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Parametric Equations

Plot a curve described by parametric equations.

Parametric Equations and Their Graphs

Consider the orbit of Earth around the Sun. Our year lasts approximately 365.25 days, but for this discussion we will use 365 days. On January 1 of each year, the physical location of Earth with respect to the Sun is nearly the same, except for leap years, when the lag introduced by the extra \(\frac{1}{4}\) day of orbiting time is built into the calendar. We call January 1 “day 1” of the year. Then, for example, day 31 is January 31, day 59 is February 28, and so on.

The number of the day in a year can be considered a variable that determines Earth’s position in its orbit. As Earth revolves around the Sun, its physical location changes relative to the Sun. After one full year, we are back where we started, and a new year begins. According to Kepler’s laws of planetary motion, the shape of the orbit is elliptical, with the Sun at one focus of the ellipse. We study this idea in more detail in Conic Sections.

depicts Earth’s orbit around the Sun during one year. The point labeled \({F}_{2}\) is one of the foci of the ellipse; the other focus is occupied by the Sun. If we superimpose coordinate axes over this graph, then we can assign ordered pairs to each point on the ellipse (). Then each x value on the graph is a value of position as a function of time, and each y value is also a value of position as a function of time. Therefore, each point on the graph corresponds to a value of Earth’s position as a function of time.

We can determine the functions for \(x(t)\) and \(y(t),\) thereby parameterizing the orbit of Earth around the Sun. The variable \(t\) is called an independent parameter and, in this context, represents time relative to the beginning of each year.

A curve in the \((x,y)\) plane can be represented parametrically. The equations that are used to define the curve are called parametric equations.

Condensed — the full section is in OpenStax Calculus Volume 3.

Eliminating the Parameter

To better understand the graph of a curve represented parametrically, it is useful to rewrite the two equations as a single equation relating the variables x and y. Then we can apply any previous knowledge of equations of curves in the plane to identify the curve. For example, the equations describing the plane curve in b. are

\[x(t)={t}^{2}-3,\ y(t)=2t+1,\ -2\le t\le 3.\]

Solving the second equation for t gives

\[t=\frac{y-1}{2}.\]

This can be substituted into the first equation:

\[x={(\frac{y-1}{2})}^{2}-3=\frac{{y}^{2}-2y+1}{4}-3=\frac{{y}^{2}-2y-11}{4}.\]

This equation describes x as a function of y. These steps give an example of eliminating the parameter. The graph of this function is a parabola opening to the right. Recall that the plane curve started at \((1,-3)\) and ended at \((6,7).\) These terminations were due to the restriction on the parameter t.

So far we have seen the method of eliminating the parameter, assuming we know a set of parametric equations that describe a plane curve. What if we would like to start with the equation of a curve and determine a pair of parametric equations for that curve? This is certainly possible, and in fact it is possible to do so in many different ways for a given curve. The process is known as parameterization of a curve.

Example

Try it.

Find two different pairs of parametric equations to represent the graph of \(y=2{x}^{2}-3.\)

Solution

First, it is always possible to parameterize a curve by defining \(x(t)=t,\) then replacing x with t in the equation for \(y(t).\) This gives the parameterization

\[x(t)=t,\ y(t)=2{t}^{2}-3.\]

Since there is no restriction on the domain in the original graph, there is no restriction on the values of t.

We have complete freedom in the choice for the second parameterization. For example, we can choose \(x(t)=3t-2.\) The only thing we need to check is that there are no restrictions imposed on x; that is, the range of \(x(t)\) is all real numbers. This is the case for \(x(t)=3t-2.\) Now since \(y=2{x}^{2}-3,\) we can substitute \(x(t)=3t-2\) for x. This gives

\[\begin{array}{ll}y(t) & =2{(3t-2)}^{2}-3 \\ & =2(9{t}^{2}-12t+4)-3 \\ & =18{t}^{2}-24t+8-3 \\ & =18{t}^{2}-24t+5.\end{array}\]

Therefore, a second parameterization of the curve can be written as

\[x(t)=3t-2\ \text{and}\ y(t)=18{t}^{2}-24t+5.\]

Condensed — the full section is in OpenStax Calculus Volume 3.

