would move down from the upper right, to the vertex at $(.5,-6.25)$ and then up to the left.

Solution 3: We can also parameterize any $y=f(x)$ by setting $t=\frac{\mathrm{d}y}{\mathrm{d}x}$. That is, $t=a$ corresponds to the point on the graph whose tangent line has a slope $a$. Computing $\frac{\mathrm{d}y}{\mathrm{d}x}$, $f^{\prime }(x)=2x-1$ we set $t=2x-1$. Solving for $x$ we find $x=\frac{t+1}{2}$ and by substitution $y=\frac{1}{4}{t}^2-\frac{25}{4}$. Thus $f(x)$ can be represented by the parametric equations

The graph of this parameterization is shown in Figure 10.2.3(c). To find the point where the tangent line has a slope of $0$, we set $t=0$. This gives us the point $(.5,-6.25)$ which is the vertex of $f(x)$.

Find parametric equations for the circle $x^2+y^2=4$.

SolutionWe will present three different approaches:

Solution 1: Consider the equivalent equation ${\left({\displaystyle \frac{x}{2}}\right)}^2+{\left({\displaystyle \frac{y}{2}}\right)}^2=1$ and the Pythagorean Identity, ${\sin }^{2}t+{\cos }^{2}t=1$. We set $\cos t=\frac{x}{2}$ and $\sin t=\frac{y}{2}$, which gives $x=2\cos t$ and $y=2\sin t$. To trace the circle once, we must have $0\le t\le 2\pi$. Note that when $t=0$ a particle tracing the curve would be at the point $(2,0)$ and would move in a counterclockwise direction.

Solution 2: Another parameterization of the same circle would be $x=2\sin t$ and $y=2\cos t$ for $0\le t\le 2\pi$. When $t=0$ a particle would be at the point $(0,2)$ and would move in a clockwise direction.

Solution 3: We could let $x=-2\sin t$ and $y=2\cos t$ for $0\le t\le 2\pi$. Also note that we could use $x=2\cos 2t$ and $y=2\sin 2t$ for $0\le t\le \pi$.

Find a parameterization that traces the ellipse ${\displaystyle \frac{{(x-2)}^2}{9}}+{\displaystyle \frac{{(y+3)}^2}{4}}=1$ starting at the point $(-1,-3)$ in a clockwise direction.

SolutionApplying the Pythagorean Identity, ${\cos }^{2}t+{\sin }^{2}t=1$, we set ${\cos }^{2}t={\displaystyle \frac{{(x-2)}^2}{9}}$ and ${\sin }^{2}t={\displaystyle \frac{{(y+3)}^2}{4}}$. Solving these equations for $x$ and $y$ we set $x=-3\cos t+2$ and $y=2\sin t-3$ for $0\le t\le 2\pi$.

Find a parameterization for the hyperbola ${\displaystyle \frac{{(x-2)}^2}{9}}-{\displaystyle \frac{{(y-3)}^2}{4}}=1$.

SolutionWe use the form of the Pythagorean Identity ${\sec }^{2}t-{\tan }^{2}t=1$. We let ${\sec }^{2}t={\displaystyle \frac{{(x-2)}^2}{9}}$ and ${\tan }^{2}t={\displaystyle \frac{{(y-3)}^2}{4}}$. Solving these equations for $x$ and $y$ we have $x=3\sec t+2$ and $y=2\tan t+3$ for $0\le t\le 2\pi$ and $t\ne \frac{\pi }{2},\frac{3\pi }{2}$.

An object is fired from a height of 0ft and lands 6 seconds later, 192ft away. Assuming ideal projectile motion, the height, in feet, of the object can be described by $h(x)=-x^2/64+3x$, where $x$ is the distance in feet from the initial location. (Thus $h(0)=h(192)=0$ft.) Find parametric equations $x=f(t)$, $y=g(t)$ for the path of the projectile where $x$ is the horizontal distance the object has traveled at time $t$ (in seconds) and $y$ is the height at time $t$.

SolutionPhysics tells us that the horizontal motion of the projectile is linear; that is, the horizontal speed of the projectile is constant. Since the object travels 192ft in 6s, we deduce that the object is moving horizontally at a rate of 32ft/s, giving the equation $x=32t$. As $y=-x^2/64+3x$, we find $y=-16t^2+96t$. We can quickly verify that $y^{\prime \prime }=-32$ft/s${}^2$, the acceleration due to gravity, and that the projectile reaches its maximum at $t=3$, halfway along its path.

These parametric equations make certain determinations about the object’s location easy: 2 seconds into the flight the object is at the point $(x(2),y(2))=(64,128)$. That is, it has traveled horizontally 64ft and is at a height of 128ft, as shown in Figure 10.2.4.

