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Parametric Equations
Parameterize a curve.
Parameterizing a Curve
When an object moves along a curve—or curvilinear path—in a given direction and in a given amount of time, the position of the object in the plane is given by the x-coordinate and the y-coordinate. However, both \(x\) and \(y\) vary over time and so are functions of time. For this reason, we add another variable, the parameter, upon which both \(x\) and \(y\) are dependent functions. In the example in the section opener, the parameter is time, \(t.\) The \(x\) position of the moon at time, \(t,\) is represented as the function \(x(t),\) and the \(y\) position of the moon at time, \(t,\) is represented as the function \(y(t).\) Together, \(x(t)\) and \(y(t)\) are called parametric equations, and generate an ordered pair \((x(t),\ y(t)).\) Parametric equations primarily describe motion and direction.
When we parameterize a curve, we are translating a single equation in two variables, such as \(x\) and \(y,\) into an equivalent pair of equations in three variables, \(x,y,\) and \(t.\) One of the reasons we parameterize a curve is because the parametric equations yield more information: specifically, the direction of the object’s motion over time.
When we graph parametric equations, we can observe the individual behaviors of \(x\) and of \(y.\) There are a number of shapes that cannot be represented in the form \(y=f(x),\) meaning that they are not functions. For example, consider the graph of a circle, given as \({r}^{2}={x}^{2}+{y}^{2}.\) Solving for \(y\) gives \(y=\pm \sqrt{{r}^{2}-{x}^{2}},\) or two equations: \({y}_{1}=\sqrt{{r}^{2}-{x}^{2}}\) and \({y}_{2}=-\sqrt{{r}^{2}-{x}^{2}}.\) If we graph \({y}_{1}\) and \({y}_{2}\) together, the graph will not pass the vertical line test, as shown in . Thus, the equation for the graph of a circle is not a function.
However, if we were to graph each equation on its own, each one would pass the vertical line test and therefore would represent a function. In some instances, the concept of breaking up the equation for a circle into two functions is similar to the concept of creating parametric equations, as we use two functions to produce a non-function. This will become clearer as we move forward.
Condensed — the full section is in OpenStax Algebra and Trigonometry 2e.
Eliminating the Parameter
In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as \(x\) and \(y.\) Eliminating the parameter is a method that may make graphing some curves easier. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate the orientation of the curve as well. There are various methods for eliminating the parameter \(t\) from a set of parametric equations; not every method works for every type of equation. Here we will review the methods for the most common types of equations.
For polynomial, exponential, or logarithmic equations expressed as two parametric equations, we choose the equation that is most easily manipulated and solve for \(t.\) We substitute the resulting expression for \(t\) into the second equation. This gives one equation in \(x\) and \(y.\)
Example
Try it.
Given \(x(t)={t}^{2}+1\) and \(y(t)=2+t,\) eliminate the parameter, and write the parametric equations as a Cartesian equation.
Solution
We will begin with the equation for \(y\) because the linear equation is easier to solve for \(t.\)
\[\begin{array}{l}y=2+t \\ y-2=t\end{array}\]Next, substitute \(y-2\) for \(t\) in \(x(t).\)
\[\begin{array}{ll}x={t}^{2}+1 & \\ x={(y-2)}^{2}+1 & \text{Substitute the expression for }t\text{ into }x. \\ x={y}^{2}-4y+4+1 & \\ x={y}^{2}-4y+5 & \\ x={y}^{2}-4y+5 & \end{array}\]The Cartesian form is \(x={y}^{2}-4y+5.\)
Example
Try it.
Eliminate the parameter and write as a Cartesian equation: \(x(t)={e}^{-t}\) and \(y(t)=3{e}^{t}\)
Solution
Isolate \({e}^{t}.\)
\[\begin{array}{l}x={e}^{-t} \\ {e}^{t}=\frac{1}{x}\end{array}\]Substitute the expression into \(y(t).\)
\[\begin{array}{l}y=3{e}^{t} \\ y=3(\frac{1}{x}) \\ y=\frac{3}{x}\end{array}\]The Cartesian form is \(y=\frac{3}{x}.\)
Example
Try it.