Cycloids and Other Parametric Curves

Imagine going on a bicycle ride through the country. The tires stay in contact with the road and rotate in a predictable pattern. Now suppose a very determined ant is tired after a long day and wants to get home. So he hangs onto the side of the tire and gets a free ride. The path that this ant travels down a straight road is called a cycloid (). A cycloid generated by a circle (or bicycle wheel) of radius a is given by the parametric equations

\[x(t)=a(t-\text{sin}\ t),\ y(t)=a(1-\text{cos}\ t).\]

To see why this is true, consider the path that the center of the wheel takes. The center moves along the x-axis at a constant height equal to the radius of the wheel. If the radius is a, then the coordinates of the center can be given by the equations

\[x(t)=at,\ y(t)=a\]

for any value of \(t.\) Next, consider the ant, which rotates around the center along a circular path. If the bicycle is moving from left to right then the wheels are rotating in a clockwise direction. A possible parameterization of the circular motion of the ant (relative to the center of the wheel) is given by

\[x(t)=\text{-}a\ \text{sin}\ t,\ y(t)=\text{-}a\ \text{cos}\ t.\]

(The negative sign is needed to reverse the orientation of the curve. If the negative sign were not there, we would have to imagine the wheel rotating counterclockwise.) Adding these equations together gives the equations for the cycloid.

\[x(t)=a(t-\text{sin}\ t),\ y(t)=a(1-\text{cos}\ t).\]

Now suppose that the bicycle wheel doesn’t travel along a straight road but instead moves along the inside of a larger wheel, as in . In this graph, the green circle is traveling around the blue circle in a counterclockwise direction. A point on the edge of the green circle traces out the red graph, which is called a hypocycloid.

The general parametric equations for a hypocycloid are

\[\begin{array}{l} \\ \\ x(t)=(a-b)\ \text{cos}\ t+b\ \text{cos}(\frac{a-b}{b})\ t \\ y(t)=(a-b)\ \text{sin}\ t-b\ \text{sin}(\frac{a-b}{b})\ t.\end{array}\]

These equations are a bit more complicated, but the derivation is somewhat similar to the equations for the cycloid. In this case we assume the radius of the larger circle is a and the radius of the smaller circle is b. Then the center of the wheel travels along a circle of radius \(a-b.\) This fact explains the first term in each equation above. The period of the second trigonometric function in both \(x(t)\) and \(y(t)\) is equal to \(\frac{2\pi b}{a-b}.\)

Condensed — the full section is in OpenStax Calculus Volume 3.

Key Concepts

Parametric equations provide a convenient way to describe a curve. A parameter can represent time or some other meaningful quantity.
It is often possible to eliminate the parameter in a parameterized curve to obtain a function or relation describing that curve.
There is always more than one way to parameterize a curve.
Parametric equations can describe complicated curves that are difficult or perhaps impossible to describe using rectangular coordinates.

Parametric Equations

For the following exercises, sketch the curves below by eliminating the parameter t. Give the orientation of the curve.

For the following exercises, eliminate the parameter and sketch the graphs.

For the following exercises, use technology (CAS or calculator) to sketch the parametric equations.

For the following exercises, sketch the parametric equations by eliminating the parameter. Indicate any asymptotes of the graph.

For the following exercises, convert the parametric equations of a curve into rectangular form. No sketch is necessary. State the domain of the rectangular form.

For the following exercises, the pairs of parametric equations represent lines, parabolas, circles, ellipses, or hyperbolas. Name the type of basic curve that each pair of equations represents.

For the following exercises, use a graphing utility to graph the curve represented by the parametric equations and identify the curve from its equation.

Parametric Equations and Their Graphs

A curve in the \((x,y)\) plane can be represented parametrically. The equations that are used to define the curve are called parametric equations.

Condensed — the full section is in OpenStax Calculus Volume 2.

Eliminating the Parameter

\[x(t)={t}^{2}-3,\ y(t)=2t+1,\ -2\le t\le 3.\]

Solving the second equation for t gives

\[t=\frac{y-1}{2}.\]

This can be substituted into the first equation:

\[x={(\frac{y-1}{2})}^{2}-3=\frac{{y}^{2}-2y+1}{4}-3=\frac{{y}^{2}-2y-11}{4}.\]

Example

Try it.

