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# Parametric Curves – Definition, Graphs, and Examples

Learning about**parametric curves**will give us one more with special attributes (time to be specific). There are instances when modeling quantities using parametric curves is more helpful than graphing them in the coordinate systems that we know – rectangular and polar coordinate systems. This is why itâ€™s important that we learn how to graph and interpret parametric curves.

**In this article, weâ€™ll take a look back at what we know of parametric equations so far. Weâ€™ll also establish the fundamentals needed to parametrize and graph parametric curves. By the end of this discussion, youâ€™ll have the essential tool kits to graph common plane curves to their parametric form. Weâ€™ll also show you how some conics in a planar curve can be defined by simpler parametric curves.**

*Parametric curves allow us to graph relationships between two or more quantities and at the same time represent each quantityâ€™s directions or orientations.***What is a parametric curve?**

**The parametric curve**is

**defined by its corresponding parametric equations**: $x = f(t)$ and $y = g(t)$ within a given interval. Parametric curves highlight the

**orientation of each set of quantities with respect to time**. In the rectangular coordinate system, we are limited to defining functions, $y = f(x)$, that pass the vertical line test. If youâ€™ve noticed, weâ€™ve been working with graphs such as the circle or ellipse that clearly do not pass the vertical line test.

**parametric curve**. Now how does $t$ affect our curve? The new parameter, $\boldsymbol{t}$,

**determines the curveâ€™s direction**.

Parametric Curves and Its Endpoints |

\begin{aligned}x &= f(t)\\y &= g(t)\end{aligned}
If we have the two plane curves redefined as parametric curves within the interval, $[a, b]$, the parametric curve will have initial points and terminal points at $(x(a),y(a))$ and $(x(b), y(b))$, respectively. |

- It is a great model that highlights the direction and orientation of $(x, y)$ based on $t$. This means that we now have a way to model physical representations of quantities that depend on time.
- It is much easier to adjust the parametric curve if we want to change the value of $t$ or we want to switch the orientations.
- There are instances when curves such as implicit curves can only be modeled by parametric equations (ie modeling both the volume of a container and its spillage simultaneously with respect to time).

**How to parametrize a curve?**

When given a plane curve, we can parametric it to a parametric curve by redefining $x$ and $y$ as a set of parametric equations defined by $t$. There are infinitely many ways to parametrize a given curve â€“ what matters is that their definition as a plane curve remains the same. The simplest way to parametrize a curve is by setting $x = t$.
PARAMETRIZING A PLANE CURVE
The most straightforward option when parametrizing a plane curve defined by $y = f(x)$ is to use the following parametric equations:
\begin{aligned}x &= t\\y&= f(t)\end{aligned}
Keep in mind that $t$ has to be within the domains of $f(x)$. |

PARAMETRIZING CIRCLES
We can parametrize a circle centered at the origin using the following parametric equations:
\begin{aligned} x &= r\cos t \\ y &= r \sin t \end{aligned}
Circles centered at $\boldsymbol{(h, k)}$ using the following parametric equations:
\begin{aligned} x &= h + r\cos t \\ y &= k + r \sin t\end{aligned}
For both cases, keep in mind that $0 \leq t \leq 2\pi$. |

**eliminating the parameters.**We simply apply our algebraic technique to

**eliminate**$\boldsymbol{t}$

**in the equations**, $x = f(t)$ and $y = g(t)$.

- Isolate $t$ from one of the parametric equations.
- Use the resulting expression (either in terms of $x$ or $y$) to eliminate the parameter in the remaining parametric equation.