Eliminate the parameter and write as a Cartesian equation: \(x(t)=\sqrt{t}+2\) and \(y(t)=\log (t).\)
Solution
Solve the first equation for \(t.\)
\[\begin{array}{ll}\ x=\sqrt{t}+2 & \\ x-2=\sqrt{t} & \\ {(x-2)}^{2}=t & \text{Square both sides}.\end{array}\]Then, substitute the expression for \(t\) into the \(y\) equation.
\[\begin{array}{l}y=\log (t) \\ y=\log {(x-2)}^{2}\end{array}\]The Cartesian form is \(y=\log {(x-2)}^{2}.\)
Condensed — the full section is in OpenStax Algebra and Trigonometry 2e.
Finding Cartesian Equations from Curves Defined Parametrically
When we are given a set of parametric equations and need to find an equivalent Cartesian equation, we are essentially “eliminating the parameter.” However, there are various methods we can use to rewrite a set of parametric equations as a Cartesian equation. The simplest method is to set one equation equal to the parameter, such as \(x(t)=t.\) In this case, \(y(t)\) can be any expression. For example, consider the following pair of equations.
\[\begin{array}{l}x(t)=t \\ y(t)={t}^{2}-3\end{array}\]Rewriting this set of parametric equations is a matter of substituting \(x\) for \(t.\) Thus, the Cartesian equation is \(y={x}^{2}-3.\)
Example
Try it.
Use two different methods to find the Cartesian equation equivalent to the given set of parametric equations.
\[\begin{array}{l}x(t)=3t-2 \\ y(t)=t+1\end{array}\]Solution
Method 1. First, let’s solve the \(x\) equation for \(t.\) Then we can substitute the result into the \(y\) equation.
\[\begin{array}{l}x=3t-2 \\ x+2=3t \\ \frac{x+2}{3}=t\end{array}\]Now substitute the expression for \(t\) into the \(y\) equation.
\[\begin{array}{l}y=t+1 \\ y=(\frac{x+2}{3})+1 \\ y=\frac{x}{3}+\frac{2}{3}+1 \\ y=\frac{1}{3}x+\frac{5}{3}\end{array}\]Method 2. Solve the \(y\) equation for \(t\) and substitute this expression in the \(x\) equation.
\[\begin{array}{l}\ y=t+1 \\ y-1=t\end{array}\]Make the substitution and then solve for \(y.\)
\[\begin{array}{l}\ x=3(y-1)-2 \\ x=3y-3-2 \\ x=3y-5 \\ x+5=3y \\ \frac{x+5}{3}=y \\ y=\frac{1}{3}x+\frac{5}{3}\end{array}\]Finding Parametric Equations for Curves Defined by Rectangular Equations
Although we have just shown that there is only one way to interpret a set of parametric equations as a rectangular equation, there are multiple ways to interpret a rectangular equation as a set of parametric equations. Any strategy we may use to find the parametric equations is valid if it produces equivalency. In other words, if we choose an expression to represent \(x,\) and then substitute it into the \(y\) equation, and it produces the same graph over the same domain as the rectangular equation, then the set of parametric equations is valid. If the domain becomes restricted in the set of parametric equations, and the function does not allow the same values for \(x\) as the domain of the rectangular equation, then the graphs will be different.
Example
Try it.
Find a set of equivalent parametric equations for \(y={(x+3)}^{2}+1.\)
Solution
An obvious choice would be to let \(x(t)=t.\) Then \(y(t)={(t+3)}^{2}+1.\) But let’s try something more interesting. What if we let \(x=t+3?\) Then we have
\[\begin{array}{l}y={(x+3)}^{2}+1 \\ y={((t+3)+3)}^{2}+1 \\ y={(t+6)}^{2}+1\end{array}\]The set of parametric equations is
\[\begin{array}{l}x(t)=t+3 \\ y(t)={(t+6)}^{2}+1\end{array}\]See .
Key Concepts
- Parameterizing a curve involves translating a rectangular equation in two variables, \(x\) and \(y,\) into two equations in three variables, x, y, and t. Often, more information is obtained from a set of parametric equations. See , , and .
- Sometimes equations are simpler to graph when written in rectangular form. By eliminating \(t,\) an equation in \(x\) and \(y\) is the result.
- To eliminate \(t,\) solve one of the equations for \(t,\) and substitute the expression into the second equation. See , , , and .
- Finding the rectangular equation for a curve defined parametrically is basically the same as eliminating the parameter. Solve for \(t\) in one of the equations, and substitute the expression into the second equation. See .
- There are an infinite number of ways to choose a set of parametric equations for a curve defined as a rectangular equation.