Find two different pairs of parametric equations to represent the graph of \(y=2{x}^{2}-3.\)

Solution

First, it is always possible to parameterize a curve by defining \(x(t)=t,\) then replacing x with t in the equation for \(y(t).\) This gives the parameterization

\[x(t)=t,\ y(t)=2{t}^{2}-3.\]

Since there is no restriction on the domain in the original graph, there is no restriction on the values of t.

\[\begin{array}{ll}y(t) & =2{(3t-2)}^{2}-3 \\ & =2(9{t}^{2}-12t+4)-3 \\ & =18{t}^{2}-24t+8-3 \\ & =18{t}^{2}-24t+5.\end{array}\]

Therefore, a second parameterization of the curve can be written as

\[x(t)=3t-2\ \text{and}\ y(t)=18{t}^{2}-24t+5.\]

Condensed — the full section is in OpenStax Calculus Volume 2.

Cycloids and Other Parametric Curves

\[x(t)=a(t-\text{sin}\ t),\ y(t)=a(1-\text{cos}\ t).\]

\[x(t)=at,\ y(t)=a\]

\[x(t)=\text{-}a\ \text{sin}\ t,\ y(t)=\text{-}a\ \text{cos}\ t.\]

\[x(t)=a(t-\text{sin}\ t),\ y(t)=a(1-\text{cos}\ t).\]

The general parametric equations for a hypocycloid are

\[\begin{array}{l} \\ \\ x(t)=(a-b)\ \text{cos}\ t+b\ \text{cos}(\frac{a-b}{b})\ t \\ y(t)=(a-b)\ \text{sin}\ t-b\ \text{sin}(\frac{a-b}{b})\ t.\end{array}\]

Condensed — the full section is in OpenStax Calculus Volume 2.

Key Concepts

Parametric equations provide a convenient way to describe a curve. A parameter can represent time or some other meaningful quantity.
It is often possible to eliminate the parameter in a parameterized curve to obtain a function or relation describing that curve.
There is always more than one way to parameterize a curve.
Parametric equations can describe complicated curves that are difficult or perhaps impossible to describe using rectangular coordinates.

Parametric Equations

For the following exercises, sketch the curves below by eliminating the parameter t. Give the orientation of the curve.

For the following exercises, eliminate the parameter and sketch the graphs.

For the following exercises, use technology (CAS or calculator) to sketch the parametric equations.

For the following exercises, sketch the parametric equations by eliminating the parameter. Indicate any asymptotes of the graph.

For the following exercises, convert the parametric equations of a curve into rectangular form. No sketch is necessary. State the domain of the rectangular form.

For the following exercises, the pairs of parametric equations represent lines, parabolas, circles, ellipses, or hyperbolas. Name the type of basic curve that each pair of equations represents.

For the following exercises, use a graphing utility to graph the curve represented by the parametric equations and identify the curve from its equation.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Sketch the curves described by the following parametric equations:

\(x(t)=t-1,\ y(t)=2t+4,\ -3\le t\le 2\)
\(x(t)={t}^{2}-3,\ y(t)=2t+1,\ -2\le t\le 3\)
\(x(t)=4\ \text{cos}\ t,\ y(t)=4\ \text{sin}\ t,\ 0\le t\le 2\pi\)

Жавобни кўрсатиш

To create a graph of this curve, first set up a table of values. Since the independent variable in both \(x(t)\) and \(y(t)\) is t, let t appear in the first column. Then \(x(t)\) and \(y(t)\) will appear in the second and third columns of the table.

t \(x(t)\) \(y(t)\)
−3 −4 −2
−2 −3 0
−1 −2 2
0 −1 4
1 0 6
2 1 8

The second and third columns in this table provide a set of points to be plotted. The graph of these points appears in . The arrows on the graph indicate the orientation of the graph, that is, the direction that a point moves on the graph as t varies from −3 to 2.
To create a graph of this curve, again set up a table of values.

t \(x(t)\) \(y(t)\)
−2 1 −3
−1 −2 −1
0 −3 1
1 −2 3
2 1 5
3 6 7

The second and third columns in this table give a set of points to be plotted (). The first point on the graph (corresponding to \(t=-2)\) has coordinates \((1,-3),\) and the last point (corresponding to \(t=3)\) has coordinates \((6,7).\) As t progresses from −2 to 3, the point on the curve travels along a parabola. The direction the point moves is again called the orientation and is indicated on the graph.