**How to graph a parametric curve?**

Now that we know how to parametrize curves, itâ€™s time that we establish the steps we need to graph parametric curves. Below are helpful guidelines for you to remember when y graphing parametric curves given their parametric equations:
- Assign some key values of $t$ within the given interval.
- Use the parametric equations, $x = f(t)$ and $y= g(t)$, to find the corresponding ordered pairs, $(x, y)$.
- Plot the ordered pairs in order â€“ start with the ordered pair returned by the lowest value of $t$ up to the highest value of $t$.

\begin{aligned}\boldsymbol{t}\end{aligned} | \begin{aligned}\boldsymbol{x = t^2 – 4}\end{aligned} | \begin{aligned}\boldsymbol{y = 4t}\end{aligned} | \begin{aligned}\boldsymbol{(, y)}\end{aligned} |

\begin{aligned}-4\end{aligned} | \begin{aligned}x= (-4)^2 â€“ 4 = 12\end{aligned} | \begin{aligned}y = 4(-4) = -16\end{aligned} | \begin{aligned} (12, -16)\end{aligned} |

\begin{aligned}-2\end{aligned} | \begin{aligned}x= (-2)^2 â€“ 4 = 0\end{aligned} | \begin{aligned}y = 4(-2) = -8\end{aligned} | \begin{aligned} (0, -8)\end{aligned} |

\begin{aligned}0\end{aligned} | \begin{aligned}x = 0^2 â€“ 4 = -4\end{aligned} | \begin{aligned}y = 4(0) = 0\end{aligned} | \begin{aligned} (-4, 0)\end{aligned} |

\begin{aligned}2\end{aligned} | \begin{aligned}x = 2^2 â€“ 4 = 0\end{aligned} | \begin{aligned}y = 4(2) = 8\end{aligned} | \begin{aligned} (0, 8)\end{aligned} |

\begin{aligned}4\end{aligned} | \begin{aligned}x = 4^2 â€“ 4 = -12\end{aligned} | \begin{aligned}y = 4(4) = 16\end{aligned} | \begin{aligned} (-12, 16)\end{aligned} |

**How to find the derivative of parametric curves?**

We can find the parametric curvesâ€™ derivative by applying the chain rule and other derivative rules. Now that we have explored the basic concepts of parametric curves, letâ€™s take a deeper understanding by seeing parametric curves in the context of calculus.
We begin by assuming that $\dfrac{dx}{dt} = f^{\prime}(t) $ and $\dfrac{dy}{dt} = g^{\prime}(t)$ exist and that $f^{\prime}(t) \neq 0$. This means that we can express $\dfrac{d}{dx} g(x) = \dfrac{dy}{dx}$ as shown below.
\begin{aligned}\dfrac{dy}{dx} &= \dfrac{dy/ dt}{dx/ dt}\\&= \dfrac{g^{\prime}(t) }{f^{\prime}(t)}\end{aligned}
DERIVATIVE OF PARAMETRIC CURVE
\begin{aligned}x & = f(t)\\ y&= g(t) \end{aligned}
We can find the derivative of the parametric curve represented by the parametric equations shown above by dividing the derivative of $g(t)$ by the derivative of $f(t)$- both with respect to $t$.
\begin{aligned}\dfrac{dy}{dx} &= \dfrac{g^{\prime}(t)}{f^{\prime}(t)}\end{aligned} |

**where**$\boldsymbol{\dfrac{dy}{dx}}$

**is equal to zero or are undefined**. We can use the derivative of parametric curves as their critical points to graph parametric curves faster. Why donâ€™t we try to find the derivative of the parametric curve , $x = t^2 â€“ 4$ and $y = 4t$? Â Take their individual derivatives with respect to $t$ then use thhe formula, $\dfrac{dy}{dx} = \dfrac{g^{\prime}(t)}{f^{\prime}(t)}$.

\begin{aligned} f^{\prime}(t) &= \dfrac{d}{dt} (t^2 – 4) \\ &= 2t^{2 – 1} – 0\\&= 2t \\g^{\prime}(t) &= \dfrac{d}{dt} (4t) \\&= 4(1) \\&= 4 \end{aligned} | \begin{aligned} \dfrac{dy}{dx} &= \dfrac{g^{\prime}(t)}{f^{\prime}(t)} \\&= \dfrac{4}{2t}\\ &= \dfrac{2}{t}\end{aligned} |