- Find an expression for \(x\) such that the domain of the set of parametric equations remains the same as the original rectangular equation. See .
Parameterizing a Curve
When an object moves along a curve—or curvilinear path—in a given direction and in a given amount of time, the position of the object in the plane is given by the x-coordinate and the y-coordinate. However, both \(x\) and \(y\) vary over time and so are functions of time. For this reason, we add another variable, the parameter, upon which both \(x\) and \(y\) are dependent functions. In the example in the section opener, the parameter is time, \(t.\) The \(x\) position of the moon at time, \(t,\) is represented as the function \(x(t),\) and the \(y\) position of the moon at time, \(t,\) is represented as the function \(y(t).\) Together, \(x(t)\) and \(y(t)\) are called parametric equations, and generate an ordered pair \((x(t),\ y(t)).\) Parametric equations primarily describe motion and direction.
When we parameterize a curve, we are translating a single equation in two variables, such as \(x\) and \(y,\) into an equivalent pair of equations in three variables, \(x,y,\) and \(t.\) One of the reasons we parameterize a curve is because the parametric equations yield more information: specifically, the direction of the object’s motion over time.
When we graph parametric equations, we can observe the individual behaviors of \(x\) and of \(y.\) There are a number of shapes that cannot be represented in the form \(y=f(x),\) meaning that they are not functions. For example, consider the graph of a circle, given as \({r}^{2}={x}^{2}+{y}^{2}.\) Solving for \(y\) gives \(y=\pm \sqrt{{r}^{2}-{x}^{2}},\) or two equations: \({y}_{1}=\sqrt{{r}^{2}-{x}^{2}}\) and \({y}_{2}=-\sqrt{{r}^{2}-{x}^{2}}.\) If we graph \({y}_{1}\) and \({y}_{2}\) together, the graph will not pass the vertical line test, as shown in . Thus, the equation for the graph of a circle is not a function.
However, if we were to graph each equation on its own, each one would pass the vertical line test and therefore would represent a function. In some instances, the concept of breaking up the equation for a circle into two functions is similar to the concept of creating parametric equations, as we use two functions to produce a non-function. This will become clearer as we move forward.
Condensed — the full section is in OpenStax Precalculus 2e.
Eliminating the Parameter
In many cases, we may have a pair of parametric equations but find that it is simpler to draw a curve if the equation involves only two variables, such as \(x\) and \(y.\) Eliminating the parameter is a method that may make graphing some curves easier. However, if we are concerned with the mapping of the equation according to time, then it will be necessary to indicate the orientation of the curve as well. There are various methods for eliminating the parameter \(t\) from a set of parametric equations; not every method works for every type of equation. Here we will review the methods for the most common types of equations.
For polynomial, exponential, or logarithmic equations expressed as two parametric equations, we choose the equation that is most easily manipulated and solve for \(t.\) We substitute the resulting expression for \(t\) into the second equation. This gives one equation in \(x\) and \(y.\)
Example
Try it.
Given \(x(t)={t}^{2}+1\) and \(y(t)=2+t,\) eliminate the parameter, and write the parametric equations as a Cartesian equation.
Solution
We will begin with the equation for \(y\) because the linear equation is easier to solve for \(t.\)
\[\begin{array}{l}y=2+t \\ y-2=t\end{array}\]Next, substitute \(y-2\) for \(t\) in \(x(t).\)
\[\begin{array}{ll}x={t}^{2}+1 & \\ x={(y-2)}^{2}+1 & \text{Substitute the expression for }t\text{ into }x. \\ x={y}^{2}-4y+4+1 & \\ x={y}^{2}-4y+5 & \\ x={y}^{2}-4y+5 & \end{array}\]The Cartesian form is \(x={y}^{2}-4y+5.\)
Example
Try it.
Eliminate the parameter and write as a Cartesian equation: \(x(t)={e}^{-t}\) and \(y(t)=3{e}^{t}\)
Solution
Isolate \({e}^{t}.\)
\[\begin{array}{l}x={e}^{-t} \\ {e}^{t}=\frac{1}{x}\end{array}\]Substitute the expression into \(y(t).\)
\[\begin{array}{l}y=3{e}^{t} \\ y=3(\frac{1}{x}) \\ y=\frac{3}{x}\end{array}\]The Cartesian form is \(y=\frac{3}{x}.\)
Example
Try it.