In this case, use multiples of \(\pi \text{/}6\) for t and create another table of values:

t	\(x(t)\)	\(y(t)\)		t	\(x(t)\)	\(y(t)\)
0	4	0		\(\frac{7\pi }{6}\)	\(-2\sqrt{3}\approx -3.5\)	2
\(\frac{\pi }{6}\)	\(2\sqrt{3}\approx 3.5\)	\(2\)	\(\frac{4\pi }{3}\)	−2	\(-2\sqrt{3}\approx -3.5\)
\(\frac{\pi }{3}\)	\(2\)	\(2\sqrt{3}\approx 3.5\)	\(\frac{3\pi }{2}\)	0	−4
\(\frac{\pi }{2}\)	0	4	\(\frac{5\pi }{3}\)	2	\(-2\sqrt{3}\approx -3.5\)
\(\frac{2\pi }{3}\)	−2	\(2\sqrt{3}\approx 3.5\)	\(\frac{11\pi }{6}\)	\(2\sqrt{3}\approx 3.5\)	2
\(\frac{5\pi }{6}\)	\(-2\sqrt{3}\approx -3.5\)	2	\(2\pi\)	4	0
\(\pi\)	−4	0

The graph of this plane curve appears in the following graph.

This is the graph of a circle with radius 4 centered at the origin, with a counterclockwise orientation. The starting point and ending points of the curve both have coordinates \((4,0).\)

Sketch the curve described by the parametric equations
\[x(t)=3t+2,\ y(t)={t}^{2}-1,\ -3\le t\le 2.\]

Жавобни кўрсатиш
Eliminate the parameter for each of the plane curves described by the following parametric equations and describe the resulting graph.
1. \(x(t)=\sqrt{2t+4},\ y(t)=2t+1,\ -2\le t\le 6\)
2. \(x(t)=4\ \text{cos}\ t,\ y(t)=3\ \text{sin}\ t,\ 0\le t\le 2\pi\)
Жавобни кўрсатиш
1. To eliminate the parameter, we can solve either of the equations for t. For example, solving the first equation for t gives
  \[\begin{array}{lll}x & = & \sqrt{2t+4} \\ {x}^{2} & = & 2t+4 \\ {x}^{2}-4 & = & 2t \\ t & = & \frac{{x}^{2}-4}{2}.\end{array}\]
  Note that when we square both sides it is important to observe that \(x\ge 0.\) Substituting \(t=\frac{{x}^{2}-4}{2}\) this into \(y(t)\) yields
  \[\begin{array}{lll}y(t) & = & 2t+1 \\ y & = & 2(\frac{{x}^{2}-4}{2})+1 \\ y & = & {x}^{2}-4+1 \\ y & = & {x}^{2}-3.\end{array}\]
  This is the equation of a parabola opening upward. There is, however, a domain restriction because of the limits on the parameter t. When \(t=-2,\) \(x=\sqrt{2(-2)+4}=0,\) and when \(t=6,\) \(x=\sqrt{2(6)+4}=4.\) The graph of this plane curve follows.
2. Sometimes it is necessary to be a bit creative in eliminating the parameter. The parametric equations for this example are
  \[x(t)=4\ \text{cos}\ t\ \text{and}\ y(t)=3\ \text{sin}\ t.\]
  Solving either equation for t directly is not advisable because sine and cosine are not one-to-one functions. However, dividing the first equation by 4 and the second equation by 3 (and suppressing the t) gives us
  \[\text{cos}\ t=\frac{x}{4}\ \text{and}\ \text{sin}\ t=\frac{y}{3}.\]
  Now use the Pythagorean identity \({\text{cos}}^{2}t+{\text{sin}}^{2}t=1\) and replace the expressions for \(\text{sin}\ t\) and \(\text{cos}\ t\) with the equivalent expressions in terms of x and y. This gives
  \[\begin{array}{lll}{(\frac{x}{4})}^{2}+{(\frac{y}{3})}^{2} & = & 1 \\ \frac{{x}^{2}}{16}+\frac{{y}^{2}}{9} & = & 1.\end{array}\]
  This is the equation of a horizontal ellipse centered at the origin, with semimajor axis 4 and semiminor axis 3 as shown in the following graph.
  