**Graph the curve defined by the parametric equations $x = 2\sqrt{t}$ and $y = t- 2$, where $t \geq 0$. Include arrows showing the orientation of the resulting parametric curve.**

*Example 1*__Solution__We begin by assigning integral values of $t$ that are greater than or equal to $0$. Evaluate each of the values at $x = 2\sqrt{t}$ and $y = t- 2$ to find some ordered pairs that will help us in plotting the parametric curve. Hereâ€™s the summary of our calculations when $t = \{0, 1, 4, 9, 16\}$.

\begin{aligned}\boldsymbol{t}\end{aligned} | \begin{aligned}\boldsymbol{x = 2\sqrt{t}}\end{aligned} | \begin{aligned}\boldsymbol{y = t- 2}\end{aligned} | \begin{aligned}\boldsymbol{(, y)}\end{aligned} |

\begin{aligned} 0\end{aligned} | \begin{aligned}x = 2\sqrt{0} = 0\end{aligned} | \begin{aligned}y = 0 -2= -2\end{aligned} | \begin{aligned} (0, -2)\end{aligned} |

\begin{aligned}1\end{aligned} | \begin{aligned}x = 2\sqrt{1} = 2\end{aligned} | \begin{aligned}y = 1 -2= -1\end{aligned} | \begin{aligned} (2, -1)\end{aligned} |

\begin{aligned}4\end{aligned} | \begin{aligned}x = 2\sqrt{4} = 4\end{aligned} | \begin{aligned}y = 4 -2= 2\end{aligned} | \begin{aligned} (4, 2)\end{aligned} |

\begin{aligned}9\end{aligned} | \begin{aligned}x = 2\sqrt{9} = 6\end{aligned} | \begin{aligned}y = 9 -2= 7\end{aligned} | \begin{aligned} (6, 7)\end{aligned} |

\begin{aligned}16\end{aligned} | \begin{aligned}x = 2\sqrt{16} = 8\end{aligned} | \begin{aligned}y = 16 -2= 14\end{aligned} | \begin{aligned} (8, 14)\end{aligned} |

**Eliminate the parameter for the parametric equations, $x= \sec t$ and $y = \tan t$. Use the resulting rectangular equation to graph the plane curve. Â Include arrows showing the orientation of the resulting parametric curve.**

*Example 2*__Solution__Itâ€™ll be easier for us to eliminate the parameter if we square both parametric equations. By doing so, we can express $\sec^2 t$ in terms of $\tan^2 t$ and vice-versa.

\begin{aligned}(x)^2 &= (\sec t)^2\\x^2 &= \sec^2 t\end{aligned} | \begin{aligned}(y)^2 &= (\tan t)^2\\y^2 &= \tan^2 t\end{aligned} |

\begin{aligned}\boldsymbol{t}\end{aligned} | \begin{aligned}\boldsymbol{x = \sec t}\end{aligned} | \begin{aligned}\boldsymbol{y = t- 2}\end{aligned} | \begin{aligned}\boldsymbol{(, y)}\end{aligned} |

\begin{aligned} 0\end{aligned} | \begin{aligned}x = \sec 0 = 1\end{aligned} | \begin{aligned}y = \tan 0= 0\end{aligned} | \begin{aligned} (1, 0)\end{aligned} |

\begin{aligned}\dfrac{\pi}{3}\end{aligned} | \begin{aligned}x = \sec \dfrac{\pi}{3} =2\end{aligned} | \begin{aligned}y = \tan \dfrac{\pi}{3}= \sqrt{3}\end{aligned} | \begin{aligned} (2, 1.73)\end{aligned} |

\begin{aligned}\dfrac{2\pi}{3}\end{aligned} | \begin{aligned}x = \sec \dfrac{2\pi}{3} =-2\end{aligned} | \begin{aligned}y = \tan \dfrac{2\pi}{3}= -\sqrt{3}\end{aligned} | \begin{aligned} (-2, -1.73)\end{aligned} |