Eliminate the parameter and write as a Cartesian equation: \(x(t)=\sqrt{t}+2\) and \(y(t)=\log (t).\)
Solution
Solve the first equation for \(t.\)
\[\begin{array}{ll}\ x=\sqrt{t}+2 & \\ x-2=\sqrt{t} & \\ {(x-2)}^{2}=t & \text{Square both sides}.\end{array}\]Then, substitute the expression for \(t\) into the \(y\) equation.
\[\begin{array}{l}y=\log (t) \\ y=\log {(x-2)}^{2}\end{array}\]The Cartesian form is \(y=\log {(x-2)}^{2}.\)
Condensed — the full section is in OpenStax Precalculus 2e.
Finding Cartesian Equations from Curves Defined Parametrically
When we are given a set of parametric equations and need to find an equivalent Cartesian equation, we are essentially “eliminating the parameter.” However, there are various methods we can use to rewrite a set of parametric equations as a Cartesian equation. The simplest method is to set one equation equal to the parameter, such as \(x(t)=t.\) In this case, \(y(t)\) can be any expression. For example, consider the following pair of equations.
\[\begin{array}{l}x(t)=t \\ y(t)={t}^{2}-3\end{array}\]Rewriting this set of parametric equations is a matter of substituting \(x\) for \(t.\) Thus, the Cartesian equation is \(y={x}^{2}-3.\)
Example
Try it.
Use two different methods to find the Cartesian equation equivalent to the given set of parametric equations.
\[\begin{array}{l}x(t)=3t-2 \\ y(t)=t+1\end{array}\]Solution
Method 1. First, let’s solve the \(x\) equation for \(t.\) Then we can substitute the result into the \(y\) equation.
\[\begin{array}{l}x=3t-2 \\ x+2=3t \\ \frac{x+2}{3}=t\end{array}\]Now substitute the expression for \(t\) into the \(y\) equation.
\[\begin{array}{l}y=t+1 \\ y=(\frac{x+2}{3})+1 \\ y=\frac{x}{3}+\frac{2}{3}+1 \\ y=\frac{1}{3}x+\frac{5}{3}\end{array}\]Method 2. Solve the \(y\) equation for \(t\) and substitute this expression in the \(x\) equation.
\[\begin{array}{l}\ y=t+1 \\ y-1=t\end{array}\]Make the substitution and then solve for \(y.\)
\[\begin{array}{l}\ x=3(y-1)-2 \\ x=3y-3-2 \\ x=3y-5 \\ x+5=3y \\ \frac{x+5}{3}=y \\ y=\frac{1}{3}x+\frac{5}{3}\end{array}\]Finding Parametric Equations for Curves Defined by Rectangular Equations
Although we have just shown that there is only one way to interpret a set of parametric equations as a rectangular equation, there are multiple ways to interpret a rectangular equation as a set of parametric equations. Any strategy we may use to find the parametric equations is valid if it produces equivalency. In other words, if we choose an expression to represent \(x,\) and then substitute it into the \(y\) equation, and it produces the same graph over the same domain as the rectangular equation, then the set of parametric equations is valid. If the domain becomes restricted in the set of parametric equations, and the function does not allow the same values for \(x\) as the domain of the rectangular equation, then the graphs will be different.
Example
Try it.
Find a set of equivalent parametric equations for \(y={(x+3)}^{2}+1.\)
Solution
An obvious choice would be to let \(x(t)=t.\) Then \(y(t)={(t+3)}^{2}+1.\) But let’s try something more interesting. What if we let \(x=t+3?\) Then we have
\[\begin{array}{l}y={(x+3)}^{2}+1 \\ y={((t+3)+3)}^{2}+1 \\ y={(t+6)}^{2}+1\end{array}\]The set of parametric equations is
\[\begin{array}{l}x(t)=t+3 \\ y(t)={(t+6)}^{2}+1\end{array}\]See .
Key Concepts
- Parameterizing a curve involves translating a rectangular equation in two variables, \(x\) and \(y,\) into two equations in three variables, x, y, and t. Often, more information is obtained from a set of parametric equations. See , , and .
- Sometimes equations are simpler to graph when written in rectangular form. By eliminating \(t,\) an equation in \(x\) and \(y\) is the result.
- To eliminate \(t,\) solve one of the equations for \(t,\) and substitute the expression into the second equation. See , , , and .
- Finding the rectangular equation for a curve defined parametrically is basically the same as eliminating the parameter. Solve for \(t\) in one of the equations, and substitute the expression into the second equation. See .