  As t progresses from \(0\) to \(2\pi ,\) a point on the curve traverses the ellipse once, in a counterclockwise direction. Recall from the section opener that the orbit of Earth around the Sun is also elliptical. This is a perfect example of using parameterized curves to model a real-world phenomenon.
Eliminate the parameter for the plane curve defined by the following parametric equations and describe the resulting graph.
\[x(t)=2+\frac{3}{t},\ y(t)=t-1,\ 2\le t\le 6\]

Жавобни кўрсатиш

\(x=2+\frac{3}{y+1},\) or \(y=-1+\frac{3}{x-2}.\) This equation describes a portion of a rectangular hyperbola centered at \((2,-1).\)
Find two different pairs of parametric equations to represent the graph of \(y=2{x}^{2}-3.\)

Жавобни кўрсатиш

First, it is always possible to parameterize a curve by defining \(x(t)=t,\) then replacing x with t in the equation for \(y(t).\) This gives the parameterization
\[x(t)=t,\ y(t)=2{t}^{2}-3.\]
Since there is no restriction on the domain in the original graph, there is no restriction on the values of t.
We have complete freedom in the choice for the second parameterization. For example, we can choose \(x(t)=3t-2.\) The only thing we need to check is that there are no restrictions imposed on x; that is, the range of \(x(t)\) is all real numbers. This is the case for \(x(t)=3t-2.\) Now since \(y=2{x}^{2}-3,\) we can substitute \(x(t)=3t-2\) for x. This gives
\[\begin{array}{ll}y(t) & =2{(3t-2)}^{2}-3 \\ & =2(9{t}^{2}-12t+4)-3 \\ & =18{t}^{2}-24t+8-3 \\ & =18{t}^{2}-24t+5.\end{array}\]
Therefore, a second parameterization of the curve can be written as
\[x(t)=3t-2\ \text{and}\ y(t)=18{t}^{2}-24t+5.\]
Find two different sets of parametric equations to represent the graph of \(y={x}^{2}+2x.\)

Жавобни кўрсатиш

One possibility is \(x(t)=t,\ y(t)={t}^{2}+2t.\) Another possibility is \(x(t)=2t-3,\ y(t)={(2t-3)}^{2}+2(2t-3)=4{t}^{2}-8t+3.\)
There are, in fact, an infinite number of possibilities.
\(x={t}^{2}+2t,\) \(y=t+1\)

Жавобни кўрсатиш

orientation: bottom to top
\(x=\text{cos}(t),y=\text{sin}(t),(0,2\pi ]\)
\(x=2t+4,y=t-1\)

Жавобни кўрсатиш

orientation: left to right
\(x=3-t,y=2t-3,1.5\le t\le 3\)
\(x=2{t}^{2},\ y={t}^{4}+1\)

Жавобни кўрсатиш

\(y=\frac{{x}^{2}}{4}+1\)
[T] \(\begin{array}{ll}x={t}^{2}+t, & y={t}^{2}-1\end{array}\)
[T] \(\begin{array}{ll}x={e}^{\text{-}t}, & y={e}^{2t}-1\end{array}\)

Жавобни кўрсатиш
[T] \(\begin{array}{ll}x=3\ \text{cos}\ t, & y=4\ \text{sin}\ t\end{array}\)
[T] \(\begin{array}{ll}x=\text{sec}\ t, & y=\text{cos}\ t\end{array}\)

Жавобни кўрсатиш
\(x={e}^{t},\ y={e}^{2t}+1\)
\(x=6\ \text{sin}(2\theta ),y=4\ \text{cos}(2\theta )\)

Жавобни кўрсатиш
\(\begin{array}{ll}x=\text{cos}\ \theta , & y=2\ \text{sin}(2\theta )\end{array}\)
\(\begin{array}{ll}x=3-2\ \text{cos}\ \theta , & y=-5+3\ \text{sin}\ \theta \end{array}\)

Жавобни кўрсатиш
\(\begin{array}{ll}x=4+2\ \text{cos}\ \theta , & y=-1+\text{sin}\ \theta \end{array}\)
\(\begin{array}{ll}x=\text{sec}\ t, & y=\text{tan}\ t\end{array}\)