\begin{aligned} \pi\end{aligned} | \begin{aligned}x = \sec \pi = -1\end{aligned} | \begin{aligned}y = \tan \pi= 0\end{aligned} | \begin{aligned} (-1, 0)\end{aligned} |

\begin{aligned}\dfrac{4\pi}{3}\end{aligned} | \begin{aligned}x = \sec \dfrac{4\pi}{3} =-2\end{aligned} | \begin{aligned}y = \tan \dfrac{4\pi}{3}= \sqrt{3}\end{aligned} | \begin{aligned} (-2, 1.73)\end{aligned} |

\begin{aligned}\dfrac{5\pi}{3}\end{aligned} | \begin{aligned}x = \sec \dfrac{5\pi}{3} =2\end{aligned} | \begin{aligned}y = \tan \dfrac{5\pi}{3}= -\sqrt{3}\end{aligned} | \begin{aligned} (2, -1.73)\end{aligned} |

**Write down two different sets of parametric equations to parametrize the equation, $y = 3x + 5$.**

*Example 3*__Solution__Before we begin, remember that rectangular equations can be represented by multiple sets of parametric equations. This means that we may show different sets of parametric equations than what you might write down. What matters is that your set of parametric equations return $y = 3x + 5$ when we eliminate the parameter, $t$. For our two sets of parametric equations, weâ€™ll let 1)$x$ be equal to $t$ and 2)$x$ be equal to $ t +1$. Weâ€™ll substitute each of this equation into the rectangular equation.

\begin{aligned}\boldsymbol{x = t }\end{aligned} | \begin{aligned}\boldsymbol{x = t + 1 }\end{aligned} |

\begin{aligned}y &= 3(t) + 5\\&= 3t + 5 \end{aligned} | \begin{aligned}y &= 3(tÂ +1) + 5\\&= 3t + 8\end{aligned} |

**Find the derivative of the plane curve defined by the equations, $x = 2t + 1$ and $y = t^3 â€“ 27t$ where $t$ is within $[-5, 10]$, then use the result to find the plane curveâ€™s critical points.**

*Example 4*__Solution__Take the derivative of each parametric equation with respect to $t$.

\begin{aligned}\boldsymbol{x = 2t + 1 }\end{aligned} | \begin{aligned}\boldsymbol{y = t^3 -27t}\end{aligned} |

\begin{aligned}\dfrac{dx}{dt} &= 2(1) \\&= 2\end{aligned} | \begin{aligned}\dfrac{dy}{dt} &= 3t^2 â€“ 27(1) \\&= 3t^2 – 27\end{aligned} |

\begin{aligned}\boldsymbol{t=-3 }\end{aligned} | \begin{aligned}\boldsymbol{t = 3}\end{aligned} |

\begin{aligned}(x, y) &= (-5, 6) \end{aligned} | \begin{aligned}(x, y) &= (7, 18) \end{aligned} |

### Practice Questions

1. Graph the curve defined by the following parametric equations using point-plotting. Include arrows showing the orientation of the resulting parametric curve. a. $x = t + 4$, $y = 4- 2t$; $-2 \leq t \leq 4$ b. $x = t^2-3$, $y = 2t -1$; $-3 \leq t \leq 4$ c. $x =\sqrt{t}$, $y = t +4$; $ t \geq 0$ 2. Eliminate the parameter for the parametric equations, $x= 5\cos t$ and $y = 4\sin t$. Use the resulting rectangular equation to graph the plane curve. Include arrows showing the orientation of the resulting parametric curve. 3. Write down two different sets of parametric equations to parametrize the following equations. a. $y = 2x – 9$ b. $y = x^2+1$ c. $y = x^2â€“5$### Answer Key

1. a.*Images/mathematical drawings are created with GeoGebra.*

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