- There are an infinite number of ways to choose a set of parametric equations for a curve defined as a rectangular equation.
- Find an expression for \(x\) such that the domain of the set of parametric equations remains the same as the original rectangular equation. See .
Practice (40)
Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.
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Parameterize the curve \(y={x}^{2}-1\) letting \(x(t)=t.\) Graph both equations.
جواب رو نشون بده
If \(x(t)=t,\) then to find \(y(t)\) we replace the variable \(x\) with the expression given in \(x(t).\) In other words, \(y(t)={t}^{2}-1.\) Make a table of values similar to , and sketch the graph.
\(t\) \(x(t)\) \(y(t)\) \(-4\) \(-4\) \(y(-4)={(-4)}^{2}-1=15\) \(-3\) \(-3\) \(y(-3)={(-3)}^{2}-1=8\) \(-2\) \(-2\) \(y(-2)={(-2)}^{2}-1=3\) \(-1\) \(-1\) \(y(-1)={(-1)}^{2}-1=0\) \(0\) \(0\) \(y(0)={(0)}^{2}-1=-1\) \(1\) \(1\) \(y(1)={(1)}^{2}-1=0\) \(2\) \(2\) \(y(2)={(2)}^{2}-1=3\) \(3\) \(3\) \(y(3)={(3)}^{2}-1=8\) \(4\) \(4\) \(y(4)={(4)}^{2}-1=15\) See the graphs in . It may be helpful to use the TRACE feature of a graphing calculator to see how the points are generated as \(t\) increases.
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Construct a table of values and plot the parametric equations: \(x(t)=t-3,\) \(y(t)=2t+4;\ -1\le t\le 2.\)
جواب رو نشون بده
\(t\) \(x(t)\) \(y(t)\) \(-1\) \(-4\) \(2\) \(0\) \(-3\) \(4\) \(1\) \(-2\) \(6\) \(2\) \(-1\) \(8\) -
Find a pair of parametric equations that models the graph of \(y=1-{x}^{2},\) using the parameter \(x(t)=t.\) Plot some points and sketch the graph.
جواب رو نشون بده
If \(x(t)=t\) and we substitute \(t\) for \(x\) into the \(y\) equation, then \(y(t)=1-{t}^{2}.\) Our pair of parametric equations is
\[\begin{array}{l}x(t)=t \\ y(t)=1-{t}^{2}\end{array}\]To graph the equations, first we construct a table of values like that in . We can choose values around \(t=0,\) from \(t=-3\) to \(t=3.\) The values in the \(x(t)\) column will be the same as those in the \(t\) column because \(x(t)=t.\) Calculate values for the column \(y(t).\)
\(t\) \(x(t)=t\) \(y(t)=1-{t}^{2}\) \(-3\) \(-3\) \(y(-3)=1-{(-3)}^{2}=-8\) \(-2\) \(-2\) \(y(-2)=1-{(-2)}^{2}=-3\) \(-1\) \(-1\) \(y(-1)=1-{(-1)}^{2}=0\) \(0\) \(0\) \(y(0)=1-0=1\) \(1\) \(1\) \(y(1)=1-{(1)}^{2}=0\) \(2\) \(2\) \(y(2)=1-{(2)}^{2}=-3\) \(3\) \(3\) \(y(3)=1-{(3)}^{2}=-8\) The graph of \(y=1-{t}^{2}\) is a parabola facing downward, as shown in . We have mapped the curve over the interval \([-3,\ 3],\) shown as a solid line with arrows indicating the orientation of the curve according to \(t.\) Orientation refers to the path traced along the curve in terms of increasing values of \(t.\) As this parabola is symmetric with respect to the line \(x=0,\) the values of \(x\) are reflected across the y-axis.
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Parameterize the curve given by \(x={y}^{3}-2y.\)
جواب رو نشون بده
\(\begin{array}{l}x(t)={t}^{3}-2t \\ y(t)=t\end{array}\)
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An object travels at a steady rate along a straight path \((-5,\ 3)\) to \((3,\ -1)\) in the same plane in four seconds. The coordinates are measured in meters. Find parametric equations for the position of the object.