Жавобни кўрсатиш

Asymptotes are \(y=x\) and \(y=\text{-}x\)
\(\begin{array}{ll}x=\text{ln}(2t), & y={t}^{2}\end{array}\)
\(\begin{array}{ll}x={e}^{t}, & y={e}^{2t}\end{array}\)

Жавобни кўрсатиш
\(\begin{array}{ll}x={e}^{-2t}, & y={e}^{3t}\end{array}\)
\(\begin{array}{ll}x={t}^{3}, & y=3\ \text{ln}\ t\end{array}\)

Жавобни кўрсатиш
\(\begin{array}{ll}x=4\ \text{sec}\ \theta , & y=3\ \text{tan}\ \theta \end{array}\)
\(\begin{array}{ll}x={t}^{2}-1, & y=\frac{t}{2}\end{array}\)

Жавобни кўрсатиш

\(y=\frac{\sqrt{x+1}}{2}\); domain: \(x\in [-1,\infty ).\)
\(\begin{array}{ll}x=\frac{1}{\sqrt{t+1}}, & y=\frac{t}{1+t},t>-1\end{array}\)
\(x=4\ \text{cos}\ \theta ,y=3\ \text{sin}\ \theta ,\theta \in (0,2\pi ]\)

Жавобни кўрсатиш

\(\frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=1;\) domain \(x\in [-4,4].\)
\(\begin{array}{ll}x=\text{cosh}\ t, & y=\text{sinh}\ t\end{array}\)
\(\begin{array}{ll}x=2t-3, & y=6t-7\end{array}\)

Жавобни кўрсатиш

\(y=3x+2;\) domain: all real numbers.
\(\begin{array}{ll}x={t}^{2}, & y={t}^{3}\end{array}\)
\(\begin{array}{ll}x=1+\text{cos}\ t, & y=3-\text{sin}\ t\end{array}\)

Жавобни кўрсатиш

\({(x-1)}^{2}+{(y-3)}^{2}=1;\) domain: \(x\in [0,2].\)
\(\begin{array}{ll}x=\sqrt{t}, & y=2t+4\end{array}\)
\(\begin{array}{ll}x=\text{sec}\ t, & y=\text{tan}\ t,\pi \le t<\frac{3\pi }{2}\end{array}\)

Жавобни кўрсатиш

\(y=\sqrt{{x}^{2}-1};\) domain: \(x\in (-\infty ,-1].\)
\(\begin{array}{ll}x=2\ \text{cosh}\ t, & y=4\ \text{sinh}\ t\end{array}\)
\(\begin{array}{ll}x=\text{cos}(2t), & y=\text{sin}\ t\end{array}\)

Жавобни кўрсатиш

\({y}^{2}=\frac{1-x}{2};\) domain: \(x\in [-1,1].\)
\(x=4t+3,y=16{t}^{2}-9\)
\(\begin{array}{ll}x={t}^{2}, & y=2\ \text{ln}\ t,t\ge 1\end{array}\)

Жавобни кўрсатиш

\(y=\text{ln}\ x;\) domain: \(x\in [1,\infty ).\)
\(\begin{array}{ll}x={t}^{3}, & y=3\ \text{ln}\ t,t\ge 1\end{array}\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

\theta

theta
The usual name for an angle.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\frac{\partial f}{\partial x}

partial derivative
Derivative with respect to x, holding the other variables fixed.

\nabla f

gradient (nabla, del)
Vector of partial derivatives; points uphill.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

\iint,\ \oint

double / contour integral
Integral over a region of the plane; integral around a closed curve.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|

dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: Parametric Equations

Plot a curve described by parametric equations.
Convert the parametric equations of a curve into the form
Recognize the parametric equations of basic curves, such as a line and a circle.
Recognize the parametric equations of a cycloid.
To create a graph of this curve, first set up a table of values. Since the independent variable in both
To create a graph of this curve, again set up a table of values.
In this case, use multiples of
To eliminate the parameter, we can solve either of the equations for

Questions people ask

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

Ўзингизни синаб кўринг

Parts of this page are adapted from OpenStax Calculus Volume 2 (CC BY-NC-SA 4.0), OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Кўпроқ Multivariable Calculus

Functions of several variables Partial derivatives and the gradient Optimisation in several variables Double and triple integrals Vector fields, line integrals and the big theorems