Find a rectangular equation for the curve described by

SolutionThere is not a set way to eliminate a parameter. One method is to solve for $t$ in one equation and then substitute that value in the second. We use that technique here, then show a second, simpler method.

Starting with $x=1/(t^2+1)$, solve for $t$: $t=\pm \sqrt{1/x-1}$. Substitute this value for $t$ in the equation for $y$:

Thus $y=1-x$. One may have recognized this earlier by manipulating the equation for $y$:

This is a shortcut that is very specific to this problem; sometimes shortcuts exist and are worth looking for.

We should be careful to limit the domain of the function $y=1-x$. The parametric equations limit $x$ to values in $(0,1]$, thus to produce the same graph we should limit the domain of $y=1-x$ to the same.

The graphs of these functions are given in Figure 10.2.5. The portion of the graph defined by the parametric equations is given in a thick line; the graph defined by $y=1-x$ with unrestricted domain is given in a thin line.

Eliminate the parameter in $x=4\cos t+3$, $y=2\sin t+1$

SolutionWe should not try to solve for $t$ in this situation as the resulting algebra/trig would be messy. Rather, we solve for $\cos t$ and $\sin t$ in each equation, respectively. This gives

The Pythagorean Theorem gives ${\cos }^{2}t+{\sin }^{2}t=1$, so:

This final equation should look familiar — it is the equation of an ellipse. Figure 10.2.6 plots the parametric equations, demonstrating that the graph is indeed of an ellipse with a horizontal major axis and center at $(3,1)$.

Sketch the graph of the parametric equations $x=t^2+t$, $y=t^2-t$. Find new parametric equations that shift this graph to the right 3 units and down 2. ^††margin: (a) (b) Figure 10.2.7: Illustrating how to shift graphs in Example 10.2.10. Λ

SolutionWe see the graph in Figure 10.2.7(a). It is a parabola with an axis of symmetry along the line $y=x$; the vertex is at $(0,0)$. It should be noted that finding the vertex is not a trivial matter and not something you will be asked to do in this text.

In order to shift the graph to the right 3 units, we need to increase the $x$-value by 3 for every point. The straightforward way to accomplish this is simply to add 3 to the function defining $x$: $x=t^2+t+3$. To shift the graph down by 2 units, we wish to decrease each $y$-value by 2, so we subtract 2 from the function defining $y$: $y=t^2-t-2$. Thus our parametric equations for the shifted graph are $x=t^2+t+3$, $y=t^2-t-2$. This is graphed in Figure 10.2.7 (b). Notice how the vertex is now at $(3,-2)$.

Plot the parametric functions $x=t^3-5t^2+3t+11$ and $y=t^2-2t+3$ and determine the $t$-values where the graph crosses itself.

SolutionUsing the methods developed in this section, we again plot points and graph the parametric equations as shown in Figure 10.2.8. It appears that the graph crosses itself at the point $(2,6)$, but we’ll need to analytically determine this.

We are looking for two different values, say, $s$ and $t$, where $x(s)=x(t)$ and $y(s)=y(t)$. That is, the $x$-values are the same precisely when the $y$-values are the same. This gives us a system of 2 equations with 2 unknowns: ^††margin: Figure 10.2.8: A graph of the parametric equations from Example 10.2.11. Λ

Solving this system is not trivial but involves only algebra. Using the quadratic formula, one can solve for $t$ in the second equation and find that $t=1\pm \sqrt{s^2-2s+1}$. This can be substituted into the first equation, revealing that the graph crosses itself at $t=-1$ and $t=3$. We confirm our result by computing $x(-1)=x(3)=2$ and $y(-1)=y(3)=6$.

Let a curve $C$ be defined by the parametric equations $x=t^3-12t+17$ and $y=t^2-4t+8$. Determine the points, if any, where it is not smooth.

SolutionWe begin by taking derivatives. ^††margin: Figure 10.2.9: Graphing the curve in Example 10.2.12; note it is not smooth at $(1,4)$. Λ

We set each equal to 0:

We consider only the value of $t=2$ since both $x^{\prime }$ and $y^{\prime }$ must be 0. Thus $C$ is not smooth at $t=2$, corresponding to the point $(1,4)$. The curve is graphed in Figure 10.2.9, illustrating the cusp at $(1,4)$.

A well-known parametric curve is the cycloid. Fix $r$, and let $x=r(t-\sin t)$, $y=r(1-\cos t)$. This represents the path traced out by a point on a wheel of radius $r$ as starts rolling to the right. We can think of $t$ as the angle through which the point has rotated.

Figure 10.2.10 shows a cycloid sketched out with the wheel shown at various places. The dot on the rim is the point on the wheel that we’re using to trace out the curve.

From this sketch we can see that one arch of the cycloid is traced out in the range $0\le t\le 2\pi$. This makes sense when you consider that the point will be back on the ground after it has rotated through an angle of $2\pi$.