جواب رو نشون بده
The parametric equations are simple linear expressions, but we need to view this problem in a step-by-step fashion. The x-value of the object starts at \(-5\) meters and goes to 3 meters. This means the distance x has changed by 8 meters in 4 seconds, which is a rate of \(\frac{\text{8 m}}{4\text{ s}},\) or \(2\ \text{m}/\text{s}.\) We can write the x-coordinate as a linear function with respect to time as \(x(t)=2t-5.\) In the linear function template \(y=mx+b,2t=mx\) and \(-5=b.\)
Similarly, the y-value of the object starts at 3 and goes to \(-1,\) which is a change in the distance y of −4 meters in 4 seconds, which is a rate of \(\frac{-4\text{ m}}{4\text{ s}},\) or \(-1\text{m}/\text{s}.\) We can also write the y-coordinate as the linear function \(y(t)=-t+3.\) Together, these are the parametric equations for the position of the object, where \(x\) and \(y\) are expressed in meters and \(t\) represents time:
\[\begin{array}{l}x(t)=2t-5 \\ y(t)=-t+3\end{array}\]Using these equations, we can build a table of values for \(t,x,\) and \(y\) (see ). In this example, we limited values of \(t\) to non-negative numbers. In general, any value of \(t\) can be used.
\(t\) \(x(t)=2t-5\) \(y(t)=-t+3\) \(0\) \(x=2(0)-5=-5\) \(y=-(0)+3=3\) \(1\) \(x=2(1)-5=-3\) \(y=-(1)+3=2\) \(2\) \(x=2(2)-5=-1\) \(y=-(2)+3=1\) \(3\) \(x=2(3)-5=1\) \(y=-(3)+3=0\) \(4\) \(x=2(4)-5=3\) \(y=-(4)+3=-1\) From this table, we can create three graphs, as shown in .
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Given \(x(t)={t}^{2}+1\) and \(y(t)=2+t,\) eliminate the parameter, and write the parametric equations as a Cartesian equation.
جواب رو نشون بده
We will begin with the equation for \(y\) because the linear equation is easier to solve for \(t.\)
\[\begin{array}{l}y=2+t \\ y-2=t\end{array}\]Next, substitute \(y-2\) for \(t\) in \(x(t).\)
\[\begin{array}{ll}x={t}^{2}+1 & \\ x={(y-2)}^{2}+1 & \text{Substitute the expression for }t\text{ into }x. \\ x={y}^{2}-4y+4+1 & \\ x={y}^{2}-4y+5 & \\ x={y}^{2}-4y+5 & \end{array}\]The Cartesian form is \(x={y}^{2}-4y+5.\)
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Given the equations below, eliminate the parameter and write as a rectangular equation for \(y\) as a function of \(x.\)
\[\begin{array}{l}x(t)=2{t}^{2}+6 \\ y(t)=5-t\end{array}\]جواب رو نشون بده
\(y=5-\sqrt{\frac{1}{2}x-3}\)
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Eliminate the parameter and write as a Cartesian equation: \(x(t)={e}^{-t}\) and \(y(t)=3{e}^{t}\)
جواب رو نشون بده
Isolate \({e}^{t}.\)
\[\begin{array}{l}x={e}^{-t} \\ {e}^{t}=\frac{1}{x}\end{array}\]Substitute the expression into \(y(t).\)
\[\begin{array}{l}y=3{e}^{t} \\ y=3(\frac{1}{x}) \\ y=\frac{3}{x}\end{array}\]The Cartesian form is \(y=\frac{3}{x}.\)
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Eliminate the parameter and write as a Cartesian equation: \(x(t)=\sqrt{t}+2\) and \(y(t)=\log (t).\)
جواب رو نشون بده
Solve the first equation for \(t.\)
\[\begin{array}{ll}\ x=\sqrt{t}+2 & \\ x-2=\sqrt{t} & \\ {(x-2)}^{2}=t & \text{Square both sides}.\end{array}\]Then, substitute the expression for \(t\) into the \(y\) equation.
\[\begin{array}{l}y=\log (t) \\ y=\log {(x-2)}^{2}\end{array}\]The Cartesian form is \(y=\log {(x-2)}^{2}.\)
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Eliminate the parameter and write as a rectangular equation.
\[\begin{array}{l}x(t)={t}^{2} \\ y(t)=\ln \ t\ t>0\end{array}\]جواب رو نشون بده
\(y=\ln \sqrt{x}\)
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Eliminate the parameter from the given pair of trigonometric equations where \(0\le t\le 2\pi\) and sketch the graph.
\[\begin{array}{l}x(t)=4\cos \ t \\ y(t)=3\sin \ t\end{array}\]جواب رو نشون بده
Solving for \(\cos \ t\) and \(\sin \ t,\) we have
\[\begin{array}{l}x=4\cos \ t \\ \frac{x}{4}=\cos \ t \\ y=3\sin \ t \\ \frac{y}{3}=\sin \ t\end{array}\]Next, use the Pythagorean identity and make the substitutions.
\[\begin{array}{l}{\cos }^{2}t+{\sin }^{2}t=1 \\ {(\frac{x}{4})}^{2}+{(\frac{y}{3})}^{2}=1 \\ \frac{{x}^{2}}{16}+\frac{{y}^{2}}{9}=1\end{array}\]The graph for the equation is shown in .
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Eliminate the parameter from the given pair of parametric equations and write as a Cartesian equation: \(x(t)=2\cos \ t\) and \(y(t)=3\sin \ t.\)
جواب رو نشون بده
\(\frac{{x}^{2}}{4}+\frac{{y}^{2}}{9}=1\)
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Use two different methods to find the Cartesian equation equivalent to the given set of parametric equations.
\[\begin{array}{l}x(t)=3t-2 \\ y(t)=t+1\end{array}\]جواب رو نشون بده
Method 1. First, let’s solve the \(x\) equation for \(t.\) Then we can substitute the result into the \(y\) equation.
\[\begin{array}{l}x=3t-2 \\ x+2=3t \\ \frac{x+2}{3}=t\end{array}\]Now substitute the expression for \(t\) into the \(y\) equation.
\[\begin{array}{l}y=t+1 \\ y=(\frac{x+2}{3})+1 \\ y=\frac{x}{3}+\frac{2}{3}+1 \\ y=\frac{1}{3}x+\frac{5}{3}\end{array}\]Method 2. Solve the \(y\) equation for \(t\) and substitute this expression in the \(x\) equation.
\[\begin{array}{l}\ y=t+1 \\ y-1=t\end{array}\]Make the substitution and then solve for \(y.\)
\[\begin{array}{l}\ x=3(y-1)-2 \\ x=3y-3-2 \\ x=3y-5 \\ x+5=3y \\ \frac{x+5}{3}=y \\ y=\frac{1}{3}x+\frac{5}{3}\end{array}\] -
Write the given parametric equations as a Cartesian equation: \(x(t)={t}^{3}\) and \(y(t)={t}^{6}.\)
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\(y={x}^{2}\)
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Find a set of equivalent parametric equations for \(y={(x+3)}^{2}+1.\)
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An obvious choice would be to let \(x(t)=t.\) Then \(y(t)={(t+3)}^{2}+1.\) But let’s try something more interesting. What if we let \(x=t+3?\) Then we have
\[\begin{array}{l}y={(x+3)}^{2}+1 \\ y={((t+3)+3)}^{2}+1 \\ y={(t+6)}^{2}+1\end{array}\]The set of parametric equations is
\[\begin{array}{l}x(t)=t+3 \\ y(t)={(t+6)}^{2}+1\end{array}\]See .
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What is a system of parametric equations?
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A pair of functions that is dependent on an external factor. The two functions are written in terms of the same parameter. For example, \(x=f(t)\) and \(y=f(t).\)
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Some examples of a third parameter are time, length, speed, and scale. Explain when time is used as a parameter.
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Explain how to eliminate a parameter given a set of parametric equations.
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Choose one equation to solve for \(t,\) substitute into the other equation and simplify.
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What is a benefit of writing a system of parametric equations as a Cartesian equation?
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What is a benefit of using parametric equations?
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Some equations cannot be written as functions, like a circle. However, when written as two parametric equations, separately the equations are functions.
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Why are there many sets of parametric equations to represent on Cartesian function?
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\(\{\begin{array}{l}x(t)=5-t \\ y(t)=8-2t\end{array}\)
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\(y=-2+2x\)
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\(\{\begin{array}{l}x(t)=6-3t \\ y(t)=10-t\end{array}\)
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\(\{\begin{array}{l}x(t)=2t+1 \\ y(t)=3\sqrt{t}\end{array}\)
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\(y=3\sqrt{\frac{x-1}{2}}\)
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\(\{\begin{array}{l}x(t)=3t-1 \\ y(t)=2{t}^{2}\end{array}\)
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\(\{\begin{array}{l}x(t)=2{e}^{t} \\ y(t)=1-5t\end{array}\)
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\(x=2{e}^{\frac{1-y}{5}}\) or \(y=1-5ln(\frac{x}{2})\)
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\(\{\begin{array}{l}x(t)={e}^{-2t} \\ y(t)=2{e}^{-t}\end{array}\)
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\(\{\begin{array}{l}x(t)=4\text{log}(t) \\ y(t)=3+2t\end{array}\)
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\(x=4\log (\frac{y-3}{2})\)
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\(\{\begin{array}{l}x(t)=\text{log}(2t) \\ y(t)=\sqrt{t-1}\end{array}\)
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\(\{\begin{array}{l}x(t)={t}^{3}-t \\ y(t)=2t\end{array}\)
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\(x={(\frac{y}{2})}^{3}-\frac{y}{2}\)
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\(\{\begin{array}{l}x(t)=t-{t}^{4} \\ y(t)=t+2\end{array}\)
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\(\{\begin{array}{l}x(t)={e}^{2t} \\ y(t)={e}^{6t}\end{array}\)
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\(y={x}^{3}\)
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\(\{\begin{array}{l}x(t)={t}^{5} \\ y(t)={t}^{10}\end{array}\)
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\(\{\begin{array}{l}x(t)=4\text{cos}\ t \\ y(t)=5\sin \ t\ \end{array}\)
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\({(\frac{x}{4})}^{2}+{(\frac{y}{5})}^{2}=1\)
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\(\{\begin{array}{l}x(t)=3\sin \ t \\ y(t)=6\cos \ t\end{array}\)
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\(\{\begin{array}{l}x(t)=2{\text{cos}}^{2}t \\ y(t)=-\sin \ t\ \end{array}\)
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\({y}^{2}=1-\frac{1}{2}x\)
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\(\{\begin{array}{l}x(t)=\cos \ t+4 \\ y(t)=2{\sin }^{2}t\end{array}\)
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\(\{\begin{array}{l}x(t)=t-1 \\ y(t)={t}^{2}\end{array}\)
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\(y={x}^{2}+2x+1\)
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\(\{\begin{array}{l}x(t)=-t \\ y(t)={t}^{3}+1\end{array}\)
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\(\{\begin{array}{l}x(t)=2t-1 \\ y(t)={t}^{3}-2\end{array}\)
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\(y={(\frac{x+1}{2})}^{3}-2\)
Symbols used here
The non-negative number whose square (n-th power) is x.
Ratio of a circle's circumference to its diameter, 3.14159…
The exponent b must be raised to for x; ln uses base e.
Ratios of sides in a right triangle; coordinates on the unit circle.
Both signs at once: x = 3 ± 2 means 5 and 1.
Inequalities that allow equality; < and > exclude it.
The usual name for an angle.
1/360 of a full turn. 180° = π radians.
The angle whose sine is the given value (and likewise arccos, arctan).
How to: Parametric Equations
- Parameterize a curve.
- Eliminate the parameter.
- Find a rectangular equation for a curve defined parametrically.
- Find parametric equations for curves defined by rectangular equations.
- Parameterizing a curve involves translating a rectangular equation in two variables,
- Sometimes equations are simpler to graph when written in rectangular form. By eliminating
- To eliminate
- Finding the rectangular equation for a curve defined parametrically is basically the same as eliminating the parameter. Solve for
Questions people ask
Why radians instead of degrees?
A radian is the angle whose arc equals the radius, so it is a pure ratio rather than an arbitrary 1/360 of a turn. With radians the derivative of sin x is exactly cos x; with degrees a factor of π/180 appears everywhere.
Why does sin x = 1/2 have infinitely many solutions?
Sine repeats every full turn, and within one turn it reaches 1/2 twice (at π/6 and 5π/6). Add any whole number of turns to either and it is still true.
How do I remember the exact values?
Two triangles: the 45-45-90 with sides 1, 1, √2 and the 30-60-90 with sides 1, √3, 2. Every value for 30°, 45° and 60° is a ratio of those sides; symmetry gives the rest of the circle.
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Parts of this page are adapted from OpenStax Algebra and Trigonometry 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.
بیشتر در Trigonometry
The unit circleTrigonometric equationsTrigonometric identitiesDegrees and radiansRight-triangle trigonometry (SOH-CAH-TOA)Law of sines and law of cosinesGraphs of sine, cosine and tangentInverse trigonometric functions